| blob | 114aaae44c8a1992b2ef954e3e3c395c5724f918 |
1 /* Fold a constant sub-tree into a single node for C-compiler
2 Copyright (C) 1987, 1988, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999,
3 2000, 2001, 2002, 2003 Free Software Foundation, Inc.
5 This file is part of GCC.
7 GCC is free software; you can redistribute it and/or modify it under
8 the terms of the GNU General Public License as published by the Free
9 Software Foundation; either version 2, or (at your option) any later
10 version.
12 GCC is distributed in the hope that it will be useful, but WITHOUT ANY
13 WARRANTY; without even the implied warranty of MERCHANTABILITY or
14 FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
15 for more details.
17 You should have received a copy of the GNU General Public License
18 along with GCC; see the file COPYING. If not, write to the Free
19 Software Foundation, 59 Temple Place - Suite 330, Boston, MA
20 02111-1307, USA. */
22 /*@@ This file should be rewritten to use an arbitrary precision
23 @@ representation for "struct tree_int_cst" and "struct tree_real_cst".
24 @@ Perhaps the routines could also be used for bc/dc, and made a lib.
25 @@ The routines that translate from the ap rep should
26 @@ warn if precision et. al. is lost.
27 @@ This would also make life easier when this technology is used
28 @@ for cross-compilers. */
30 /* The entry points in this file are fold, size_int_wide, size_binop
31 and force_fit_type.
33 fold takes a tree as argument and returns a simplified tree.
35 size_binop takes a tree code for an arithmetic operation
36 and two operands that are trees, and produces a tree for the
37 result, assuming the type comes from `sizetype'.
39 size_int takes an integer value, and creates a tree constant
40 with type from `sizetype'.
42 force_fit_type takes a constant and prior overflow indicator, and
43 forces the value to fit the type. It returns an overflow indicator. */
93 tree);
111 /* The following constants represent a bit based encoding of GCC's
112 comparison operators. This encoding simplifies transformations
113 on relational comparison operators, such as AND and OR. */
114 #define COMPCODE_FALSE 0
115 #define COMPCODE_LT 1
116 #define COMPCODE_EQ 2
117 #define COMPCODE_LE 3
118 #define COMPCODE_GT 4
119 #define COMPCODE_NE 5
120 #define COMPCODE_GE 6
121 #define COMPCODE_TRUE 7
123 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
124 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
125 and SUM1. Then this yields nonzero if overflow occurred during the
126 addition.
128 Overflow occurs if A and B have the same sign, but A and SUM differ in
129 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
130 sign. */
131 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
132 \f
133 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
134 We do that by representing the two-word integer in 4 words, with only
135 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
136 number. The value of the word is LOWPART + HIGHPART * BASE. */
138 #define LOWPART(x) \
139 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
140 #define HIGHPART(x) \
141 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
142 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
144 /* Unpack a two-word integer into 4 words.
145 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
146 WORDS points to the array of HOST_WIDE_INTs. */
148 static void
150 {
155 }
157 /* Pack an array of 4 words into a two-word integer.
158 WORDS points to the array of words.
159 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
161 static void
164 {
167 }
168 \f
169 /* Make the integer constant T valid for its type by setting to 0 or 1 all
170 the bits in the constant that don't belong in the type.
172 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
173 nonzero, a signed overflow has already occurred in calculating T, so
174 propagate it. */
176 int
178 {
180 HOST_WIDE_INT high;
184 {
185 /* ??? Used to check for overflow here via CHECK_FLOAT_TYPE.
186 Consider doing it via real_convert now. */
188 }
199 else
202 /* First clear all bits that are beyond the type's precision. */
205 ;
209 else
210 {
214 }
216 /* Unsigned types do not suffer sign extension or overflow unless they
217 are a sizetype. */
223 /* If the value's sign bit is set, extend the sign. */
231 {
232 /* Value is negative:
233 set to 1 all the bits that are outside this type's precision. */
237 else
238 {
242 }
243 }
245 /* Return nonzero if signed overflow occurred. */
246 return
249 }
250 \f
251 /* Add two doubleword integers with doubleword result.
252 Each argument is given as two `HOST_WIDE_INT' pieces.
253 One argument is L1 and H1; the other, L2 and H2.
254 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
256 int
260 {
262 HOST_WIDE_INT h;
270 }
272 /* Negate a doubleword integer with doubleword result.
273 Return nonzero if the operation overflows, assuming it's signed.
274 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
275 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
277 int
280 {
282 {
286 }
287 else
288 {
292 }
293 }
294 \f
295 /* Multiply two doubleword integers with doubleword result.
296 Return nonzero if the operation overflows, assuming it's signed.
297 Each argument is given as two `HOST_WIDE_INT' pieces.
298 One argument is L1 and H1; the other, L2 and H2.
299 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
301 int
305 {
320 {
323 {
325 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
327 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
331 }
333 }
337 /* Check for overflow by calculating the top half of the answer in full;
338 it should agree with the low half's sign bit. */
341 {
344 }
346 {
349 }
351 }
352 \f
353 /* Shift the doubleword integer in L1, H1 left by COUNT places
354 keeping only PREC bits of result.
355 Shift right if COUNT is negative.
356 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
357 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
359 void
363 {
367 {
370 }
372 #ifdef SHIFT_COUNT_TRUNCATED
375 #endif
378 {
379 /* Shifting by the host word size is undefined according to the
380 ANSI standard, so we must handle this as a special case. */
383 }
385 {
388 }
389 else
390 {
394 }
396 /* Sign extend all bits that are beyond the precision. */
404 ;
406 {
409 }
410 else
411 {
415 }
416 }
418 /* Shift the doubleword integer in L1, H1 right by COUNT places
419 keeping only PREC bits of result. COUNT must be positive.
420 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
421 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
423 void
428 {
431 signmask = (arith
435 #ifdef SHIFT_COUNT_TRUNCATED
438 #endif
441 {
442 /* Shifting by the host word size is undefined according to the
443 ANSI standard, so we must handle this as a special case. */
446 }
448 {
451 }
452 else
453 {
457 }
459 /* Zero / sign extend all bits that are beyond the precision. */
462 {
465 }
467 ;
469 {
472 }
473 else
474 {
478 }
479 }
480 \f
481 /* Rotate the doubleword integer in L1, H1 left by COUNT places
482 keeping only PREC bits of result.
483 Rotate right if COUNT is negative.
484 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
486 void
490 {
502 }
504 /* Rotate the doubleword integer in L1, H1 left by COUNT places
505 keeping only PREC bits of result. COUNT must be positive.
506 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
508 void
512 {
524 }
525 \f
526 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
527 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
528 CODE is a tree code for a kind of division, one of
529 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
530 or EXACT_DIV_EXPR
531 It controls how the quotient is rounded to an integer.
532 Return nonzero if the operation overflows.
533 UNS nonzero says do unsigned division. */
535 int
538 HOST_WIDE_INT hnum_orig,
540 HOST_WIDE_INT hden_orig,
544 {
560 /* calculate quotient sign and convert operands to unsigned. */
562 {
564 {
566 /* (minimum integer) / (-1) is the only overflow case. */
570 }
572 {
575 }
576 }
581 /* This unsigned division rounds toward zero. */
584 }
588 /* hden != 0 already checked. */
593 }
603 /* Special code for when the divisor < BASE. */
605 {
606 /* hnum != 0 already checked. */
608 {
612 }
613 }
614 else
615 {
616 /* Full double precision division,
617 with thanks to Don Knuth's "Seminumerical Algorithms". */
621 /* Find the highest nonzero divisor digit. */
624 {
627 }
629 /* Insure that the first digit of the divisor is at least BASE/2.
630 This is required by the quotient digit estimation algorithm. */
637 {
641 }
646 {
651 }
652 }
656 /* Main loop */
658 {
659 /* Guess the next quotient digit, quo_est, by dividing the first
660 two remaining dividend digits by the high order quotient digit.
661 quo_est is never low and is at most 2 high. */
668 else
671 /* Refine quo_est so it's usually correct, and at most one high. */
676 quo_est--;
678 /* Try QUO_EST as the quotient digit, by multiplying the
679 divisor by QUO_EST and subtracting from the remaining dividend.
680 Keep in mind that QUO_EST is the I - 1st digit. */
684 {
690 }
692 /* If quo_est was high by one, then num[i] went negative and
693 we need to correct things. */
695 {
696 quo_est--;
699 {
703 }
706 }
708 /* Store the quotient digit. */
710 }
711 }
715 finish_up:
716 /* If result is negative, make it so. */
720 /* compute trial remainder: rem = num - (quo * den) */
726 {
735 {
736 /* quo = quo - 1; */
739 }
740 else
747 {
750 }
751 else
757 {
763 /* Get absolute values. */
769 /* If (2 * abs (lrem) >= abs (lden)) */
778 {
780 /* quo = quo - 1; */
783 else
784 /* quo = quo + 1; */
787 }
788 else
790 }
795 }
797 /* compute true remainder: rem = num - (quo * den) */
802 }
803 \f
804 /* Determine whether an expression T can be cheaply negated using
805 the function negate_expr. */
807 static bool
809 {
812 tree type;
821 {
826 /* Check that -CST will not overflow type. */
829 {
834 }
835 else
848 }
850 }
852 /* Given T, an expression, return the negation of T. Allow for T to be
853 null, in which case return null. */
855 static tree
857 {
858 tree type;
859 tree tem;
868 {
881 /* - (A - B) -> B - A */
891 }
894 }
895 \f
896 /* Split a tree IN into a constant, literal and variable parts that could be
897 combined with CODE to make IN. "constant" means an expression with
898 TREE_CONSTANT but that isn't an actual constant. CODE must be a
899 commutative arithmetic operation. Store the constant part into *CONP,
900 the literal in *LITP and return the variable part. If a part isn't
901 present, set it to null. If the tree does not decompose in this way,
902 return the entire tree as the variable part and the other parts as null.
904 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
905 case, we negate an operand that was subtracted. Except if it is a
906 literal for which we use *MINUS_LITP instead.
908 If NEGATE_P is true, we are negating all of IN, again except a literal
909 for which we use *MINUS_LITP instead.
911 If IN is itself a literal or constant, return it as appropriate.
913 Note that we do not guarantee that any of the three values will be the
914 same type as IN, but they will have the same signedness and mode. */
916 static tree
919 {
926 /* Strip any conversions that don't change the machine mode or signedness. */
933 /* We can associate addition and subtraction together (even
934 though the C standard doesn't say so) for integers because
935 the value is not affected. For reals, the value might be
936 affected, so we can't. */
939 {
945 /* First see if either of the operands is a literal, then a constant. */
956 /* If we haven't dealt with either operand, this is not a case we can
957 decompose. Otherwise, VAR is either of the ones remaining, if any. */
962 else
965 /* Now do any needed negations. */
972 }
975 else
979 {
986 }
989 }
991 /* Re-associate trees split by the above function. T1 and T2 are either
992 expressions to associate or null. Return the new expression, if any. If
993 we build an operation, do it in TYPE and with CODE. */
995 static tree
997 {
1003 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
1004 try to fold this since we will have infinite recursion. But do
1005 deal with any NEGATE_EXPRs. */
1008 {
1010 {
1017 }
1019 }
1022 }
1023 \f
1024 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1025 to produce a new constant.
1027 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1029 static tree
1031 {
1035 HOST_WIDE_INT hi;
1037 HOST_WIDE_INT garbageh;
1038 tree t;
1041 int is_sizetype
1052 {
1072 /* It's unclear from the C standard whether shifts can overflow.
1073 The following code ignores overflow; perhaps a C standard
1074 interpretation ruling is needed. */
1104 /* This is a shortcut for a common special case. */
1109 {
1115 }
1117 /* ... fall through ... */
1121 {
1124 }
1127 {
1130 }
1137 /* This is a shortcut for a common special case. */
1142 {
1147 }
1149 /* ... fall through ... */
1165 else
1171 else
1177 }
1179 /* If this is for a sizetype, can be represented as one (signed)
1180 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1181 constants. */
1187 else
1188 {
1191 }
1194 = ((notrunc
1201 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1202 So check if force_fit_type truncated the value. */
1213 }
1215 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1216 constant. We assume ARG1 and ARG2 have the same data type, or at least
1217 are the same kind of constant and the same machine mode.
1219 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1221 static tree
1223 {
1231 {
1233 REAL_VALUE_TYPE d1;
1234 REAL_VALUE_TYPE d2;
1235 REAL_VALUE_TYPE value;
1244 /* Don't perform operation if we honor signaling NaNs and
1245 either operand is a NaN. */
1250 /* Don't perform operation if it would raise a division
1251 by zero exception. */
1257 /* If either operand is a NaN, just return it. Otherwise, set up
1258 for floating-point trap; we return an overflow. */
1276 }
1278 {
1284 tree t;
1287 {
1307 notrunc),
1313 notrunc));
1317 {
1318 tree magsquared
1322 notrunc);
1325 const_binop
1330 notrunc),
1332 notrunc),
1333 notrunc),
1335 const_binop
1340 notrunc),
1342 notrunc),
1343 notrunc),
1345 }
1350 }
1352 }
1354 }
1356 /* These are the hash table functions for the hash table of INTEGER_CST
1357 nodes of a sizetype. */
1359 /* Return the hash code code X, an INTEGER_CST. */
1361 static hashval_t
1363 {
1369 }
1371 /* Return nonzero if the value represented by *X (an INTEGER_CST tree node)
1372 is the same as that given by *Y, which is the same. */
1374 static int
1376 {
1384 }
1385 \f
1386 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1387 bits are given by NUMBER and of the sizetype represented by KIND. */
1389 tree
1391 {
1393 }
1395 /* Likewise, but the desired type is specified explicitly. */
1399 htab_t size_htab;
1401 tree
1403 {
1407 {
1410 }
1412 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1413 hash table, we return the value from the hash table. Otherwise, we
1414 place that in the hash table and make a new node for the next time. */
1423 {
1429 }
1430 else
1432 }
1434 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1435 is a tree code. The type of the result is taken from the operands.
1436 Both must be the same type integer type and it must be a size type.
1437 If the operands are constant, so is the result. */
1439 tree
1441 {
1448 /* Handle the special case of two integer constants faster. */
1450 {
1451 /* And some specific cases even faster than that. */
1460 /* Handle general case of two integer constants. */
1462 }
1468 }
1470 /* Given two values, either both of sizetype or both of bitsizetype,
1471 compute the difference between the two values. Return the value
1472 in signed type corresponding to the type of the operands. */
1474 tree
1476 {
1478 tree ctype;
1484 /* If the type is already signed, just do the simple thing. */
1491 /* If either operand is not a constant, do the conversions to the signed
1492 type and subtract. The hardware will do the right thing with any
1493 overflow in the subtraction. */
1498 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1499 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1500 overflow) and negate (which can't either). Special-case a result
1501 of zero while we're here. */
1506 else
1509 }
1510 \f
1512 /* Given T, a tree representing type conversion of ARG1, a constant,
1513 return a constant tree representing the result of conversion. */
1515 static tree
1517 {
1522 {
1524 {
1525 /* If we would build a constant wider than GCC supports,
1526 leave the conversion unfolded. */
1530 /* If we are trying to make a sizetype for a small integer, use
1531 size_int to pick up cached types to reduce duplicate nodes. */
1537 /* Given an integer constant, make new constant with new type,
1538 appropriately sign-extended or truncated. */
1542 /* Indicate an overflow if (1) ARG1 already overflowed,
1543 or (2) force_fit_type indicates an overflow.
1544 Tell force_fit_type that an overflow has already occurred
1545 if ARG1 is a too-large unsigned value and T is signed.
1546 But don't indicate an overflow if converting a pointer. */
1556 }
1558 {
1559 /* Don't initialize these, use assignments.
1560 Initialized local aggregates don't work on old compilers. */
1561 REAL_VALUE_TYPE x;
1562 REAL_VALUE_TYPE l;
1563 REAL_VALUE_TYPE u;
1574 /* See if X will be in range after truncation towards 0.
1575 To compensate for truncation, move the bounds away from 0,
1576 but reject if X exactly equals the adjusted bounds. */
1580 /* If X is a NaN, use zero instead and show we have an overflow.
1581 Otherwise, range check. */
1585 && !no_upper_bound
1589 {
1593 }
1599 }
1601 }
1603 {
1607 {
1609 {
1610 /* We make a copy of ARG1 so that we don't modify an
1611 existing constant tree. */
1615 }
1626 }
1627 }
1630 }
1631 \f
1632 /* Return an expr equal to X but certainly not valid as an lvalue. */
1634 tree
1636 {
1637 tree result;
1639 /* These things are certainly not lvalues. */
1650 }
1652 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1653 Zero means allow extended lvalues. */
1657 /* When pedantic, return an expr equal to X but certainly not valid as a
1658 pedantic lvalue. Otherwise, return X. */
1660 tree
1662 {
1665 else
1667 }
1668 \f
1669 /* Given a tree comparison code, return the code that is the logical inverse
1670 of the given code. It is not safe to do this for floating-point
1671 comparisons, except for NE_EXPR and EQ_EXPR. */
1673 static enum tree_code
1675 {
1677 {
1692 }
1693 }
1695 /* Similar, but return the comparison that results if the operands are
1696 swapped. This is safe for floating-point. */
1698 static enum tree_code
1700 {
1702 {
1716 }
1717 }
1720 /* Convert a comparison tree code from an enum tree_code representation
1721 into a compcode bit-based encoding. This function is the inverse of
1722 compcode_to_comparison. */
1724 static int
1726 {
1728 {
1743 }
1744 }
1746 /* Convert a compcode bit-based encoding of a comparison operator back
1747 to GCC's enum tree_code representation. This function is the
1748 inverse of comparison_to_compcode. */
1750 static enum tree_code
1752 {
1754 {
1769 }
1770 }
1772 /* Return nonzero if CODE is a tree code that represents a truth value. */
1774 static int
1776 {
1781 }
1782 \f
1783 /* Return nonzero if two operands are necessarily equal.
1784 If ONLY_CONST is nonzero, only return nonzero for constants.
1785 This function tests whether the operands are indistinguishable;
1786 it does not test whether they are equal using C's == operation.
1787 The distinction is important for IEEE floating point, because
1788 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1789 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1791 int
1793 {
1794 /* If both types don't have the same signedness, then we can't consider
1795 them equal. We must check this before the STRIP_NOPS calls
1796 because they may change the signedness of the arguments. */
1804 /* This is needed for conversions and for COMPONENT_REF.
1805 Might as well play it safe and always test this. */
1811 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1812 We don't care about side effects in that case because the SAVE_EXPR
1813 takes care of that for us. In all other cases, two expressions are
1814 equal if they have no side effects. If we have two identical
1815 expressions with side effects that should be treated the same due
1816 to the only side effects being identical SAVE_EXPR's, that will
1817 be detected in the recursive calls below. */
1823 /* Next handle constant cases, those for which we can return 1 even
1824 if ONLY_CONST is set. */
1827 {
1840 {
1850 {
1855 }
1858 }
1862 only_const)
1864 only_const));
1877 }
1883 {
1885 /* Two conversions are equal only if signedness and modes match. */
1901 /* For commutative ops, allow the other order. */
1914 /* If either of the pointer (or reference) expressions we are
1915 dereferencing contain a side effect, these cannot be equal. */
1921 {
1943 }
1947 {
1957 /* If the CALL_EXPRs call different functions, then they
1958 clearly can not be equal. */
1963 /* Only consider const functions equivalent. */
1965 {
1969 }
1970 else
1973 /* Now see if all the arguments are the same. operand_equal_p
1974 does not handle TREE_LIST, so we walk the operands here
1975 feeding them to operand_equal_p. */
1979 {
1985 }
1987 /* If we get here and both argument lists are exhausted
1988 then the CALL_EXPRs are equal. */
1993 }
1996 /* Consider __builtin_sqrt equal to sqrt. */
2004 }
2005 }
2006 \f
2007 /* Similar to operand_equal_p, but see if ARG0 might have been made by
2008 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
2010 When in doubt, return 0. */
2012 static int
2014 {
2026 /* Discard any conversions that don't change the modes of ARG0 and ARG1
2027 and see if the inner values are the same. This removes any
2028 signedness comparison, which doesn't matter here. */
2035 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
2036 actual comparison operand, ARG0.
2038 First throw away any conversions to wider types
2039 already present in the operands. */
2048 {
2051 /* Make sure shorter operand is extended the right way
2052 to match the longer operand. */
2058 }
2061 }
2062 \f
2063 /* See if ARG is an expression that is either a comparison or is performing
2064 arithmetic on comparisons. The comparisons must only be comparing
2065 two different values, which will be stored in *CVAL1 and *CVAL2; if
2066 they are nonzero it means that some operands have already been found.
2067 No variables may be used anywhere else in the expression except in the
2068 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
2069 the expression and save_expr needs to be called with CVAL1 and CVAL2.
2071 If this is true, return 1. Otherwise, return zero. */
2073 static int
2075 {
2079 /* We can handle some of the 'e' cases here. */
2089 {
2090 /* If we've already found a CVAL1 or CVAL2, this expression is
2091 two complex to handle. */
2097 }
2100 {
2123 /* First see if we can handle the first operand, then the second. For
2124 the second operand, we know *CVAL1 can't be zero. It must be that
2125 one side of the comparison is each of the values; test for the
2126 case where this isn't true by failing if the two operands
2127 are the same. */
2136 ;
2140 ;
2141 else
2145 ;
2149 ;
2150 else
2157 }
2158 }
2159 \f
2160 /* ARG is a tree that is known to contain just arithmetic operations and
2161 comparisons. Evaluate the operations in the tree substituting NEW0 for
2162 any occurrence of OLD0 as an operand of a comparison and likewise for
2163 NEW1 and OLD1. */
2165 static tree
2167 {
2172 /* We can handle some of the 'e' cases here. */
2180 {
2195 {
2212 }
2213 /* Fall through - ??? */
2216 {
2220 /* We need to check both for exact equality and tree equality. The
2221 former will be true if the operand has a side-effect. In that
2222 case, we know the operand occurred exactly once. */
2235 }
2239 }
2240 }
2241 \f
2242 /* Return a tree for the case when the result of an expression is RESULT
2243 converted to TYPE and OMITTED was previously an operand of the expression
2244 but is now not needed (e.g., we folded OMITTED * 0).
2246 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2247 the conversion of RESULT to TYPE. */
2249 tree
2251 {
2258 }
2260 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2262 static tree
2264 {
2271 }
2272 \f
2273 /* Return a simplified tree node for the truth-negation of ARG. This
2274 never alters ARG itself. We assume that ARG is an operation that
2275 returns a truth value (0 or 1). */
2277 tree
2279 {
2286 /* If this is a comparison, we can simply invert it, except for
2287 floating-point non-equality comparisons, in which case we just
2288 enclose a TRUTH_NOT_EXPR around what we have. */
2291 {
2293 && !flag_unsafe_math_optimizations
2297 else
2300 }
2303 {
2318 /* Here we can invert either operand. We invert the first operand
2319 unless the second operand is a TRUTH_NOT_EXPR in which case our
2320 result is the XOR of the first operand with the inside of the
2321 negation of the second operand. */
2326 else
2381 }
2385 }
2387 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2388 operands are another bit-wise operation with a common input. If so,
2389 distribute the bit operations to save an operation and possibly two if
2390 constants are involved. For example, convert
2391 (A | B) & (A | C) into A | (B & C)
2392 Further simplification will occur if B and C are constants.
2394 If this optimization cannot be done, 0 will be returned. */
2396 static tree
2398 {
2399 tree common;
2409 {
2413 }
2415 {
2419 }
2421 {
2425 }
2427 {
2431 }
2432 else
2437 }
2438 \f
2439 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2440 starting at BITPOS. The field is unsigned if UNSIGNEDP is nonzero. */
2442 static tree
2445 {
2452 }
2454 /* Optimize a bit-field compare.
2456 There are two cases: First is a compare against a constant and the
2457 second is a comparison of two items where the fields are at the same
2458 bit position relative to the start of a chunk (byte, halfword, word)
2459 large enough to contain it. In these cases we can avoid the shift
2460 implicit in bitfield extractions.
2462 For constants, we emit a compare of the shifted constant with the
2463 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2464 compared. For two fields at the same position, we do the ANDs with the
2465 similar mask and compare the result of the ANDs.
2467 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2468 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2469 are the left and right operands of the comparison, respectively.
2471 If the optimization described above can be done, we return the resulting
2472 tree. Otherwise we return zero. */
2474 static tree
2477 {
2486 tree mask;
2487 tree offset;
2489 /* Get all the information about the extractions being done. If the bit size
2490 if the same as the size of the underlying object, we aren't doing an
2491 extraction at all and so can do nothing. We also don't want to
2492 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2493 then will no longer be able to replace it. */
2501 {
2502 /* If this is not a constant, we can only do something if bit positions,
2503 sizes, and signedness are the same. */
2511 }
2513 /* See if we can find a mode to refer to this field. We should be able to,
2514 but fail if we can't. */
2523 /* Set signed and unsigned types of the precision of this mode for the
2524 shifts below. */
2528 /* Compute the bit position and size for the new reference and our offset
2529 within it. If the new reference is the same size as the original, we
2530 won't optimize anything, so return zero. */
2540 /* Make the mask to be used against the extracted field. */
2550 /* If not comparing with constant, just rework the comparison
2551 and return. */
2556 mask),
2560 mask));
2562 /* Otherwise, we are handling the constant case. See if the constant is too
2563 big for the field. Warn and return a tree of for 0 (false) if so. We do
2564 this not only for its own sake, but to avoid having to test for this
2565 error case below. If we didn't, we might generate wrong code.
2567 For unsigned fields, the constant shifted right by the field length should
2568 be all zero. For signed fields, the high-order bits should agree with
2569 the sign bit. */
2572 {
2576 {
2582 }
2583 }
2584 else
2585 {
2589 {
2595 }
2596 }
2598 /* Single-bit compares should always be against zero. */
2600 {
2603 }
2605 /* Make a new bitfield reference, shift the constant over the
2606 appropriate number of bits and mask it with the computed mask
2607 (in case this was a signed field). If we changed it, make a new one. */
2610 {
2613 }
2623 rhs);
2624 }
2625 \f
2626 /* Subroutine for fold_truthop: decode a field reference.
2628 If EXP is a comparison reference, we return the innermost reference.
2630 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2631 set to the starting bit number.
2633 If the innermost field can be completely contained in a mode-sized
2634 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2636 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2637 otherwise it is not changed.
2639 *PUNSIGNEDP is set to the signedness of the field.
2641 *PMASK is set to the mask used. This is either contained in a
2642 BIT_AND_EXPR or derived from the width of the field.
2644 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2646 Return 0 if this is not a component reference or is one that we can't
2647 do anything with. */
2649 static tree
2654 {
2658 tree unsigned_type;
2661 /* All the optimizations using this function assume integer fields.
2662 There are problems with FP fields since the type_for_size call
2663 below can fail for, e.g., XFmode. */
2667 /* We are interested in the bare arrangement of bits, so strip everything
2668 that doesn't affect the machine mode. However, record the type of the
2669 outermost expression if it may matter below. */
2677 {
2683 }
2692 /* If the number of bits in the reference is the same as the bitsize of
2693 the outer type, then the outer type gives the signedness. Otherwise
2694 (in case of a small bitfield) the signedness is unchanged. */
2698 /* Compute the mask to access the bitfield. */
2708 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2716 }
2718 /* Return nonzero if MASK represents a mask of SIZE ones in the low-order
2719 bit positions. */
2721 static int
2723 {
2726 tree tmask;
2731 return
2738 }
2740 /* Subroutine for fold: determine if VAL is the INTEGER_CONST that
2741 represents the sign bit of EXP's type. If EXP represents a sign
2742 or zero extension, also test VAL against the unextended type.
2743 The return value is the (sub)expression whose sign bit is VAL,
2744 or NULL_TREE otherwise. */
2746 static tree
2748 {
2752 tree t;
2754 /* Tree EXP must have an integral type. */
2759 /* Tree VAL must be an integer constant. */
2766 {
2773 }
2774 else
2775 {
2782 }
2784 /* We mask off those bits beyond TREE_TYPE (exp) so that we can
2785 treat VAL as if it were unsigned. */
2790 /* Handle extension from a narrower type. */
2796 }
2798 /* Subroutine for fold_truthop: determine if an operand is simple enough
2799 to be evaluated unconditionally. */
2801 static int
2803 {
2804 /* Strip any conversions that don't change the machine mode. */
2816 /* Don't regard global variables as simple. They may be
2817 allocated in ways unknown to the compiler (shared memory,
2818 #pragma weak, etc). */
2821 /* Loading a static variable is unduly expensive, but global
2822 registers aren't expensive. */
2824 }
2825 \f
2826 /* The following functions are subroutines to fold_range_test and allow it to
2827 try to change a logical combination of comparisons into a range test.
2829 For example, both
2830 X == 2 || X == 3 || X == 4 || X == 5
2831 and
2832 X >= 2 && X <= 5
2833 are converted to
2834 (unsigned) (X - 2) <= 3
2836 We describe each set of comparisons as being either inside or outside
2837 a range, using a variable named like IN_P, and then describe the
2838 range with a lower and upper bound. If one of the bounds is omitted,
2839 it represents either the highest or lowest value of the type.
2841 In the comments below, we represent a range by two numbers in brackets
2842 preceded by a "+" to designate being inside that range, or a "-" to
2843 designate being outside that range, so the condition can be inverted by
2844 flipping the prefix. An omitted bound is represented by a "-". For
2845 example, "- [-, 10]" means being outside the range starting at the lowest
2846 possible value and ending at 10, in other words, being greater than 10.
2847 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2848 always false.
2850 We set up things so that the missing bounds are handled in a consistent
2851 manner so neither a missing bound nor "true" and "false" need to be
2852 handled using a special case. */
2854 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2855 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2856 and UPPER1_P are nonzero if the respective argument is an upper bound
2857 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2858 must be specified for a comparison. ARG1 will be converted to ARG0's
2859 type if both are specified. */
2861 static tree
2864 {
2865 tree tem;
2869 /* If neither arg represents infinity, do the normal operation.
2870 Else, if not a comparison, return infinity. Else handle the special
2871 comparison rules. Note that most of the cases below won't occur, but
2872 are handled for consistency. */
2875 {
2880 }
2885 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2886 for neither. In real maths, we cannot assume open ended ranges are
2887 the same. But, this is computer arithmetic, where numbers are finite.
2888 We can therefore make the transformation of any unbounded range with
2889 the value Z, Z being greater than any representable number. This permits
2890 us to treat unbounded ranges as equal. */
2894 {
2915 }
2918 }
2919 \f
2920 /* Given EXP, a logical expression, set the range it is testing into
2921 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2922 actually being tested. *PLOW and *PHIGH will be made of the same type
2923 as the returned expression. If EXP is not a comparison, we will most
2924 likely not be returning a useful value and range. */
2926 static tree
2928 {
2935 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2936 and see if we can refine the range. Some of the cases below may not
2937 happen, but it doesn't seem worth worrying about this. We "continue"
2938 the outer loop when we've changed something; otherwise we "break"
2939 the switch, which will "break" the while. */
2944 {
2948 {
2960 }
2962 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2963 lose a cast by accident. */
2968 {
2975 /* We can only do something if the range is testing for zero
2976 and if the second operand is an integer constant. Note that
2977 saying something is "in" the range we make is done by
2978 complementing IN_P since it will set in the initial case of
2979 being not equal to zero; "out" is leaving it alone. */
2986 {
3007 }
3011 /* If this is an unsigned comparison, we also know that EXP is
3012 greater than or equal to zero. We base the range tests we make
3013 on that fact, so we record it here so we can parse existing
3014 range tests. */
3016 {
3019 NULL_TREE))
3024 /* If the high bound is missing, but we
3025 have a low bound, reverse the range so
3026 it goes from zero to the low bound minus 1. */
3028 {
3033 }
3034 }
3038 /* (-x) IN [a,b] -> x in [-b, -a] */
3048 /* ~ X -> -X - 1 */
3057 /* If EXP is signed, any overflow in the computation is undefined,
3058 so we don't worry about it so long as our computations on
3059 the bounds don't overflow. For unsigned, overflow is defined
3060 and this is exactly the right thing. */
3069 /* Check for an unsigned range which has wrapped around the maximum
3070 value thus making n_high < n_low, and normalize it. */
3072 {
3078 /* If the range is of the form +/- [ x+1, x ], we won't
3079 be able to normalize it. But then, it represents the
3080 whole range or the empty set, so make it
3081 +/- [ -, - ]. */
3085 else
3087 }
3088 else
3111 /* If we're converting from an unsigned to a signed type,
3112 we will be doing the comparison as unsigned. The tests above
3113 have already verified that LOW and HIGH are both positive.
3115 So we have to make sure that the original unsigned value will
3116 be interpreted as positive. */
3118 {
3121 tree high_positive;
3123 /* A range without an upper bound is, naturally, unbounded.
3124 Since convert would have cropped a very large value, use
3125 the max value for the destination type. */
3126 high_positive
3135 /* If the low bound is specified, "and" the range with the
3136 range for which the original unsigned value will be
3137 positive. */
3139 {
3143 high_positive))
3147 }
3148 else
3149 {
3150 /* Otherwise, "or" the range with the range of the input
3151 that will be interpreted as negative. */
3155 high_positive))
3159 }
3160 }
3168 }
3171 }
3173 /* If EXP is a constant, we can evaluate whether this is true or false. */
3175 {
3182 }
3186 }
3187 \f
3188 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3189 type, TYPE, return an expression to test if EXP is in (or out of, depending
3190 on IN_P) the range. */
3192 static tree
3194 {
3196 tree value;
3215 {
3217 {
3221 }
3223 }
3225 /* Optimize (c>=1) && (c<=127) into (signed char)c > 0. */
3227 {
3229 HOST_WIDE_INT hi;
3234 {
3237 }
3238 else
3239 {
3242 }
3245 {
3247 {
3250 }
3253 }
3254 }
3263 }
3264 \f
3265 /* Given two ranges, see if we can merge them into one. Return 1 if we
3266 can, 0 if we can't. Set the output range into the specified parameters. */
3268 static int
3271 {
3275 tree tem;
3285 /* Make range 0 be the range that starts first, or ends last if they
3286 start at the same value. Swap them if it isn't. */
3289 || (lowequal
3292 {
3296 }
3298 /* Now flag two cases, whether the ranges are disjoint or whether the
3299 second range is totally subsumed in the first. Note that the tests
3300 below are simplified by the ones above. */
3306 /* We now have four cases, depending on whether we are including or
3307 excluding the two ranges. */
3309 {
3310 /* If they don't overlap, the result is false. If the second range
3311 is a subset it is the result. Otherwise, the range is from the start
3312 of the second to the end of the first. */
3317 else
3319 }
3322 {
3323 /* If they don't overlap, the result is the first range. If they are
3324 equal, the result is false. If the second range is a subset of the
3325 first, and the ranges begin at the same place, we go from just after
3326 the end of the first range to the end of the second. If the second
3327 range is not a subset of the first, or if it is a subset and both
3328 ranges end at the same place, the range starts at the start of the
3329 first range and ends just before the second range.
3330 Otherwise, we can't describe this as a single range. */
3336 {
3340 }
3342 {
3346 }
3347 else
3349 }
3352 {
3353 /* If they don't overlap, the result is the second range. If the second
3354 is a subset of the first, the result is false. Otherwise,
3355 the range starts just after the first range and ends at the
3356 end of the second. */
3361 else
3362 {
3366 }
3367 }
3369 else
3370 {
3371 /* The case where we are excluding both ranges. Here the complex case
3372 is if they don't overlap. In that case, the only time we have a
3373 range is if they are adjacent. If the second is a subset of the
3374 first, the result is the first. Otherwise, the range to exclude
3375 starts at the beginning of the first range and ends at the end of the
3376 second. */
3378 {
3385 else
3387 }
3390 else
3392 }
3396 }
3397 \f
3398 #ifndef RANGE_TEST_NON_SHORT_CIRCUIT
3399 #define RANGE_TEST_NON_SHORT_CIRCUIT (BRANCH_COST >= 2)
3400 #endif
3402 /* EXP is some logical combination of boolean tests. See if we can
3403 merge it into some range test. Return the new tree if so. */
3405 static tree
3407 {
3414 tree tem;
3416 /* If this is an OR operation, invert both sides; we will invert
3417 again at the end. */
3421 /* If both expressions are the same, if we can merge the ranges, and we
3422 can build the range test, return it or it inverted. If one of the
3423 ranges is always true or always false, consider it to be the same
3424 expression as the other. */
3434 /* On machines where the branch cost is expensive, if this is a
3435 short-circuited branch and the underlying object on both sides
3436 is the same, make a non-short-circuit operation. */
3442 {
3443 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3444 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3445 which cases we can't do this. */
3454 {
3466 }
3467 }
3470 }
3471 \f
3472 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3473 bit value. Arrange things so the extra bits will be set to zero if and
3474 only if C is signed-extended to its full width. If MASK is nonzero,
3475 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3477 static tree
3479 {
3482 tree temp;
3487 /* We work by getting just the sign bit into the low-order bit, then
3488 into the high-order bit, then sign-extend. We then XOR that value
3489 with C. */
3493 /* We must use a signed type in order to get an arithmetic right shift.
3494 However, we must also avoid introducing accidental overflows, so that
3495 a subsequent call to integer_zerop will work. Hence we must
3496 do the type conversion here. At this point, the constant is either
3497 zero or one, and the conversion to a signed type can never overflow.
3498 We could get an overflow if this conversion is done anywhere else. */
3506 /* If necessary, convert the type back to match the type of C. */
3511 }
3512 \f
3513 /* Find ways of folding logical expressions of LHS and RHS:
3514 Try to merge two comparisons to the same innermost item.
3515 Look for range tests like "ch >= '0' && ch <= '9'".
3516 Look for combinations of simple terms on machines with expensive branches
3517 and evaluate the RHS unconditionally.
3519 For example, if we have p->a == 2 && p->b == 4 and we can make an
3520 object large enough to span both A and B, we can do this with a comparison
3521 against the object ANDed with the a mask.
3523 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3524 operations to do this with one comparison.
3526 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3527 function and the one above.
3529 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3530 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3532 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3533 two operands.
3535 We return the simplified tree or 0 if no optimization is possible. */
3537 static tree
3539 {
3540 /* If this is the "or" of two comparisons, we can do something if
3541 the comparisons are NE_EXPR. If this is the "and", we can do something
3542 if the comparisons are EQ_EXPR. I.e.,
3543 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3545 WANTED_CODE is this operation code. For single bit fields, we can
3546 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3547 comparison for one-bit fields. */
3567 /* Start by getting the comparison codes. Fail if anything is volatile.
3568 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3569 it were surrounded with a NE_EXPR. */
3594 /* Simplify (x<y) && (x==y) into (x<=y) and related optimizations. */
3598 {
3603 {
3611 }
3614 {
3623 }
3624 else
3634 }
3636 /* If the RHS can be evaluated unconditionally and its operands are
3637 simple, it wins to evaluate the RHS unconditionally on machines
3638 with expensive branches. In this case, this isn't a comparison
3639 that can be merged. Avoid doing this if the RHS is a floating-point
3640 comparison since those can trap. */
3646 {
3647 /* Convert (a != 0) || (b != 0) into (a | b) != 0. */
3655 integer_zero_node);
3657 /* Convert (a == 0) && (b == 0) into (a | b) == 0. */
3665 integer_zero_node);
3668 }
3670 /* See if the comparisons can be merged. Then get all the parameters for
3671 each side. */
3695 /* It must be true that the inner operation on the lhs of each
3696 comparison must be the same if we are to be able to do anything.
3697 Then see if we have constants. If not, the same must be true for
3698 the rhs's. */
3709 else
3712 /* If either comparison code is not correct for our logical operation,
3713 fail. However, we can convert a one-bit comparison against zero into
3714 the opposite comparison against that bit being set in the field. */
3718 {
3720 {
3721 /* Make the left operand unsigned, since we are only interested
3722 in the value of one bit. Otherwise we are doing the wrong
3723 thing below. */
3726 }
3727 else
3729 }
3731 /* This is analogous to the code for l_const above. */
3733 {
3735 {
3738 }
3739 else
3741 }
3743 /* After this point all optimizations will generate bit-field
3744 references, which we might not want. */
3748 /* See if we can find a mode that contains both fields being compared on
3749 the left. If we can't, fail. Otherwise, update all constants and masks
3750 to be relative to a field of that size. */
3755 volatilep);
3765 {
3768 }
3776 {
3784 {
3788 wanted_code == NE_EXPR
3790 }
3791 }
3793 {
3801 {
3805 wanted_code == NE_EXPR
3807 }
3808 }
3810 /* If the right sides are not constant, do the same for it. Also,
3811 disallow this optimization if a size or signedness mismatch occurs
3812 between the left and right sides. */
3814 {
3817 /* Make sure the two fields on the right
3818 correspond to the left without being swapped. */
3826 volatilep);
3836 {
3839 }
3846 /* Make a mask that corresponds to both fields being compared.
3847 Do this for both items being compared. If the operands are the
3848 same size and the bits being compared are in the same position
3849 then we can do this by masking both and comparing the masked
3850 results. */
3854 {
3866 }
3868 /* There is still another way we can do something: If both pairs of
3869 fields being compared are adjacent, we may be able to make a wider
3870 field containing them both.
3872 Note that we still must mask the lhs/rhs expressions. Furthermore,
3873 the mask must be shifted to account for the shift done by
3874 make_bit_field_ref. */
3879 {
3880 tree type;
3892 /* Convert to the smaller type before masking out unwanted bits. */
3895 {
3897 {
3901 }
3903 {
3907 }
3908 }
3917 }
3920 }
3922 /* Handle the case of comparisons with constants. If there is something in
3923 common between the masks, those bits of the constants must be the same.
3924 If not, the condition is always false. Test for this to avoid generating
3925 incorrect code below. */
3930 {
3932 {
3935 }
3936 else
3937 {
3940 }
3941 }
3943 /* Construct the expression we will return. First get the component
3944 reference we will make. Unless the mask is all ones the width of
3945 that field, perform the mask operation. Then compare with the
3946 merged constant. */
3956 }
3957 \f
3958 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3959 constant. */
3961 static tree
3963 {
3968 tree minmax_const;
3970 tree inner;
3980 /* If something does not permit us to optimize, return the original tree. */
3988 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3989 and GT_EXPR, doing the rest with recursive calls using logical
3990 simplifications. */
3992 {
3994 return
3998 return
4000 optimize_minmax_comparison
4002 optimize_minmax_comparison
4007 /* MAX (X, 0) == 0 -> X <= 0 */
4011 /* MAX (X, 0) == 5 -> X == 5 */
4015 /* MAX (X, 0) == -1 -> false */
4019 /* MIN (X, 0) == 0 -> X >= 0 */
4023 /* MIN (X, 0) == 5 -> false */
4026 else
4027 /* MIN (X, 0) == -1 -> X == -1 */
4032 /* MAX (X, 0) > 0 -> X > 0
4033 MAX (X, 0) > 5 -> X > 5 */
4037 /* MAX (X, 0) > -1 -> true */
4041 /* MIN (X, 0) > 0 -> false
4042 MIN (X, 0) > 5 -> false */
4045 else
4046 /* MIN (X, 0) > -1 -> X > -1 */
4051 }
4052 }
4053 \f
4054 /* T is an integer expression that is being multiplied, divided, or taken a
4055 modulus (CODE says which and what kind of divide or modulus) by a
4056 constant C. See if we can eliminate that operation by folding it with
4057 other operations already in T. WIDE_TYPE, if non-null, is a type that
4058 should be used for the computation if wider than our type.
4060 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
4061 (X * 2) + (Y * 4). We must, however, be assured that either the original
4062 expression would not overflow or that overflow is undefined for the type
4063 in the language in question.
4065 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
4066 the machine has a multiply-accumulate insn or that this is part of an
4067 addressing calculation.
4069 If we return a non-null expression, it is an equivalent form of the
4070 original computation, but need not be in the original type. */
4072 static tree
4074 {
4075 /* To avoid exponential search depth, refuse to allow recursion past
4076 three levels. Beyond that (1) it's highly unlikely that we'll find
4077 something interesting and (2) we've probably processed it before
4078 when we built the inner expression. */
4081 tree ret;
4086 depth++;
4088 depth--;
4091 }
4093 static tree
4095 {
4105 /* Don't deal with constants of zero here; they confuse the code below. */
4115 /* Note that we need not handle conditional operations here since fold
4116 already handles those cases. So just do arithmetic here. */
4118 {
4120 /* For a constant, we can always simplify if we are a multiply
4121 or (for divide and modulus) if it is a multiple of our constant. */
4128 /* If op0 is an expression ... */
4133 /* ... and is unsigned, and its type is smaller than ctype,
4134 then we cannot pass through as widening. */
4140 /* ... or its type is larger than ctype,
4141 then we cannot pass through this truncation. */
4144 /* ... or signedness changes for division or modulus,
4145 then we cannot pass through this conversion. */
4151 /* Pass the constant down and see if we can make a simplification. If
4152 we can, replace this expression with the inner simplification for
4153 possible later conversion to our or some other type. */
4158 code == MULT_EXPR
4169 /* If widening the type changes the signedness, then we can't perform
4170 this optimization as that changes the result. */
4174 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
4177 {
4183 }
4193 /* If the second operand is constant, this is a multiplication
4194 or floor division, by a power of two, so we can treat it that
4195 way unless the multiplier or divisor overflows. */
4197 /* const_binop may not detect overflow correctly,
4198 so check for it explicitly here. */
4212 /* See if we can eliminate the operation on both sides. If we can, we
4213 can return a new PLUS or MINUS. If we can't, the only remaining
4214 cases where we can do anything are if the second operand is a
4215 constant. */
4220 /* If not multiplication, we can only do this if both operands
4221 are divisible by c. */
4227 /* If this was a subtraction, negate OP1 and set it to be an addition.
4228 This simplifies the logic below. */
4235 /* If either OP1 or C are negative, this optimization is not safe for
4236 some of the division and remainder types while for others we need
4237 to change the code. */
4239 {
4247 }
4249 /* If it's a multiply or a division/modulus operation of a multiple
4250 of our constant, do the operation and verify it doesn't overflow. */
4253 {
4257 }
4258 else
4261 /* If we have an unsigned type is not a sizetype, we cannot widen
4262 the operation since it will change the result if the original
4263 computation overflowed. */
4269 /* If we were able to eliminate our operation from the first side,
4270 apply our operation to the second side and reform the PLUS. */
4274 /* The last case is if we are a multiply. In that case, we can
4275 apply the distributive law to commute the multiply and addition
4276 if the multiplication of the constants doesn't overflow. */
4281 op1));
4286 /* We have a special case here if we are doing something like
4287 (C * 8) % 4 since we know that's zero. */
4294 /* ... fall through ... */
4298 /* If we can extract our operation from the LHS, do so and return a
4299 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4300 do something only if the second operand is a constant. */
4312 /* If these are the same operation types, we can associate them
4313 assuming no overflow. */
4320 /* If these operations "cancel" each other, we have the main
4321 optimizations of this pass, which occur when either constant is a
4322 multiple of the other, in which case we replace this with either an
4323 operation or CODE or TCODE.
4325 If we have an unsigned type that is not a sizetype, we cannot do
4326 this since it will change the result if the original computation
4327 overflowed. */
4330 && ! flag_wrapv
4335 {
4346 }
4351 }
4354 }
4355 \f
4356 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4357 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4358 that we may sometimes modify the tree. */
4360 static tree
4362 {
4365 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4370 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4371 don't bother handling any other types. */
4373 {
4377 }
4382 {
4385 }
4388 }
4389 \f
4390 /* Return a node which has the indicated constant VALUE (either 0 or
4391 1), and is of the indicated TYPE. */
4393 static tree
4395 {
4400 integer_zero_node);
4401 else
4402 {
4407 }
4408 }
4410 /* Utility function for the following routine, to see how complex a nesting of
4411 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4412 we don't care (to avoid spending too much time on complex expressions.). */
4414 static int
4416 {
4427 }
4429 /* Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
4430 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4431 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4432 expression, and ARG to `a'. If COND_FIRST_P is nonzero, then the
4433 COND is the first argument to CODE; otherwise (as in the example
4434 given here), it is the second argument. TYPE is the type of the
4435 original expression. */
4437 static tree
4440 {
4444 /* In the end, we'll produce a COND_EXPR. Both arms of the
4445 conditional expression will be binary operations. The left-hand
4446 side of the expression to be executed if the condition is true
4447 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4448 of the expression to be executed if the condition is true will be
4449 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4450 but apply to the expression to be executed if the conditional is
4451 false. */
4456 /* These are the codes to use for the left-hand side and right-hand
4457 side of the COND_EXPR. Normally, they are the same as CODE. */
4460 /* And these are the types of the expressions. */
4466 {
4470 }
4471 else
4472 {
4476 }
4479 {
4483 /* If this operand throws an expression, then it does not make
4484 sense to try to perform a logical or arithmetic operation
4485 involving it. Instead of building `a + throw 3' for example,
4486 we simply build `a, throw 3'. */
4488 {
4490 {
4493 }
4494 else
4496 }
4498 {
4500 {
4503 }
4504 else
4506 }
4507 }
4508 else
4509 {
4514 }
4516 /* If ARG is complex we want to make sure we only evaluate it once. Though
4517 this is only required if it is volatile, it might be more efficient even
4518 if it is not. However, if we succeed in folding one part to a constant,
4519 we do not need to make this SAVE_EXPR. Since we do this optimization
4520 primarily to see if we do end up with constant and this SAVE_EXPR
4521 interferes with later optimizations, suppressing it when we can is
4522 important.
4524 If we are not in a function, we can't make a SAVE_EXPR, so don't try to
4525 do so. Don't try to see if the result is a constant if an arm is a
4526 COND_EXPR since we get exponential behavior in that case. */
4535 {
4544 {
4548 }
4549 }
4562 else
4564 }
4566 \f
4567 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4569 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4570 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4571 ADDEND is the same as X.
4573 X + 0 and X - 0 both give X when X is NaN, infinite, or nonzero
4574 and finite. The problematic cases are when X is zero, and its mode
4575 has signed zeros. In the case of rounding towards -infinity,
4576 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4577 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4579 static bool
4581 {
4585 /* Don't allow the fold with -fsignaling-nans. */
4589 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4593 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4598 /* The mode has signed zeros, and we have to honor their sign.
4599 In this situation, there is only one case we can return true for.
4600 X - 0 is the same as X unless rounding towards -infinity is
4601 supported. */
4603 }
4605 /* Subroutine of fold() that checks comparisons of built-in math
4606 functions against real constants.
4608 FCODE is the DECL_FUNCTION_CODE of the built-in, CODE is the comparison
4609 operator: EQ_EXPR, NE_EXPR, GT_EXPR, LT_EXPR, GE_EXPR or LE_EXPR. TYPE
4610 is the type of the result and ARG0 and ARG1 are the operands of the
4611 comparison. ARG1 must be a TREE_REAL_CST.
4613 The function returns the constant folded tree if a simplification
4614 can be made, and NULL_TREE otherwise. */
4616 static tree
4619 {
4620 REAL_VALUE_TYPE c;
4625 {
4631 {
4632 /* sqrt(x) < y is always false, if y is negative. */
4636 arg);
4638 /* sqrt(x) > y is always true, if y is negative and we
4639 don't care about NaNs, i.e. negative values of x. */
4643 arg);
4645 /* sqrt(x) > y is the same as x >= 0, if y is negative. */
4648 }
4650 {
4651 REAL_VALUE_TYPE c2;
4657 {
4658 /* sqrt(x) > y is x == +Inf, when y is very large. */
4663 /* sqrt(x) > y is always false, when y is very large
4664 and we don't care about infinities. */
4667 arg);
4668 }
4670 /* sqrt(x) > c is the same as x > c*c. */
4673 }
4675 {
4676 REAL_VALUE_TYPE c2;
4682 {
4683 /* sqrt(x) < y is always true, when y is a very large
4684 value and we don't care about NaNs or Infinities. */
4688 arg);
4690 /* sqrt(x) < y is x != +Inf when y is very large and we
4691 don't care about NaNs. */
4696 /* sqrt(x) < y is x >= 0 when y is very large and we
4697 don't care about Infinities. */
4702 /* sqrt(x) < y is x >= 0 && x != +Inf, when y is large. */
4711 dconst0))),
4714 c2)))));
4715 }
4717 /* sqrt(x) < c is the same as x < c*c, if we ignore NaNs. */
4722 /* sqrt(x) < c is the same as x >= 0 && x < c*c. */
4725 {
4730 dconst0))),
4733 c2)))));
4734 }
4735 }
4736 }
4739 }
4741 /* Subroutine of fold() that optimizes comparisons against Infinities,
4742 either +Inf or -Inf.
4744 CODE is the comparison operator: EQ_EXPR, NE_EXPR, GT_EXPR, LT_EXPR,
4745 GE_EXPR or LE_EXPR. TYPE is the type of the result and ARG0 and ARG1
4746 are the operands of the comparison. ARG1 must be a TREE_REAL_CST.
4748 The function returns the constant folded tree if a simplification
4749 can be made, and NULL_TREE otherwise. */
4751 static tree
4753 {
4755 REAL_VALUE_TYPE max;
4756 tree temp;
4761 /* For negative infinity swap the sense of the comparison. */
4767 {
4769 /* x > +Inf is always false, if with ignore sNANs. */
4774 arg0);
4777 /* x <= +Inf is always true, if we don't case about NaNs. */
4781 arg0);
4783 /* x <= +Inf is the same as x == x, i.e. isfinite(x). */
4786 {
4789 }
4794 /* x == +Inf and x >= +Inf are always equal to x > DBL_MAX. */
4800 /* x < +Inf is always equal to x <= DBL_MAX. */
4806 /* x != +Inf is always equal to !(x > DBL_MAX). */
4817 }
4820 }
4822 /* If CODE with arguments ARG0 and ARG1 represents a single bit
4823 equality/inequality test, then return a simplified form of
4824 the test using shifts and logical operations. Otherwise return
4825 NULL. TYPE is the desired result type. */
4827 tree
4829 tree result_type)
4830 {
4831 /* If this is a TRUTH_NOT_EXPR, it may have a single bit test inside
4832 operand 0. */
4834 {
4839 /* Extract the arguments of the EQ/NE. */
4843 /* This requires us to invert the code. */
4845 }
4847 /* If this is testing a single bit, we can optimize the test. */
4851 {
4858 tree arg00;
4860 /* If we have (A & C) != 0 where C is the sign bit of A, convert
4861 this into A < 0. Similarly for (A & C) == 0 into A >= 0. */
4864 {
4869 }
4871 /* At this point, we know that arg0 is not testing the sign bit. */
4875 /* Otherwise we have (A & C) != 0 where C is a single bit,
4876 convert that into ((A >> C2) & 1). Where C2 = log2(C).
4877 Similarly for (A & C) == 0. */
4879 /* If INNER is a right shift of a constant and it plus BITNUM does
4880 not overflow, adjust BITNUM and INNER. */
4887 {
4890 }
4892 /* If we are going to be able to omit the AND below, we must do our
4893 operations as unsigned. If we must use the AND, we have a choice.
4894 Normally unsigned is faster, but for some machines signed is. */
4895 #ifdef LOAD_EXTEND_OP
4897 #else
4899 #endif
4912 /* Put the AND last so it can combine with more things. */
4916 /* Make sure to return the proper type. */
4921 }
4923 }
4925 /* Perform constant folding and related simplification of EXPR.
4926 The related simplifications include x*1 => x, x*0 => 0, etc.,
4927 and application of the associative law.
4928 NOP_EXPR conversions may be removed freely (as long as we
4929 are careful not to change the C type of the overall expression)
4930 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4931 but we can constant-fold them if they have constant operands. */
4933 tree
4935 {
4938 tree tem;
4944 /* WINS will be nonzero when the switch is done
4945 if all operands are constant. */
4948 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4949 Likewise for a SAVE_EXPR that's already been evaluated. */
4953 /* Return right away if a constant. */
4957 #ifdef MAX_INTEGER_COMPUTATION_MODE
4959 #endif
4962 {
4963 tree subop;
4965 /* Special case for conversion ops that can have fixed point args. */
4968 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4974 else
4979 )
4980 /* Note that TREE_CONSTANT isn't enough:
4981 static var addresses are constant but we can't
4982 do arithmetic on them. */
4984 }
4986 {
4990 {
4992 tree subop;
4998 {
4999 /* Signedness matters here. Perhaps we can refine this
5000 later. */
5002 }
5003 else
5004 /* Strip any conversions that don't change the mode. */
5009 else
5014 /* Note that TREE_CONSTANT isn't enough:
5015 static var addresses are constant but we can't
5016 do arithmetic on them. */
5023 }
5024 }
5026 /* If this is a commutative operation, and ARG0 is a constant, move it
5027 to ARG1 to reduce the number of tests below. */
5032 {
5037 }
5039 /* Now WINS is set as described above,
5040 ARG0 is the first operand of EXPR,
5041 and ARG1 is the second operand (if it has more than one operand).
5043 First check for cases where an arithmetic operation is applied to a
5044 compound, conditional, or comparison operation. Push the arithmetic
5045 operation inside the compound or conditional to see if any folding
5046 can then be done. Convert comparison to conditional for this purpose.
5047 The also optimizes non-constant cases that used to be done in
5048 expand_expr.
5050 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
5051 one of the operands is a comparison and the other is a comparison, a
5052 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
5053 code below would make the expression more complex. Change it to a
5054 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
5055 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
5067 {
5077 }
5080 {
5085 {
5095 /* If this was a conversion, and all we did was to move into
5096 inside the COND_EXPR, bring it back out. But leave it if
5097 it is a conversion from integer to integer and the
5098 result precision is no wider than a word since such a
5099 conversion is cheap and may be optimized away by combine,
5100 while it couldn't if it were outside the COND_EXPR. Then return
5101 so we don't get into an infinite recursion loop taking the
5102 conversion out and then back in. */
5114 && (INTEGRAL_TYPE_P
5125 }
5130 }
5141 {
5156 return
5170 return
5173 }
5176 {
5192 /* Other kinds of FIX are not handled properly by fold_convert. */
5197 /* Handle cases of two conversions in a row. */
5200 {
5220 /* In addition to the cases of two conversions in a row
5221 handled below, if we are converting something to its own
5222 type via an object of identical or wider precision, neither
5223 conversion is needed. */
5229 /* Likewise, if the intermediate and final types are either both
5230 float or both integer, we don't need the middle conversion if
5231 it is wider than the final type and doesn't change the signedness
5232 (for integers). Avoid this if the final type is a pointer
5233 since then we sometimes need the inner conversion. Likewise if
5234 the outer has a precision not equal to the size of its mode. */
5244 /* If we have a sign-extension of a zero-extended value, we can
5245 replace that by a single zero-extension. */
5251 /* Two conversions in a row are not needed unless:
5252 - some conversion is floating-point (overstrict for now), or
5253 - the intermediate type is narrower than both initial and
5254 final, or
5255 - the intermediate type and innermost type differ in signedness,
5256 and the outermost type is wider than the intermediate, or
5257 - the initial type is a pointer type and the precisions of the
5258 intermediate and final types differ, or
5259 - the final type is a pointer type and the precisions of the
5260 initial and intermediate types differ. */
5274 }
5278 /* Detect assigning a bitfield. */
5281 {
5282 /* Don't leave an assignment inside a conversion
5283 unless assigning a bitfield. */
5286 /* First do the assignment, then return converted constant. */
5290 }
5292 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
5293 constants (if x has signed type, the sign bit cannot be set
5294 in c). This folds extension into the BIT_AND_EXPR. */
5299 {
5309 <= HOST_BITS_PER_WIDE_INT
5311 {
5318 #ifdef LOAD_EXTEND_OP
5322 {
5326 }
5327 #endif
5328 }
5333 }
5336 {
5339 }
5351 {
5355 }
5364 {
5366 {
5368 HOST_WIDE_INT high;
5379 }
5382 }
5385 /* Convert -((double)float) into (double)(-float). */
5388 {
5393 }
5395 /* Convert - (a - b) to (b - a) for non-floating-point. */
5401 /* Convert -f(x) into f(-x) where f is sin, tan or atan. */
5403 {
5414 {
5422 }
5427 }
5432 {
5434 {
5435 /* If the value is unsigned, then the absolute value is
5436 the same as the ordinary value. */
5439 /* Similarly, if the value is non-negative. */
5442 /* If the value is negative, then the absolute value is
5443 its negation. */
5444 else
5445 {
5447 HOST_WIDE_INT high;
5458 }
5459 }
5461 {
5465 }
5466 }
5469 /* Convert fabs((double)float) into (double)fabsf(float). */
5472 {
5476 targ0)));
5477 }
5504 {
5511 }
5517 /* A + (-B) -> A - B */
5520 /* (-A) + B -> B - A */
5524 {
5528 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
5529 with a constant, and the two constants have no bits in common,
5530 we should treat this as a BIT_IOR_EXPR since this may produce more
5531 simplifications. */
5539 {
5542 }
5544 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
5545 (plus (plus (mult) (mult)) (foo)) so that we can
5546 take advantage of the factoring cases below. */
5551 {
5556 else
5577 }
5580 {
5584 /* (A * C) + (B * C) -> (A+B) * C.
5585 We are most concerned about the case where C is a constant,
5586 but other combinations show up during loop reduction. Since
5587 it is not difficult, try all four possibilities. */
5604 /* No identical multiplicands; see if we can find a common
5605 power-of-two factor in non-power-of-two multiplies. This
5606 can help in multi-dimensional array access. */
5611 {
5616 /* Move min of absolute values to int11. */
5619 {
5623 }
5626 {
5631 }
5632 }
5637 same));
5638 }
5639 }
5641 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5645 /* Likewise if the operands are reversed. */
5649 bit_rotate:
5650 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5651 is a rotate of A by C1 bits. */
5652 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5653 is a rotate of A by B bits. */
5654 {
5663 {
5682 {
5690 TYPE_PRECISION
5695 ? LROTATE_EXPR
5698 }
5700 {
5708 TYPE_PRECISION
5713 ? LROTATE_EXPR
5716 }
5717 }
5718 }
5720 associate:
5721 /* In most languages, can't associate operations on floats through
5722 parentheses. Rather than remember where the parentheses were, we
5723 don't associate floats at all. It shouldn't matter much. However,
5724 associating multiplications is only very slightly inaccurate, so do
5725 that if -funsafe-math-optimizations is specified. */
5730 {
5734 /* Split both trees into variables, constants, and literals. Then
5735 associate each group together, the constants with literals,
5736 then the result with variables. This increases the chances of
5737 literals being recombined later and of generating relocatable
5738 expressions for the sum of a constant and literal. */
5743 /* Only do something if we found more than two objects. Otherwise,
5744 nothing has changed and we risk infinite recursion. */
5749 {
5750 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5759 /* Preserve the MINUS_EXPR if the negative part of the literal is
5760 greater than the positive part. Otherwise, the multiplicative
5761 folding code (i.e extract_muldiv) may be fooled in case
5762 unsigned constants are subtracted, like in the following
5763 example: ((X*2 + 4) - 8U)/2. */
5765 {
5767 {
5771 }
5772 else
5773 {
5777 }
5778 }
5780 {
5784 else
5785 {
5790 }
5791 }
5795 }
5796 }
5798 binary:
5802 {
5803 /* The return value should always have
5804 the same type as the original expression. */
5809 }
5813 /* A - (-B) -> A + B */
5816 /* (-A) - B -> (-B) - A where B is easily negated and we can swap. */
5827 {
5833 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5834 about the case where C is a constant, just try one of the
5835 four possibilities. */
5846 /* Fold A - (A & B) into ~B & A. */
5849 {
5854 arg0));
5859 arg0));
5860 }
5861 }
5863 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5867 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5868 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5869 (-ARG1 + ARG0) reduces to -ARG1. */
5873 /* Fold &x - &x. This can happen from &x.foo - &x.
5874 This is unsafe for certain floats even in non-IEEE formats.
5875 In IEEE, it is unsafe because it does wrong for NaNs.
5876 Also note that operand_equal_p is always false if an operand
5877 is volatile. */
5886 /* (-A) * (-B) -> A * B */
5892 {
5898 /* (a * (1 << b)) is (a << b) */
5914 }
5915 else
5916 {
5917 /* Maybe fold x * 0 to 0. The expressions aren't the same
5918 when x is NaN, since x * 0 is also NaN. Nor are they the
5919 same in modes with signed zeros, since multiplying a
5920 negative value by 0 gives -0, not +0. */
5925 /* In IEEE floating point, x*1 is not equivalent to x for snans. */
5930 /* Transform x * -1.0 into -x. */
5935 /* x*2 is x+x */
5939 {
5942 }
5945 {
5949 /* Optimizations of sqrt(...)*sqrt(...). */
5953 {
5958 /* Optimize sqrt(x)*sqrt(x) as x. */
5963 /* Optimize sqrt(x)*sqrt(y) as sqrt(x*y). */
5968 }
5970 /* Optimize exp(x)*exp(y) as exp(x+y). */
5974 {
5981 }
5983 /* Optimizations of pow(...)*pow(...). */
5987 {
5995 /* Optimize pow(x,y)*pow(z,y) as pow(x*z,y). */
5997 {
6002 arg01));
6004 }
6006 /* Optimize pow(x,y)*pow(x,z) as pow(x,y+z). */
6008 {
6013 arg));
6015 }
6016 }
6018 /* Optimize tan(x)*cos(x) as sin(x). */
6027 {
6028 tree sinfn;
6031 {
6046 }
6051 }
6052 }
6053 }
6057 bit_ior:
6066 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
6068 This results in more efficient code for machines without a NAND
6069 instruction. Combine will canonicalize to the first form
6070 which will allow use of NAND instructions provided by the
6071 backend if they exist. */
6074 {
6079 }
6081 /* See if this can be simplified into a rotate first. If that
6082 is unsuccessful continue in the association code. */
6091 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
6092 with a constant, and the two constants have no bits in common,
6093 we should treat this as a BIT_IOR_EXPR since this may produce more
6094 simplifications. */
6102 {
6105 }
6107 /* See if this can be simplified into a rotate first. If that
6108 is unsuccessful continue in the association code. */
6112 bit_and:
6120 /* Simplify ((int)c & 0377) into (int)c, if c is unsigned char. */
6123 {
6124 unsigned int prec
6131 }
6133 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
6135 This results in more efficient code for machines without a NOR
6136 instruction. Combine will canonicalize to the first form
6137 which will allow use of NOR instructions provided by the
6138 backend if they exist. */
6141 {
6146 }
6156 {
6160 }
6164 /* Don't touch a floating-point divide by zero unless the mode
6165 of the constant can represent infinity. */
6171 /* (-A) / (-B) -> A / B */
6176 /* In IEEE floating point, x/1 is not equivalent to x for snans. */
6181 /* If ARG1 is a constant, we can convert this to a multiply by the
6182 reciprocal. This does not have the same rounding properties,
6183 so only do this if -funsafe-math-optimizations. We can actually
6184 always safely do it if ARG1 is a power of two, but it's hard to
6185 tell if it is or not in a portable manner. */
6187 {
6192 /* Find the reciprocal if optimizing and the result is exact. */
6194 {
6195 REAL_VALUE_TYPE r;
6198 {
6201 }
6202 }
6203 }
6204 /* Convert A/B/C to A/(B*C). */
6207 {
6210 arg1)));
6211 }
6212 /* Convert A/(B/C) to (A/B)*C. */
6215 {
6220 }
6223 {
6225 /* Optimize x/exp(y) into x*exp(-y). */
6229 {
6236 }
6238 /* Optimize x/pow(y,z) into x*pow(y,-z). */
6242 {
6251 }
6252 }
6255 {
6259 /* Optimize sin(x)/cos(x) as tan(x). */
6265 {
6266 tree tanfn;
6274 else
6280 }
6282 /* Optimize cos(x)/sin(x) as 1.0/tan(x). */
6288 {
6289 tree tanfn;
6297 else
6301 {
6306 tmp));
6307 }
6308 }
6309 }
6322 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
6323 operation, EXACT_DIV_EXPR.
6325 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
6326 At one time others generated faster code, it's not clear if they do
6327 after the last round to changes to the DIV code in expmed.c. */
6362 /* Optimize -1 >> x for arithmetic right shifts. */
6365 /* ... fall through ... */
6368 shift:
6374 /* Since negative shift count is not well-defined,
6375 don't try to compute it in the compiler. */
6378 /* Rewrite an LROTATE_EXPR by a constant into an
6379 RROTATE_EXPR by a new constant. */
6381 {
6385 = const_binop
6392 }
6394 /* If we have a rotate of a bit operation with the rotate count and
6395 the second operand of the bit operation both constant,
6396 permute the two operations. */
6409 /* Two consecutive rotates adding up to the width of the mode can
6410 be ignored. */
6441 /* Note that the operand of this must be an int
6442 and its values must be 0 or 1.
6443 ("true" is a fixed value perhaps depending on the language,
6444 but we don't handle values other than 1 correctly yet.) */
6446 /* Avoid infinite recursion. */
6448 {
6453 }
6457 /* Note that the operands of this must be ints
6458 and their values must be 0 or 1.
6459 ("true" is a fixed value perhaps depending on the language.) */
6460 /* If first arg is constant zero, return it. */
6464 /* If either arg is constant true, drop it. */
6468 /* Preserve sequence points. */
6471 /* If second arg is constant zero, result is zero, but first arg
6472 must be evaluated. */
6475 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
6476 case will be handled here. */
6480 truth_andor:
6481 /* We only do these simplifications if we are optimizing. */
6485 /* Check for things like (A || B) && (A || C). We can convert this
6486 to A || (B && C). Note that either operator can be any of the four
6487 truth and/or operations and the transformation will still be
6488 valid. Also note that we only care about order for the
6489 ANDIF and ORIF operators. If B contains side effects, this
6490 might change the truth-value of A. */
6497 {
6517 /* This case if tricky because we must either have commutative
6518 operators or else A10 must not have side-effects. */
6524 a01));
6525 }
6527 /* See if we can build a range comparison. */
6531 /* Check for the possibility of merging component references. If our
6532 lhs is another similar operation, try to merge its rhs with our
6533 rhs. Then try to merge our lhs and rhs. */
6545 /* Note that the operands of this must be ints
6546 and their values must be 0 or true.
6547 ("true" is a fixed value perhaps depending on the language.) */
6548 /* If first arg is constant true, return it. */
6552 /* If either arg is constant zero, drop it. */
6556 /* Preserve sequence points. */
6559 /* If second arg is constant true, result is true, but we must
6560 evaluate first arg. */
6563 /* Likewise for first arg, but note this only occurs here for
6564 TRUTH_OR_EXPR. */
6570 /* If either arg is constant zero, drop it. */
6575 /* If either arg is constant true, this is a logical inversion. */
6588 /* If one arg is a real or integer constant, put it last. */
6593 {
6600 }
6603 {
6611 /* Fold (double)float1 CMP (double)float2 into float1 CMP float2. */
6616 /* (-a) CMP (-b) -> b CMP a */
6623 {
6624 REAL_VALUE_TYPE cst;
6627 /* (-a) CMP CST -> a swap(CMP) (-CST) */
6629 return
6635 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
6636 /* a CMP (-0) -> a CMP 0 */
6641 /* x != NaN is always true, other ops are always false. */
6644 {
6647 }
6649 /* Fold comparisons against infinity. */
6651 {
6655 }
6656 }
6658 /* If this is a comparison of a real constant with a PLUS_EXPR
6659 or a MINUS_EXPR of a real constant, we can convert it into a
6660 comparison with a revised real constant as long as no overflow
6661 occurs when unsafe_math_optimizations are enabled. */
6673 /* Likewise, we can simplify a comparison of a real constant with
6674 a MINUS_EXPR whose first operand is also a real constant, i.e.
6675 (c1 - x) < c2 becomes x > c1-c2. */
6686 /* Fold comparisons against built-in math functions. */
6688 && flag_unsafe_math_optimizations
6690 {
6694 {
6698 }
6699 }
6700 }
6702 /* Convert foo++ == CONST into ++foo == CONST + INCR.
6703 First, see if one arg is constant; find the constant arg
6704 and the other one. */
6705 {
6715 {
6716 /* This optimization is invalid for ordered comparisons
6717 if CONST+INCR overflows or if foo+incr might overflow.
6718 This optimization is invalid for floating point due to rounding.
6719 For pointer types we assume overflow doesn't happen. */
6723 {
6724 tree newconst
6728 /* Do not overwrite the current varop to be a preincrement,
6729 create a new node so that we won't confuse our caller who
6730 might create trees and throw them away, reusing the
6731 arguments that they passed to build. This shows up in
6732 the THEN or ELSE parts of ?: being postincrements. */
6737 /* If VAROP is a reference to a bitfield, we must mask
6738 the constant by the width of the field. */
6742 {
6743 int size
6745 (TREE_OPERAND
6749 tree folded_compare;
6751 /* First check whether the comparison would come out
6752 always the same. If we don't do that we would
6753 change the meaning with the masking. */
6757 else
6760 constop));
6775 mask)));
6776 }
6782 }
6783 }
6785 {
6789 {
6790 tree newconst
6794 /* Do not overwrite the current varop to be a predecrement,
6795 create a new node so that we won't confuse our caller who
6796 might create trees and throw them away, reusing the
6797 arguments that they passed to build. This shows up in
6798 the THEN or ELSE parts of ?: being postdecrements. */
6806 {
6807 int size
6809 (TREE_OPERAND
6813 tree folded_compare;
6818 else
6821 constop));
6836 mask)));
6837 }
6843 }
6844 }
6845 }
6847 /* Change X >= C to X > (C - 1) and X < C to X <= (C - 1) if C > 0.
6848 This transformation affects the cases which are handled in later
6849 optimizations involving comparisons with non-negative constants. */
6853 {
6855 {
6870 }
6871 }
6873 /* Comparisons with the highest or lowest possible integer of
6874 the specified size will have known values. */
6875 {
6883 {
6890 {
6893 }
6894 else
6895 {
6898 }
6903 {
6907 arg0);
6915 arg0);
6921 /* The GE_EXPR and LT_EXPR cases above are not normally
6922 reached because of previous transformations. */
6926 }
6930 {
6943 }
6947 {
6951 arg0);
6960 arg0);
6968 }
6972 {
6985 }
6990 /* signed_type does not work on pointer types. */
6992 {
6993 /* The following case also applies to X < signed_max+1
6994 and X >= signed_max+1 because previous transformations. */
6996 {
7000 return fold
7004 }
7005 }
7006 }
7007 }
7009 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
7010 a MINUS_EXPR of a constant, we can convert it into a comparison with
7011 a revised constant as long as no overflow occurs. */
7023 /* Similarly for a NEGATE_EXPR. */
7032 /* If we have X - Y == 0, we can convert that to X == Y and similarly
7033 for !=. Don't do this for ordered comparisons due to overflow. */
7039 /* If we are widening one operand of an integer comparison,
7040 see if the other operand is similarly being widened. Perhaps we
7041 can do the comparison in the narrower type. */
7051 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
7052 constant, we can simplify it. */
7059 /* If we are comparing an ABS_EXPR with a constant, we can
7060 convert all the cases into explicit comparisons, but they may
7061 well not be faster than doing the ABS and one comparison.
7062 But ABS (X) <= C is a range comparison, which becomes a subtraction
7063 and a comparison, and is probably faster. */
7075 /* If this is an EQ or NE comparison with zero and ARG0 is
7076 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
7077 two operations, but the latter can be done in one less insn
7078 on machines that have only two-operand insns or on which a
7079 constant cannot be the first operand. */
7082 {
7085 return
7093 integer_one_node)),
7094 arg1));
7097 return
7105 integer_one_node)),
7106 arg1));
7107 }
7109 /* If this is an NE or EQ comparison of zero against the result of a
7110 signed MOD operation whose second operand is a power of 2, make
7111 the MOD operation unsigned since it is simpler and equivalent. */
7120 {
7127 }
7129 /* If this is an NE comparison of zero with an AND of one, remove the
7130 comparison since the AND will give the correct value. */
7136 /* If we have (A & C) == C where C is a power of 2, convert this into
7137 (A & C) != 0. Similarly for NE_EXPR. */
7145 /* If we have (A & C) != 0 or (A & C) == 0 and C is a power of
7146 2, then fold the expression into shifts and logical operations. */
7151 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
7152 and similarly for >= into !=. */
7168 return
7175 /* Simplify comparison of something with itself. (For IEEE
7176 floating-point, we can only do some of these simplifications.) */
7178 {
7180 {
7192 /* For NE, we can only do this simplification if integer
7193 or we don't honor IEEE floating point NaNs. */
7197 /* ... fall through ... */
7203 }
7204 }
7206 /* If we are comparing an expression that just has comparisons
7207 of two integer values, arithmetic expressions of those comparisons,
7208 and constants, we can simplify it. There are only three cases
7209 to check: the two values can either be equal, the first can be
7210 greater, or the second can be greater. Fold the expression for
7211 those three values. Since each value must be 0 or 1, we have
7212 eight possibilities, each of which corresponds to the constant 0
7213 or 1 or one of the six possible comparisons.
7215 This handles common cases like (a > b) == 0 but also handles
7216 expressions like ((x > y) - (y > x)) > 0, which supposedly
7217 occur in macroized code. */
7220 {
7225 /* Don't handle degenerate cases here; they should already
7226 have been handled anyway. */
7235 {
7239 /* We can't just pass T to eval_subst in case cval1 or cval2
7240 was the same as ARG1. */
7242 tree high_result
7245 arg1));
7246 tree equal_result
7249 arg1));
7250 tree low_result
7253 arg1));
7255 /* All three of these results should be 0 or 1. Confirm they
7256 are. Then use those values to select the proper code
7257 to use. */
7265 {
7266 /* Make a 3-bit mask with the high-order bit being the
7267 value for `>', the next for '=', and the low for '<'. */
7271 {
7273 /* Always false. */
7294 /* Always true. */
7296 }
7301 else
7303 }
7304 }
7305 }
7307 /* If this is a comparison of a field, we may be able to simplify it. */
7312 /* Handle the constant case even without -O
7313 to make sure the warnings are given. */
7315 {
7318 }
7320 /* If this is a comparison of complex values and either or both sides
7321 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
7322 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
7323 This may prevent needless evaluations. */
7330 {
7343 type,
7346 }
7348 /* Optimize comparisons of strlen vs zero to a compare of the
7349 first character of the string vs zero. To wit,
7350 strlen(ptr) == 0 => *ptr == 0
7351 strlen(ptr) != 0 => *ptr != 0
7352 Other cases should reduce to one of these two (or a constant)
7353 due to the return value of strlen being unsigned. */
7358 {
7360 tree arglist;
7372 integer_zero_node));
7373 }
7375 /* From here on, the only cases we handle are when the result is
7376 known to be a constant.
7378 To compute GT, swap the arguments and do LT.
7379 To compute GE, do LT and invert the result.
7380 To compute LE, swap the arguments, do LT and invert the result.
7381 To compute NE, do EQ and invert the result.
7383 Therefore, the code below must handle only EQ and LT. */
7386 {
7389 }
7391 /* Note that it is safe to invert for real values here because we
7392 will check below in the one case that it matters. */
7397 {
7400 }
7402 /* Compute a result for LT or EQ if args permit;
7403 otherwise return T. */
7405 {
7408 else
7413 }
7416 /* Assume a nonexplicit constant cannot equal an explicit one,
7417 since such code would be undefined anyway.
7418 Exception: on sysvr4, using #pragma weak,
7419 a label can come out as 0. */
7426 #endif
7427 /* Two real constants can be compared explicitly. */
7429 {
7430 /* If either operand is a NaN, the result is false with two
7431 exceptions: First, an NE_EXPR is true on NaNs, but that case
7432 is already handled correctly since we will be inverting the
7433 result for NE_EXPR. Second, if we had inverted a LE_EXPR
7434 or a GE_EXPR into a LT_EXPR, we must return true so that it
7435 will be inverted into false. */
7445 else
7449 }
7463 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
7464 so all simple results must be passed through pedantic_non_lvalue. */
7466 return pedantic_non_lvalue
7471 /* If the second operand is zero, invert the comparison and swap
7472 the second and third operands. Likewise if the second operand
7473 is constant and the third is not or if the third operand is
7474 equivalent to the first operand of the comparison. */
7482 {
7483 /* See if this can be inverted. If it can't, possibly because
7484 it was a floating-point inequality comparison, don't do
7485 anything. */
7489 {
7493 /* arg1 should be the first argument of the new T. */
7496 }
7497 }
7499 /* If we have A op B ? A : C, we may be able to convert this to a
7500 simpler expression, depending on the operation and the values
7501 of B and C. Signed zeros prevent all of these transformations,
7502 for reasons given above each one. */
7508 {
7514 /* If we have A op 0 ? A : -A, consider applying the following
7515 transformations:
7517 A == 0? A : -A same as -A
7518 A != 0? A : -A same as A
7519 A >= 0? A : -A same as abs (A)
7520 A > 0? A : -A same as abs (A)
7521 A <= 0? A : -A same as -abs (A)
7522 A < 0? A : -A same as -abs (A)
7524 None of these transformations work for modes with signed
7525 zeros. If A is +/-0, the first two transformations will
7526 change the sign of the result (from +0 to -0, or vice
7527 versa). The last four will fix the sign of the result,
7528 even though the original expressions could be positive or
7529 negative, depending on the sign of A.
7531 Note that all these transformations are correct if A is
7532 NaN, since the two alternatives (A and -A) are also NaNs. */
7539 {
7541 return
7542 pedantic_non_lvalue
7544 negate_expr
7546 arg1))));
7554 return pedantic_non_lvalue
7562 return pedantic_non_lvalue
7566 arg1)))));
7569 }
7571 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
7572 A == 0 ? A : 0 is always 0 unless A is -0. Note that
7573 both transformations are correct when A is NaN: A != 0
7574 is then true, and A == 0 is false. */
7577 {
7582 }
7584 /* Try some transformations of A op B ? A : B.
7586 A == B? A : B same as B
7587 A != B? A : B same as A
7588 A >= B? A : B same as max (A, B)
7589 A > B? A : B same as max (B, A)
7590 A <= B? A : B same as min (A, B)
7591 A < B? A : B same as min (B, A)
7593 As above, these transformations don't work in the presence
7594 of signed zeros. For example, if A and B are zeros of
7595 opposite sign, the first two transformations will change
7596 the sign of the result. In the last four, the original
7597 expressions give different results for (A=+0, B=-0) and
7598 (A=-0, B=+0), but the transformed expressions do not.
7600 The first two transformations are correct if either A or B
7601 is a NaN. In the first transformation, the condition will
7602 be false, and B will indeed be chosen. In the case of the
7603 second transformation, the condition A != B will be true,
7604 and A will be chosen.
7606 The conversions to max() and min() are not correct if B is
7607 a number and A is not. The conditions in the original
7608 expressions will be false, so all four give B. The min()
7609 and max() versions would give a NaN instead. */
7612 {
7617 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
7619 {
7623 }
7626 {
7633 /* In C++ a ?: expression can be an lvalue, so put the
7634 operand which will be used if they are equal first
7635 so that we can convert this back to the
7636 corresponding COND_EXPR. */
7638 return pedantic_non_lvalue
7648 return pedantic_non_lvalue
7657 }
7658 }
7660 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
7661 we might still be able to simplify this. For example,
7662 if C1 is one less or one more than C2, this might have started
7663 out as a MIN or MAX and been transformed by this function.
7664 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
7670 {
7672 /* We can replace A with C1 in this case. */
7679 /* If C1 is C2 + 1, this is min(A, C2). */
7684 return pedantic_non_lvalue
7689 /* If C1 is C2 - 1, this is min(A, C2). */
7694 return pedantic_non_lvalue
7699 /* If C1 is C2 - 1, this is max(A, C2). */
7704 return pedantic_non_lvalue
7709 /* If C1 is C2 + 1, this is max(A, C2). */
7714 return pedantic_non_lvalue
7721 }
7722 }
7724 /* If the second operand is simpler than the third, swap them
7725 since that produces better jump optimization results. */
7731 {
7732 /* See if this can be inverted. If it can't, possibly because
7733 it was a floating-point inequality comparison, don't do
7734 anything. */
7738 {
7742 /* arg1 should be the first argument of the new T. */
7745 }
7746 }
7748 /* Convert A ? 1 : 0 to simply A. */
7751 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
7752 call to fold will try to move the conversion inside
7753 a COND, which will recurse. In that case, the COND_EXPR
7754 is probably the best choice, so leave it alone. */
7758 /* Convert A ? 0 : 1 to !A. This prefers the use of NOT_EXPR
7759 over COND_EXPR in cases such as floating point comparisons. */
7766 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
7767 operation is simply A & 2. */
7778 /* Convert A ? B : 0 into A && B if A and B are truth values. */
7785 /* Convert A ? B : 1 into !A || B if A and B are truth values. */
7789 {
7790 /* Only perform transformation if ARG0 is easily inverted. */
7795 }
7800 /* When pedantic, a compound expression can be neither an lvalue
7801 nor an integer constant expression. */
7804 /* Don't let (0, 0) be null pointer constant. */
7846 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
7847 appropriate. */
7852 {
7856 tree arg01;
7867 {
7881 arg01));
7882 }
7885 }
7888 /* Check for a built-in function. */
7893 {
7897 }
7903 }
7905 /* Perform constant folding and related simplification of intializer
7906 expression EXPR. This behaves identically to "fold" but ignores
7907 potential run-time traps and exceptions that fold must preserve. */
7909 tree
7911 {
7915 tree result;
7928 }
7930 /* Determine if first argument is a multiple of second argument. Return 0 if
7931 it is not, or we cannot easily determined it to be.
7933 An example of the sort of thing we care about (at this point; this routine
7934 could surely be made more general, and expanded to do what the *_DIV_EXPR's
7935 fold cases do now) is discovering that
7937 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7939 is a multiple of
7941 SAVE_EXPR (J * 8)
7943 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
7945 This code also handles discovering that
7947 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7949 is a multiple of 8 so we don't have to worry about dealing with a
7950 possible remainder.
7952 Note that we *look* inside a SAVE_EXPR only to determine how it was
7953 calculated; it is not safe for fold to do much of anything else with the
7954 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
7955 at run time. For example, the latter example above *cannot* be implemented
7956 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
7957 evaluation time of the original SAVE_EXPR is not necessarily the same at
7958 the time the new expression is evaluated. The only optimization of this
7959 sort that would be valid is changing
7961 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
7963 divided by 8 to
7965 SAVE_EXPR (I) * SAVE_EXPR (J)
7967 (where the same SAVE_EXPR (J) is used in the original and the
7968 transformed version). */
7970 static int
7972 {
7980 {
7992 {
7996 /* const_binop may not detect overflow correctly,
7997 so check for it explicitly here. */
8006 }
8010 /* Can't handle conversions from non-integral or wider integral type. */
8016 /* .. fall through ... */
8032 }
8033 }
8035 /* Return true if `t' is known to be non-negative. */
8037 int
8039 {
8041 {
8050 /* These are undefined at zero. This is true even if
8051 C[LT]Z_DEFINED_VALUE_AT_ZERO is set, since what we're
8052 computing here is a user-visible property. */
8066 /* zero_extend(x) + zero_extend(y) is non-negative if x and y are
8067 both unsigned and at least 2 bits shorter than the result. */
8071 {
8076 {
8080 }
8081 }
8086 {
8087 /* x * x for floating point x is always non-negative. */
8092 }
8094 /* zero_extend(x) * zero_extend(y) is non-negative if x and y are
8095 both unsigned and their total bits is shorter than the result. */
8099 {
8106 }
8127 {
8132 {
8136 {
8140 }
8141 }
8143 {
8149 }
8150 }
8179 {
8186 {
8228 }
8229 }
8231 /* ... fall through ... */
8235 /* Truth values evaluate to 0 or 1, which is nonnegative. */
8237 }
8239 /* We don't know sign of `t', so be conservative and return false. */
8241 }
8243 /* Return true if `r' is known to be non-negative.
8244 Only handles constants at the moment. */
8246 int
8248 {
8250 {
8260 {
8262 rtx elt;
8267 {
8271 }
8274 }
8278 /* These are always nonnegative. */
8283 }
8284 }