| blob | 48c3683f8c2667de2c0967cf2aa97c61e12f6480 |
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. */
62 HOST_WIDE_INT));
65 HOST_WIDE_INT *));
90 HOST_WIDE_INT *,
112 static tree fold_binary_op_with_conditional_arg
116 /* The following constants represent a bit based encoding of GCC's
117 comparison operators. This encoding simplifies transformations
118 on relational comparison operators, such as AND and OR. */
119 #define COMPCODE_FALSE 0
120 #define COMPCODE_LT 1
121 #define COMPCODE_EQ 2
122 #define COMPCODE_LE 3
123 #define COMPCODE_GT 4
124 #define COMPCODE_NE 5
125 #define COMPCODE_GE 6
126 #define COMPCODE_TRUE 7
128 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
129 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
130 and SUM1. Then this yields nonzero if overflow occurred during the
131 addition.
133 Overflow occurs if A and B have the same sign, but A and SUM differ in
134 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
135 sign. */
136 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
137 \f
138 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
139 We do that by representing the two-word integer in 4 words, with only
140 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
141 number. The value of the word is LOWPART + HIGHPART * BASE. */
143 #define LOWPART(x) \
144 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
145 #define HIGHPART(x) \
146 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
147 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
149 /* Unpack a two-word integer into 4 words.
150 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
151 WORDS points to the array of HOST_WIDE_INTs. */
153 static void
157 HOST_WIDE_INT hi;
158 {
163 }
165 /* Pack an array of 4 words into a two-word integer.
166 WORDS points to the array of words.
167 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
169 static void
174 {
177 }
178 \f
179 /* Make the integer constant T valid for its type by setting to 0 or 1 all
180 the bits in the constant that don't belong in the type.
182 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
183 nonzero, a signed overflow has already occurred in calculating T, so
184 propagate it. */
186 int
188 tree t;
190 {
192 HOST_WIDE_INT high;
196 {
197 /* ??? Used to check for overflow here via CHECK_FLOAT_TYPE.
198 Consider doing it via real_convert now. */
200 }
210 else
213 /* First clear all bits that are beyond the type's precision. */
216 ;
220 else
221 {
225 }
227 /* Unsigned types do not suffer sign extension or overflow unless they
228 are a sizetype. */
234 /* If the value's sign bit is set, extend the sign. */
242 {
243 /* Value is negative:
244 set to 1 all the bits that are outside this type's precision. */
248 else
249 {
253 }
254 }
256 /* Return nonzero if signed overflow occurred. */
257 return
260 }
261 \f
262 /* Add two doubleword integers with doubleword result.
263 Each argument is given as two `HOST_WIDE_INT' pieces.
264 One argument is L1 and H1; the other, L2 and H2.
265 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
267 int
273 {
275 HOST_WIDE_INT h;
283 }
285 /* Negate a doubleword integer with doubleword result.
286 Return nonzero if the operation overflows, assuming it's signed.
287 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
288 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
290 int
293 HOST_WIDE_INT h1;
296 {
298 {
302 }
303 else
304 {
308 }
309 }
310 \f
311 /* Multiply two doubleword integers with doubleword result.
312 Return nonzero if the operation overflows, assuming it's signed.
313 Each argument is given as two `HOST_WIDE_INT' pieces.
314 One argument is L1 and H1; the other, L2 and H2.
315 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
317 int
323 {
338 {
341 {
343 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
345 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
349 }
351 }
355 /* Check for overflow by calculating the top half of the answer in full;
356 it should agree with the low half's sign bit. */
359 {
362 }
364 {
367 }
369 }
370 \f
371 /* Shift the doubleword integer in L1, H1 left by COUNT places
372 keeping only PREC bits of result.
373 Shift right if COUNT is negative.
374 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
375 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
377 void
385 {
389 {
392 }
394 #ifdef SHIFT_COUNT_TRUNCATED
397 #endif
400 {
401 /* Shifting by the host word size is undefined according to the
402 ANSI standard, so we must handle this as a special case. */
405 }
407 {
410 }
411 else
412 {
416 }
418 /* Sign extend all bits that are beyond the precision. */
426 ;
428 {
431 }
432 else
433 {
437 }
438 }
440 /* Shift the doubleword integer in L1, H1 right by COUNT places
441 keeping only PREC bits of result. COUNT must be positive.
442 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
443 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
445 void
453 {
456 signmask = (arith
460 #ifdef SHIFT_COUNT_TRUNCATED
463 #endif
466 {
467 /* Shifting by the host word size is undefined according to the
468 ANSI standard, so we must handle this as a special case. */
471 }
473 {
476 }
477 else
478 {
482 }
484 /* Zero / sign extend all bits that are beyond the precision. */
487 {
490 }
492 ;
494 {
497 }
498 else
499 {
503 }
504 }
505 \f
506 /* Rotate the doubleword integer in L1, H1 left by COUNT places
507 keeping only PREC bits of result.
508 Rotate right if COUNT is negative.
509 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
511 void
518 {
530 }
532 /* Rotate the doubleword integer in L1, H1 left by COUNT places
533 keeping only PREC bits of result. COUNT must be positive.
534 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
536 void
543 {
555 }
556 \f
557 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
558 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
559 CODE is a tree code for a kind of division, one of
560 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
561 or EXACT_DIV_EXPR
562 It controls how the quotient is rounded to an integer.
563 Return nonzero if the operation overflows.
564 UNS nonzero says do unsigned division. */
566 int
573 HOST_WIDE_INT hnum_orig;
575 HOST_WIDE_INT hden_orig;
578 {
594 /* calculate quotient sign and convert operands to unsigned. */
596 {
598 {
600 /* (minimum integer) / (-1) is the only overflow case. */
604 }
606 {
609 }
610 }
615 /* This unsigned division rounds toward zero. */
618 }
622 /* hden != 0 already checked. */
627 }
637 /* Special code for when the divisor < BASE. */
639 {
640 /* hnum != 0 already checked. */
642 {
646 }
647 }
648 else
649 {
650 /* Full double precision division,
651 with thanks to Don Knuth's "Seminumerical Algorithms". */
655 /* Find the highest nonzero divisor digit. */
658 {
661 }
663 /* Insure that the first digit of the divisor is at least BASE/2.
664 This is required by the quotient digit estimation algorithm. */
671 {
675 }
680 {
685 }
686 }
690 /* Main loop */
692 {
693 /* Guess the next quotient digit, quo_est, by dividing the first
694 two remaining dividend digits by the high order quotient digit.
695 quo_est is never low and is at most 2 high. */
702 else
705 /* Refine quo_est so it's usually correct, and at most one high. */
710 quo_est--;
712 /* Try QUO_EST as the quotient digit, by multiplying the
713 divisor by QUO_EST and subtracting from the remaining dividend.
714 Keep in mind that QUO_EST is the I - 1st digit. */
718 {
724 }
726 /* If quo_est was high by one, then num[i] went negative and
727 we need to correct things. */
729 {
730 quo_est--;
733 {
737 }
740 }
742 /* Store the quotient digit. */
744 }
745 }
749 finish_up:
750 /* if result is negative, make it so. */
754 /* compute trial remainder: rem = num - (quo * den) */
760 {
769 {
770 /* quo = quo - 1; */
773 }
774 else
781 {
784 }
785 else
791 {
797 /* Get absolute values */
803 /* If (2 * abs (lrem) >= abs (lden)) */
812 {
814 /* quo = quo - 1; */
817 else
818 /* quo = quo + 1; */
821 }
822 else
824 }
829 }
831 /* compute true remainder: rem = num - (quo * den) */
836 }
837 \f
838 /* Determine whether an expression T can be cheaply negated using
839 the function negate_expr. */
841 static bool
843 tree t;
844 {
847 tree type;
856 {
861 /* Check that -CST will not overflow type. */
864 {
869 }
870 else
883 }
885 }
887 /* Given T, an expression, return the negation of T. Allow for T to be
888 null, in which case return null. */
890 static tree
892 tree t;
893 {
894 tree type;
895 tree tem;
904 {
917 /* - (A - B) -> B - A */
927 }
930 }
931 \f
932 /* Split a tree IN into a constant, literal and variable parts that could be
933 combined with CODE to make IN. "constant" means an expression with
934 TREE_CONSTANT but that isn't an actual constant. CODE must be a
935 commutative arithmetic operation. Store the constant part into *CONP,
936 the literal in *LITP and return the variable part. If a part isn't
937 present, set it to null. If the tree does not decompose in this way,
938 return the entire tree as the variable part and the other parts as null.
940 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
941 case, we negate an operand that was subtracted. Except if it is a
942 literal for which we use *MINUS_LITP instead.
944 If NEGATE_P is true, we are negating all of IN, again except a literal
945 for which we use *MINUS_LITP instead.
947 If IN is itself a literal or constant, return it as appropriate.
949 Note that we do not guarantee that any of the three values will be the
950 same type as IN, but they will have the same signedness and mode. */
952 static tree
954 tree in;
958 {
965 /* Strip any conversions that don't change the machine mode or signedness. */
972 /* We can associate addition and subtraction together (even
973 though the C standard doesn't say so) for integers because
974 the value is not affected. For reals, the value might be
975 affected, so we can't. */
978 {
984 /* First see if either of the operands is a literal, then a constant. */
995 /* If we haven't dealt with either operand, this is not a case we can
996 decompose. Otherwise, VAR is either of the ones remaining, if any. */
1001 else
1004 /* Now do any needed negations. */
1011 }
1014 else
1018 {
1025 }
1028 }
1030 /* Re-associate trees split by the above function. T1 and T2 are either
1031 expressions to associate or null. Return the new expression, if any. If
1032 we build an operation, do it in TYPE and with CODE. */
1034 static tree
1038 tree type;
1039 {
1045 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
1046 try to fold this since we will have infinite recursion. But do
1047 deal with any NEGATE_EXPRs. */
1050 {
1052 {
1059 }
1061 }
1064 }
1065 \f
1066 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1067 to produce a new constant.
1069 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1071 static tree
1076 {
1080 HOST_WIDE_INT hi;
1082 HOST_WIDE_INT garbageh;
1083 tree t;
1086 int is_sizetype
1097 {
1117 /* It's unclear from the C standard whether shifts can overflow.
1118 The following code ignores overflow; perhaps a C standard
1119 interpretation ruling is needed. */
1149 /* This is a shortcut for a common special case. */
1154 {
1160 }
1162 /* ... fall through ... */
1166 {
1169 }
1172 {
1175 }
1182 /* This is a shortcut for a common special case. */
1187 {
1192 }
1194 /* ... fall through ... */
1210 else
1216 else
1222 }
1224 /* If this is for a sizetype, can be represented as one (signed)
1225 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1226 constants. */
1232 else
1233 {
1236 }
1239 = ((notrunc
1246 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1247 So check if force_fit_type truncated the value. */
1258 }
1260 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1261 constant. We assume ARG1 and ARG2 have the same data type, or at least
1262 are the same kind of constant and the same machine mode.
1264 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1266 static tree
1271 {
1279 {
1280 REAL_VALUE_TYPE d1;
1281 REAL_VALUE_TYPE d2;
1282 REAL_VALUE_TYPE value;
1283 tree t;
1288 /* If either operand is a NaN, just return it. Otherwise, set up
1289 for floating-point trap; we return an overflow. */
1299 value));
1309 }
1311 {
1317 tree t;
1320 {
1340 notrunc),
1346 notrunc));
1350 {
1351 tree magsquared
1355 notrunc);
1358 const_binop
1363 notrunc),
1365 notrunc),
1366 notrunc),
1368 const_binop
1373 notrunc),
1375 notrunc),
1376 notrunc),
1378 }
1383 }
1385 }
1387 }
1389 /* These are the hash table functions for the hash table of INTEGER_CST
1390 nodes of a sizetype. */
1392 /* Return the hash code code X, an INTEGER_CST. */
1394 static hashval_t
1397 {
1403 }
1405 /* Return nonzero if the value represented by *X (an INTEGER_CST tree node)
1406 is the same as that given by *Y, which is the same. */
1408 static int
1412 {
1420 }
1421 \f
1422 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1423 bits are given by NUMBER and of the sizetype represented by KIND. */
1425 tree
1427 HOST_WIDE_INT number;
1429 {
1431 }
1433 /* Likewise, but the desired type is specified explicitly. */
1437 htab_t size_htab;
1439 tree
1441 HOST_WIDE_INT number;
1442 tree type;
1443 {
1447 {
1450 }
1452 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1453 hash table, we return the value from the hash table. Otherwise, we
1454 place that in the hash table and make a new node for the next time. */
1463 {
1469 }
1470 else
1472 }
1474 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1475 is a tree code. The type of the result is taken from the operands.
1476 Both must be the same type integer type and it must be a size type.
1477 If the operands are constant, so is the result. */
1479 tree
1483 {
1490 /* Handle the special case of two integer constants faster. */
1492 {
1493 /* And some specific cases even faster than that. */
1502 /* Handle general case of two integer constants. */
1504 }
1510 }
1512 /* Given two values, either both of sizetype or both of bitsizetype,
1513 compute the difference between the two values. Return the value
1514 in signed type corresponding to the type of the operands. */
1516 tree
1519 {
1521 tree ctype;
1527 /* If the type is already signed, just do the simple thing. */
1534 /* If either operand is not a constant, do the conversions to the signed
1535 type and subtract. The hardware will do the right thing with any
1536 overflow in the subtraction. */
1541 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1542 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1543 overflow) and negate (which can't either). Special-case a result
1544 of zero while we're here. */
1549 else
1552 }
1553 \f
1555 /* Given T, a tree representing type conversion of ARG1, a constant,
1556 return a constant tree representing the result of conversion. */
1558 static tree
1560 tree t;
1561 tree arg1;
1562 {
1567 {
1569 {
1570 /* If we would build a constant wider than GCC supports,
1571 leave the conversion unfolded. */
1575 /* If we are trying to make a sizetype for a small integer, use
1576 size_int to pick up cached types to reduce duplicate nodes. */
1582 /* Given an integer constant, make new constant with new type,
1583 appropriately sign-extended or truncated. */
1587 /* Indicate an overflow if (1) ARG1 already overflowed,
1588 or (2) force_fit_type indicates an overflow.
1589 Tell force_fit_type that an overflow has already occurred
1590 if ARG1 is a too-large unsigned value and T is signed.
1591 But don't indicate an overflow if converting a pointer. */
1601 }
1603 {
1604 /* Don't initialize these, use assignments.
1605 Initialized local aggregates don't work on old compilers. */
1606 REAL_VALUE_TYPE x;
1607 REAL_VALUE_TYPE l;
1608 REAL_VALUE_TYPE u;
1619 /* See if X will be in range after truncation towards 0.
1620 To compensate for truncation, move the bounds away from 0,
1621 but reject if X exactly equals the adjusted bounds. */
1625 /* If X is a NaN, use zero instead and show we have an overflow.
1626 Otherwise, range check. */
1630 && !no_upper_bound
1634 {
1638 }
1644 }
1646 }
1648 {
1652 {
1654 {
1655 /* We make a copy of ARG1 so that we don't modify an
1656 existing constant tree. */
1660 }
1671 }
1672 }
1675 }
1676 \f
1677 /* Return an expr equal to X but certainly not valid as an lvalue. */
1679 tree
1681 tree x;
1682 {
1683 tree result;
1685 /* These things are certainly not lvalues. */
1696 }
1698 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1699 Zero means allow extended lvalues. */
1703 /* When pedantic, return an expr equal to X but certainly not valid as a
1704 pedantic lvalue. Otherwise, return X. */
1706 tree
1708 tree x;
1709 {
1712 else
1714 }
1715 \f
1716 /* Given a tree comparison code, return the code that is the logical inverse
1717 of the given code. It is not safe to do this for floating-point
1718 comparisons, except for NE_EXPR and EQ_EXPR. */
1720 static enum tree_code
1723 {
1725 {
1740 }
1741 }
1743 /* Similar, but return the comparison that results if the operands are
1744 swapped. This is safe for floating-point. */
1746 static enum tree_code
1749 {
1751 {
1765 }
1766 }
1769 /* Convert a comparison tree code from an enum tree_code representation
1770 into a compcode bit-based encoding. This function is the inverse of
1771 compcode_to_comparison. */
1773 static int
1776 {
1778 {
1793 }
1794 }
1796 /* Convert a compcode bit-based encoding of a comparison operator back
1797 to GCC's enum tree_code representation. This function is the
1798 inverse of comparison_to_compcode. */
1800 static enum tree_code
1803 {
1805 {
1820 }
1821 }
1823 /* Return nonzero if CODE is a tree code that represents a truth value. */
1825 static int
1828 {
1833 }
1834 \f
1835 /* Return nonzero if two operands are necessarily equal.
1836 If ONLY_CONST is nonzero, only return nonzero for constants.
1837 This function tests whether the operands are indistinguishable;
1838 it does not test whether they are equal using C's == operation.
1839 The distinction is important for IEEE floating point, because
1840 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1841 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1843 int
1847 {
1848 /* If both types don't have the same signedness, then we can't consider
1849 them equal. We must check this before the STRIP_NOPS calls
1850 because they may change the signedness of the arguments. */
1858 /* This is needed for conversions and for COMPONENT_REF.
1859 Might as well play it safe and always test this. */
1865 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1866 We don't care about side effects in that case because the SAVE_EXPR
1867 takes care of that for us. In all other cases, two expressions are
1868 equal if they have no side effects. If we have two identical
1869 expressions with side effects that should be treated the same due
1870 to the only side effects being identical SAVE_EXPR's, that will
1871 be detected in the recursive calls below. */
1877 /* Next handle constant cases, those for which we can return 1 even
1878 if ONLY_CONST is set. */
1881 {
1894 {
1904 {
1909 }
1912 }
1916 only_const)
1918 only_const));
1931 }
1937 {
1939 /* Two conversions are equal only if signedness and modes match. */
1955 /* For commutative ops, allow the other order. */
1968 /* If either of the pointer (or reference) expressions we are dereferencing
1969 contain a side effect, these cannot be equal. */
1975 {
1997 }
2006 }
2007 }
2008 \f
2009 /* Similar to operand_equal_p, but see if ARG0 might have been made by
2010 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
2012 When in doubt, return 0. */
2014 static int
2017 tree other;
2018 {
2030 /* Discard any conversions that don't change the modes of ARG0 and ARG1
2031 and see if the inner values are the same. This removes any
2032 signedness comparison, which doesn't matter here. */
2039 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
2040 actual comparison operand, ARG0.
2042 First throw away any conversions to wider types
2043 already present in the operands. */
2052 {
2055 /* Make sure shorter operand is extended the right way
2056 to match the longer operand. */
2062 }
2065 }
2066 \f
2067 /* See if ARG is an expression that is either a comparison or is performing
2068 arithmetic on comparisons. The comparisons must only be comparing
2069 two different values, which will be stored in *CVAL1 and *CVAL2; if
2070 they are nonzero it means that some operands have already been found.
2071 No variables may be used anywhere else in the expression except in the
2072 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
2073 the expression and save_expr needs to be called with CVAL1 and CVAL2.
2075 If this is true, return 1. Otherwise, return zero. */
2077 static int
2079 tree arg;
2082 {
2086 /* We can handle some of the 'e' cases here. */
2096 {
2097 /* If we've already found a CVAL1 or CVAL2, this expression is
2098 two complex to handle. */
2104 }
2107 {
2130 /* First see if we can handle the first operand, then the second. For
2131 the second operand, we know *CVAL1 can't be zero. It must be that
2132 one side of the comparison is each of the values; test for the
2133 case where this isn't true by failing if the two operands
2134 are the same. */
2143 ;
2147 ;
2148 else
2152 ;
2156 ;
2157 else
2164 }
2165 }
2166 \f
2167 /* ARG is a tree that is known to contain just arithmetic operations and
2168 comparisons. Evaluate the operations in the tree substituting NEW0 for
2169 any occurrence of OLD0 as an operand of a comparison and likewise for
2170 NEW1 and OLD1. */
2172 static tree
2174 tree arg;
2176 {
2181 /* We can handle some of the 'e' cases here. */
2189 {
2204 {
2221 }
2222 /* fall through - ??? */
2225 {
2229 /* We need to check both for exact equality and tree equality. The
2230 former will be true if the operand has a side-effect. In that
2231 case, we know the operand occurred exactly once. */
2244 }
2248 }
2249 }
2250 \f
2251 /* Return a tree for the case when the result of an expression is RESULT
2252 converted to TYPE and OMITTED was previously an operand of the expression
2253 but is now not needed (e.g., we folded OMITTED * 0).
2255 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2256 the conversion of RESULT to TYPE. */
2258 tree
2261 {
2268 }
2270 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2272 static tree
2275 {
2282 }
2283 \f
2284 /* Return a simplified tree node for the truth-negation of ARG. This
2285 never alters ARG itself. We assume that ARG is an operation that
2286 returns a truth value (0 or 1). */
2288 tree
2290 tree arg;
2291 {
2298 /* If this is a comparison, we can simply invert it, except for
2299 floating-point non-equality comparisons, in which case we just
2300 enclose a TRUTH_NOT_EXPR around what we have. */
2303 {
2305 && !flag_unsafe_math_optimizations
2309 else
2312 }
2315 {
2330 /* Here we can invert either operand. We invert the first operand
2331 unless the second operand is a TRUTH_NOT_EXPR in which case our
2332 result is the XOR of the first operand with the inside of the
2333 negation of the second operand. */
2338 else
2393 }
2397 }
2399 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2400 operands are another bit-wise operation with a common input. If so,
2401 distribute the bit operations to save an operation and possibly two if
2402 constants are involved. For example, convert
2403 (A | B) & (A | C) into A | (B & C)
2404 Further simplification will occur if B and C are constants.
2406 If this optimization cannot be done, 0 will be returned. */
2408 static tree
2411 tree type;
2413 {
2414 tree common;
2424 {
2428 }
2430 {
2434 }
2436 {
2440 }
2442 {
2446 }
2447 else
2452 }
2453 \f
2454 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2455 starting at BITPOS. The field is unsigned if UNSIGNEDP is nonzero. */
2457 static tree
2459 tree inner;
2460 tree type;
2463 {
2470 }
2472 /* Optimize a bit-field compare.
2474 There are two cases: First is a compare against a constant and the
2475 second is a comparison of two items where the fields are at the same
2476 bit position relative to the start of a chunk (byte, halfword, word)
2477 large enough to contain it. In these cases we can avoid the shift
2478 implicit in bitfield extractions.
2480 For constants, we emit a compare of the shifted constant with the
2481 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2482 compared. For two fields at the same position, we do the ANDs with the
2483 similar mask and compare the result of the ANDs.
2485 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2486 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2487 are the left and right operands of the comparison, respectively.
2489 If the optimization described above can be done, we return the resulting
2490 tree. Otherwise we return zero. */
2492 static tree
2495 tree compare_type;
2497 {
2506 tree mask;
2507 tree offset;
2509 /* Get all the information about the extractions being done. If the bit size
2510 if the same as the size of the underlying object, we aren't doing an
2511 extraction at all and so can do nothing. We also don't want to
2512 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2513 then will no longer be able to replace it. */
2521 {
2522 /* If this is not a constant, we can only do something if bit positions,
2523 sizes, and signedness are the same. */
2531 }
2533 /* See if we can find a mode to refer to this field. We should be able to,
2534 but fail if we can't. */
2543 /* Set signed and unsigned types of the precision of this mode for the
2544 shifts below. */
2548 /* Compute the bit position and size for the new reference and our offset
2549 within it. If the new reference is the same size as the original, we
2550 won't optimize anything, so return zero. */
2560 /* Make the mask to be used against the extracted field. */
2570 /* If not comparing with constant, just rework the comparison
2571 and return. */
2576 mask),
2580 mask));
2582 /* Otherwise, we are handling the constant case. See if the constant is too
2583 big for the field. Warn and return a tree of for 0 (false) if so. We do
2584 this not only for its own sake, but to avoid having to test for this
2585 error case below. If we didn't, we might generate wrong code.
2587 For unsigned fields, the constant shifted right by the field length should
2588 be all zero. For signed fields, the high-order bits should agree with
2589 the sign bit. */
2592 {
2596 {
2602 }
2603 }
2604 else
2605 {
2609 {
2615 }
2616 }
2618 /* Single-bit compares should always be against zero. */
2620 {
2623 }
2625 /* Make a new bitfield reference, shift the constant over the
2626 appropriate number of bits and mask it with the computed mask
2627 (in case this was a signed field). If we changed it, make a new one. */
2630 {
2633 }
2643 rhs);
2644 }
2645 \f
2646 /* Subroutine for fold_truthop: decode a field reference.
2648 If EXP is a comparison reference, we return the innermost reference.
2650 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2651 set to the starting bit number.
2653 If the innermost field can be completely contained in a mode-sized
2654 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2656 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2657 otherwise it is not changed.
2659 *PUNSIGNEDP is set to the signedness of the field.
2661 *PMASK is set to the mask used. This is either contained in a
2662 BIT_AND_EXPR or derived from the width of the field.
2664 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2666 Return 0 if this is not a component reference or is one that we can't
2667 do anything with. */
2669 static tree
2672 tree exp;
2678 {
2681 tree unsigned_type;
2684 /* All the optimizations using this function assume integer fields.
2685 There are problems with FP fields since the type_for_size call
2686 below can fail for, e.g., XFmode. */
2693 {
2699 }
2708 /* Compute the mask to access the bitfield. */
2718 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2726 }
2728 /* Return nonzero if MASK represents a mask of SIZE ones in the low-order
2729 bit positions. */
2731 static int
2733 tree mask;
2735 {
2738 tree tmask;
2743 return
2750 }
2752 /* Subroutine for fold: determine if VAL is the INTEGER_CONST that
2753 represents the sign bit of EXP's type. If EXP represents a sign
2754 or zero extension, also test VAL against the unextended type.
2755 The return value is the (sub)expression whose sign bit is VAL,
2756 or NULL_TREE otherwise. */
2758 static tree
2760 tree exp;
2761 tree val;
2762 {
2764 HOST_WIDE_INT hi;
2766 tree t;
2768 /* Tree EXP must have an integral type. */
2773 /* Tree VAL must be an integer constant. */
2780 {
2783 }
2784 else
2785 {
2788 }
2793 /* Handle extension from a narrower type. */
2799 }
2801 /* Subroutine for fold_truthop: determine if an operand is simple enough
2802 to be evaluated unconditionally. */
2804 static int
2806 tree exp;
2807 {
2808 /* Strip any conversions that don't change the machine mode. */
2820 /* Don't regard global variables as simple. They may be
2821 allocated in ways unknown to the compiler (shared memory,
2822 #pragma weak, etc). */
2825 /* Loading a static variable is unduly expensive, but global
2826 registers aren't expensive. */
2828 }
2829 \f
2830 /* The following functions are subroutines to fold_range_test and allow it to
2831 try to change a logical combination of comparisons into a range test.
2833 For example, both
2834 X == 2 || X == 3 || X == 4 || X == 5
2835 and
2836 X >= 2 && X <= 5
2837 are converted to
2838 (unsigned) (X - 2) <= 3
2840 We describe each set of comparisons as being either inside or outside
2841 a range, using a variable named like IN_P, and then describe the
2842 range with a lower and upper bound. If one of the bounds is omitted,
2843 it represents either the highest or lowest value of the type.
2845 In the comments below, we represent a range by two numbers in brackets
2846 preceded by a "+" to designate being inside that range, or a "-" to
2847 designate being outside that range, so the condition can be inverted by
2848 flipping the prefix. An omitted bound is represented by a "-". For
2849 example, "- [-, 10]" means being outside the range starting at the lowest
2850 possible value and ending at 10, in other words, being greater than 10.
2851 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2852 always false.
2854 We set up things so that the missing bounds are handled in a consistent
2855 manner so neither a missing bound nor "true" and "false" need to be
2856 handled using a special case. */
2858 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2859 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2860 and UPPER1_P are nonzero if the respective argument is an upper bound
2861 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2862 must be specified for a comparison. ARG1 will be converted to ARG0's
2863 type if both are specified. */
2865 static tree
2868 tree type;
2871 {
2872 tree tem;
2876 /* If neither arg represents infinity, do the normal operation.
2877 Else, if not a comparison, return infinity. Else handle the special
2878 comparison rules. Note that most of the cases below won't occur, but
2879 are handled for consistency. */
2882 {
2887 }
2892 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2893 for neither. In real maths, we cannot assume open ended ranges are
2894 the same. But, this is computer arithmetic, where numbers are finite.
2895 We can therefore make the transformation of any unbounded range with
2896 the value Z, Z being greater than any representable number. This permits
2897 us to treat unbounded ranges as equal. */
2901 {
2922 }
2925 }
2926 \f
2927 /* Given EXP, a logical expression, set the range it is testing into
2928 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2929 actually being tested. *PLOW and *PHIGH will be made of the same type
2930 as the returned expression. If EXP is not a comparison, we will most
2931 likely not be returning a useful value and range. */
2933 static tree
2935 tree exp;
2938 {
2945 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2946 and see if we can refine the range. Some of the cases below may not
2947 happen, but it doesn't seem worth worrying about this. We "continue"
2948 the outer loop when we've changed something; otherwise we "break"
2949 the switch, which will "break" the while. */
2954 {
2958 {
2969 }
2971 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2972 lose a cast by accident. */
2977 {
2984 /* We can only do something if the range is testing for zero
2985 and if the second operand is an integer constant. Note that
2986 saying something is "in" the range we make is done by
2987 complementing IN_P since it will set in the initial case of
2988 being not equal to zero; "out" is leaving it alone. */
2995 {
3016 }
3020 /* If this is an unsigned comparison, we also know that EXP is
3021 greater than or equal to zero. We base the range tests we make
3022 on that fact, so we record it here so we can parse existing
3023 range tests. */
3025 {
3028 NULL_TREE))
3033 /* If the high bound is missing, but we
3034 have a low bound, reverse the range so
3035 it goes from zero to the low bound minus 1. */
3037 {
3042 }
3043 }
3047 /* (-x) IN [a,b] -> x in [-b, -a] */
3057 /* ~ X -> -X - 1 */
3066 /* If EXP is signed, any overflow in the computation is undefined,
3067 so we don't worry about it so long as our computations on
3068 the bounds don't overflow. For unsigned, overflow is defined
3069 and this is exactly the right thing. */
3078 /* Check for an unsigned range which has wrapped around the maximum
3079 value thus making n_high < n_low, and normalize it. */
3081 {
3087 /* If the range is of the form +/- [ x+1, x ], we won't
3088 be able to normalize it. But then, it represents the
3089 whole range or the empty set, so make it
3090 +/- [ -, - ]. */
3094 else
3096 }
3097 else
3120 /* If we're converting from an unsigned to a signed type,
3121 we will be doing the comparison as unsigned. The tests above
3122 have already verified that LOW and HIGH are both positive.
3124 So we have to make sure that the original unsigned value will
3125 be interpreted as positive. */
3127 {
3130 tree high_positive;
3132 /* A range without an upper bound is, naturally, unbounded.
3133 Since convert would have cropped a very large value, use
3134 the max value for the destination type. */
3135 high_positive
3144 /* If the low bound is specified, "and" the range with the
3145 range for which the original unsigned value will be
3146 positive. */
3148 {
3152 high_positive))
3156 }
3157 else
3158 {
3159 /* Otherwise, "or" the range with the range of the input
3160 that will be interpreted as negative. */
3164 high_positive))
3168 }
3169 }
3177 }
3180 }
3182 /* If EXP is a constant, we can evaluate whether this is true or false. */
3184 {
3191 }
3195 }
3196 \f
3197 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3198 type, TYPE, return an expression to test if EXP is in (or out of, depending
3199 on IN_P) the range. */
3201 static tree
3203 tree type;
3204 tree exp;
3207 {
3209 tree value;
3228 {
3230 {
3234 }
3236 }
3238 /* Optimize (c>=1) && (c<=127) into (signed char)c > 0. */
3240 {
3242 HOST_WIDE_INT hi;
3247 {
3250 }
3251 else
3252 {
3255 }
3258 {
3260 {
3263 }
3266 }
3267 }
3276 }
3277 \f
3278 /* Given two ranges, see if we can merge them into one. Return 1 if we
3279 can, 0 if we can't. Set the output range into the specified parameters. */
3281 static int
3287 {
3291 tree tem;
3301 /* Make range 0 be the range that starts first, or ends last if they
3302 start at the same value. Swap them if it isn't. */
3305 || (lowequal
3308 {
3312 }
3314 /* Now flag two cases, whether the ranges are disjoint or whether the
3315 second range is totally subsumed in the first. Note that the tests
3316 below are simplified by the ones above. */
3322 /* We now have four cases, depending on whether we are including or
3323 excluding the two ranges. */
3325 {
3326 /* If they don't overlap, the result is false. If the second range
3327 is a subset it is the result. Otherwise, the range is from the start
3328 of the second to the end of the first. */
3333 else
3335 }
3338 {
3339 /* If they don't overlap, the result is the first range. If they are
3340 equal, the result is false. If the second range is a subset of the
3341 first, and the ranges begin at the same place, we go from just after
3342 the end of the first range to the end of the second. If the second
3343 range is not a subset of the first, or if it is a subset and both
3344 ranges end at the same place, the range starts at the start of the
3345 first range and ends just before the second range.
3346 Otherwise, we can't describe this as a single range. */
3352 {
3356 }
3358 {
3362 }
3363 else
3365 }
3368 {
3369 /* If they don't overlap, the result is the second range. If the second
3370 is a subset of the first, the result is false. Otherwise,
3371 the range starts just after the first range and ends at the
3372 end of the second. */
3377 else
3378 {
3382 }
3383 }
3385 else
3386 {
3387 /* The case where we are excluding both ranges. Here the complex case
3388 is if they don't overlap. In that case, the only time we have a
3389 range is if they are adjacent. If the second is a subset of the
3390 first, the result is the first. Otherwise, the range to exclude
3391 starts at the beginning of the first range and ends at the end of the
3392 second. */
3394 {
3401 else
3403 }
3406 else
3408 }
3412 }
3413 \f
3414 /* EXP is some logical combination of boolean tests. See if we can
3415 merge it into some range test. Return the new tree if so. */
3417 static tree
3419 tree exp;
3420 {
3427 tree tem;
3429 /* If this is an OR operation, invert both sides; we will invert
3430 again at the end. */
3434 /* If both expressions are the same, if we can merge the ranges, and we
3435 can build the range test, return it or it inverted. If one of the
3436 ranges is always true or always false, consider it to be the same
3437 expression as the other. */
3447 /* On machines where the branch cost is expensive, if this is a
3448 short-circuited branch and the underlying object on both sides
3449 is the same, make a non-short-circuit operation. */
3455 {
3456 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3457 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3458 which cases we can't do this. */
3467 {
3479 }
3480 }
3483 }
3484 \f
3485 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3486 bit value. Arrange things so the extra bits will be set to zero if and
3487 only if C is signed-extended to its full width. If MASK is nonzero,
3488 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3490 static tree
3492 tree c;
3495 tree mask;
3496 {
3499 tree temp;
3504 /* We work by getting just the sign bit into the low-order bit, then
3505 into the high-order bit, then sign-extend. We then XOR that value
3506 with C. */
3510 /* We must use a signed type in order to get an arithmetic right shift.
3511 However, we must also avoid introducing accidental overflows, so that
3512 a subsequent call to integer_zerop will work. Hence we must
3513 do the type conversion here. At this point, the constant is either
3514 zero or one, and the conversion to a signed type can never overflow.
3515 We could get an overflow if this conversion is done anywhere else. */
3523 /* If necessary, convert the type back to match the type of C. */
3528 }
3529 \f
3530 /* Find ways of folding logical expressions of LHS and RHS:
3531 Try to merge two comparisons to the same innermost item.
3532 Look for range tests like "ch >= '0' && ch <= '9'".
3533 Look for combinations of simple terms on machines with expensive branches
3534 and evaluate the RHS unconditionally.
3536 For example, if we have p->a == 2 && p->b == 4 and we can make an
3537 object large enough to span both A and B, we can do this with a comparison
3538 against the object ANDed with the a mask.
3540 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3541 operations to do this with one comparison.
3543 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3544 function and the one above.
3546 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3547 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3549 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3550 two operands.
3552 We return the simplified tree or 0 if no optimization is possible. */
3554 static tree
3558 {
3559 /* If this is the "or" of two comparisons, we can do something if
3560 the comparisons are NE_EXPR. If this is the "and", we can do something
3561 if the comparisons are EQ_EXPR. I.e.,
3562 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3564 WANTED_CODE is this operation code. For single bit fields, we can
3565 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3566 comparison for one-bit fields. */
3586 /* Start by getting the comparison codes. Fail if anything is volatile.
3587 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3588 it were surrounded with a NE_EXPR. */
3613 /* Simplify (x<y) && (x==y) into (x<=y) and related optimizations. */
3617 {
3622 {
3630 }
3633 {
3642 }
3643 else
3653 }
3655 /* If the RHS can be evaluated unconditionally and its operands are
3656 simple, it wins to evaluate the RHS unconditionally on machines
3657 with expensive branches. In this case, this isn't a comparison
3658 that can be merged. Avoid doing this if the RHS is a floating-point
3659 comparison since those can trap. */
3665 {
3666 /* Convert (a != 0) || (b != 0) into (a | b) != 0. */
3674 integer_zero_node);
3676 /* Convert (a == 0) && (b == 0) into (a | b) == 0. */
3684 integer_zero_node);
3687 }
3689 /* See if the comparisons can be merged. Then get all the parameters for
3690 each side. */
3714 /* It must be true that the inner operation on the lhs of each
3715 comparison must be the same if we are to be able to do anything.
3716 Then see if we have constants. If not, the same must be true for
3717 the rhs's. */
3728 else
3731 /* If either comparison code is not correct for our logical operation,
3732 fail. However, we can convert a one-bit comparison against zero into
3733 the opposite comparison against that bit being set in the field. */
3737 {
3739 {
3740 /* Make the left operand unsigned, since we are only interested
3741 in the value of one bit. Otherwise we are doing the wrong
3742 thing below. */
3745 }
3746 else
3748 }
3750 /* This is analogous to the code for l_const above. */
3752 {
3754 {
3757 }
3758 else
3760 }
3762 /* After this point all optimizations will generate bit-field
3763 references, which we might not want. */
3767 /* See if we can find a mode that contains both fields being compared on
3768 the left. If we can't, fail. Otherwise, update all constants and masks
3769 to be relative to a field of that size. */
3774 volatilep);
3784 {
3787 }
3795 {
3803 {
3807 wanted_code == NE_EXPR
3809 }
3810 }
3812 {
3820 {
3824 wanted_code == NE_EXPR
3826 }
3827 }
3829 /* If the right sides are not constant, do the same for it. Also,
3830 disallow this optimization if a size or signedness mismatch occurs
3831 between the left and right sides. */
3833 {
3836 /* Make sure the two fields on the right
3837 correspond to the left without being swapped. */
3845 volatilep);
3855 {
3858 }
3865 /* Make a mask that corresponds to both fields being compared.
3866 Do this for both items being compared. If the operands are the
3867 same size and the bits being compared are in the same position
3868 then we can do this by masking both and comparing the masked
3869 results. */
3873 {
3885 }
3887 /* There is still another way we can do something: If both pairs of
3888 fields being compared are adjacent, we may be able to make a wider
3889 field containing them both.
3891 Note that we still must mask the lhs/rhs expressions. Furthermore,
3892 the mask must be shifted to account for the shift done by
3893 make_bit_field_ref. */
3898 {
3899 tree type;
3911 /* Convert to the smaller type before masking out unwanted bits. */
3914 {
3916 {
3920 }
3922 {
3926 }
3927 }
3936 }
3939 }
3941 /* Handle the case of comparisons with constants. If there is something in
3942 common between the masks, those bits of the constants must be the same.
3943 If not, the condition is always false. Test for this to avoid generating
3944 incorrect code below. */
3949 {
3951 {
3954 }
3955 else
3956 {
3959 }
3960 }
3962 /* Construct the expression we will return. First get the component
3963 reference we will make. Unless the mask is all ones the width of
3964 that field, perform the mask operation. Then compare with the
3965 merged constant. */
3975 }
3976 \f
3977 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3978 constant. */
3980 static tree
3982 tree t;
3983 {
3988 tree minmax_const;
3990 tree inner;
4000 /* If something does not permit us to optimize, return the original tree. */
4008 /* Now handle all the various comparison codes. We only handle EQ_EXPR
4009 and GT_EXPR, doing the rest with recursive calls using logical
4010 simplifications. */
4012 {
4014 return
4018 return
4020 optimize_minmax_comparison
4022 optimize_minmax_comparison
4027 /* MAX (X, 0) == 0 -> X <= 0 */
4031 /* MAX (X, 0) == 5 -> X == 5 */
4035 /* MAX (X, 0) == -1 -> false */
4039 /* MIN (X, 0) == 0 -> X >= 0 */
4043 /* MIN (X, 0) == 5 -> false */
4046 else
4047 /* MIN (X, 0) == -1 -> X == -1 */
4052 /* MAX (X, 0) > 0 -> X > 0
4053 MAX (X, 0) > 5 -> X > 5 */
4057 /* MAX (X, 0) > -1 -> true */
4061 /* MIN (X, 0) > 0 -> false
4062 MIN (X, 0) > 5 -> false */
4065 else
4066 /* MIN (X, 0) > -1 -> X > -1 */
4071 }
4072 }
4073 \f
4074 /* T is an integer expression that is being multiplied, divided, or taken a
4075 modulus (CODE says which and what kind of divide or modulus) by a
4076 constant C. See if we can eliminate that operation by folding it with
4077 other operations already in T. WIDE_TYPE, if non-null, is a type that
4078 should be used for the computation if wider than our type.
4080 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
4081 (X * 2) + (Y * 4). We must, however, be assured that either the original
4082 expression would not overflow or that overflow is undefined for the type
4083 in the language in question.
4085 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
4086 the machine has a multiply-accumulate insn or that this is part of an
4087 addressing calculation.
4089 If we return a non-null expression, it is an equivalent form of the
4090 original computation, but need not be in the original type. */
4092 static tree
4094 tree t;
4095 tree c;
4097 tree wide_type;
4098 {
4099 /* To avoid exponential search depth, refuse to allow recursion past
4100 three levels. Beyond that (1) it's highly unlikely that we'll find
4101 something interesting and (2) we've probably processed it before
4102 when we built the inner expression. */
4105 tree ret;
4110 depth++;
4112 depth--;
4115 }
4117 static tree
4119 tree t;
4120 tree c;
4122 tree wide_type;
4123 {
4133 /* Don't deal with constants of zero here; they confuse the code below. */
4143 /* Note that we need not handle conditional operations here since fold
4144 already handles those cases. So just do arithmetic here. */
4146 {
4148 /* For a constant, we can always simplify if we are a multiply
4149 or (for divide and modulus) if it is a multiple of our constant. */
4156 /* If op0 is an expression ... */
4161 /* ... and is unsigned, and its type is smaller than ctype,
4162 then we cannot pass through as widening. */
4168 /* ... or its type is larger than ctype,
4169 then we cannot pass through this truncation. */
4174 /* Pass the constant down and see if we can make a simplification. If
4175 we can, replace this expression with the inner simplification for
4176 possible later conversion to our or some other type. */
4188 /* If widening the type changes the signedness, then we can't perform
4189 this optimization as that changes the result. */
4193 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
4196 {
4202 }
4212 /* If this has not been evaluated and the operand has no side effects,
4213 we can see if we can do something inside it and make a new one.
4214 Note that this test is overly conservative since we can do this
4215 if the only reason it had side effects is that it was another
4216 similar SAVE_EXPR, but that isn't worth bothering with. */
4219 wide_type)))
4220 {
4227 }
4231 /* If the second operand is constant, this is a multiplication
4232 or floor division, by a power of two, so we can treat it that
4233 way unless the multiplier or divisor overflows. */
4235 /* const_binop may not detect overflow correctly,
4236 so check for it explicitly here. */
4250 /* See if we can eliminate the operation on both sides. If we can, we
4251 can return a new PLUS or MINUS. If we can't, the only remaining
4252 cases where we can do anything are if the second operand is a
4253 constant. */
4258 /* If not multiplication, we can only do this if both operands
4259 are divisible by c. */
4265 /* If this was a subtraction, negate OP1 and set it to be an addition.
4266 This simplifies the logic below. */
4273 /* If either OP1 or C are negative, this optimization is not safe for
4274 some of the division and remainder types while for others we need
4275 to change the code. */
4277 {
4285 }
4287 /* If it's a multiply or a division/modulus operation of a multiple
4288 of our constant, do the operation and verify it doesn't overflow. */
4291 {
4295 }
4296 else
4299 /* If we have an unsigned type is not a sizetype, we cannot widen
4300 the operation since it will change the result if the original
4301 computation overflowed. */
4307 /* If we were able to eliminate our operation from the first side,
4308 apply our operation to the second side and reform the PLUS. */
4312 /* The last case is if we are a multiply. In that case, we can
4313 apply the distributive law to commute the multiply and addition
4314 if the multiplication of the constants doesn't overflow. */
4319 op1));
4324 /* We have a special case here if we are doing something like
4325 (C * 8) % 4 since we know that's zero. */
4332 /* Arrange for the code below to simplify two constants first. */
4334 {
4338 }
4340 /* ... fall through ... */
4344 /* If we can extract our operation from the LHS, do so and return a
4345 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4346 do something only if the second operand is a constant. */
4358 /* If these are the same operation types, we can associate them
4359 assuming no overflow. */
4366 /* If these operations "cancel" each other, we have the main
4367 optimizations of this pass, which occur when either constant is a
4368 multiple of the other, in which case we replace this with either an
4369 operation or CODE or TCODE.
4371 If we have an unsigned type that is not a sizetype, we cannot do
4372 this since it will change the result if the original computation
4373 overflowed. */
4380 {
4391 }
4396 }
4399 }
4400 \f
4401 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4402 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4403 that we may sometimes modify the tree. */
4405 static tree
4407 tree t;
4408 tree s;
4409 {
4412 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4417 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4418 don't bother handling any other types. */
4420 {
4424 }
4429 {
4432 }
4435 }
4436 \f
4437 /* Return a node which has the indicated constant VALUE (either 0 or
4438 1), and is of the indicated TYPE. */
4440 static tree
4443 tree type;
4444 {
4449 integer_zero_node);
4450 else
4451 {
4456 }
4457 }
4459 /* Utility function for the following routine, to see how complex a nesting of
4460 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4461 we don't care (to avoid spending too much time on complex expressions.). */
4463 static int
4465 tree expr;
4467 {
4478 }
4480 /* Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
4481 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4482 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4483 expression, and ARG to `a'. If COND_FIRST_P is nonzero, then the
4484 COND is the first argument to CODE; otherwise (as in the example
4485 given here), it is the second argument. TYPE is the type of the
4486 original expression. */
4488 static tree
4491 tree type;
4492 tree cond;
4493 tree arg;
4495 {
4499 /* In the end, we'll produce a COND_EXPR. Both arms of the
4500 conditional expression will be binary operations. The left-hand
4501 side of the expression to be executed if the condition is true
4502 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4503 of the expression to be executed if the condition is true will be
4504 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4505 but apply to the expression to be executed if the conditional is
4506 false. */
4511 /* These are the codes to use for the left-hand side and right-hand
4512 side of the COND_EXPR. Normally, they are the same as CODE. */
4515 /* And these are the types of the expressions. */
4521 {
4525 }
4526 else
4527 {
4531 }
4534 {
4538 /* If this operand throws an expression, then it does not make
4539 sense to try to perform a logical or arithmetic operation
4540 involving it. Instead of building `a + throw 3' for example,
4541 we simply build `a, throw 3'. */
4543 {
4545 {
4548 }
4549 else
4551 }
4553 {
4555 {
4558 }
4559 else
4561 }
4562 }
4563 else
4564 {
4569 }
4571 /* If ARG is complex we want to make sure we only evaluate
4572 it once. Though this is only required if it is volatile, it
4573 might be more efficient even if it is not. However, if we
4574 succeed in folding one part to a constant, we do not need
4575 to make this SAVE_EXPR. Since we do this optimization
4576 primarily to see if we do end up with constant and this
4577 SAVE_EXPR interferes with later optimizations, suppressing
4578 it when we can is important.
4580 If we are not in a function, we can't make a SAVE_EXPR, so don't
4581 try to do so. Don't try to see if the result is a constant
4582 if an arm is a COND_EXPR since we get exponential behavior
4583 in that case. */
4592 {
4601 {
4605 }
4606 }
4619 else
4621 }
4623 \f
4624 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4626 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4627 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4628 ADDEND is the same as X.
4630 X + 0 and X - 0 both give X when X is NaN, infinite, or nonzero
4631 and finite. The problematic cases are when X is zero, and its mode
4632 has signed zeros. In the case of rounding towards -infinity,
4633 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4634 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4636 static bool
4640 {
4644 /* Don't allow the fold with -fsignaling-nans. */
4648 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4652 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4657 /* The mode has signed zeros, and we have to honor their sign.
4658 In this situation, there is only one case we can return true for.
4659 X - 0 is the same as X unless rounding towards -infinity is
4660 supported. */
4662 }
4665 /* Perform constant folding and related simplification of EXPR.
4666 The related simplifications include x*1 => x, x*0 => 0, etc.,
4667 and application of the associative law.
4668 NOP_EXPR conversions may be removed freely (as long as we
4669 are careful not to change the C type of the overall expression)
4670 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4671 but we can constant-fold them if they have constant operands. */
4673 tree
4675 tree expr;
4676 {
4679 tree tem;
4685 /* WINS will be nonzero when the switch is done
4686 if all operands are constant. */
4689 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4690 Likewise for a SAVE_EXPR that's already been evaluated. */
4694 /* Return right away if a constant. */
4698 #ifdef MAX_INTEGER_COMPUTATION_MODE
4700 #endif
4703 {
4704 tree subop;
4706 /* Special case for conversion ops that can have fixed point args. */
4709 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4715 else
4720 )
4721 /* Note that TREE_CONSTANT isn't enough:
4722 static var addresses are constant but we can't
4723 do arithmetic on them. */
4725 }
4727 {
4731 {
4733 tree subop;
4739 {
4740 /* Signedness matters here. Perhaps we can refine this
4741 later. */
4743 }
4744 else
4745 /* Strip any conversions that don't change the mode. */
4750 else
4755 /* Note that TREE_CONSTANT isn't enough:
4756 static var addresses are constant but we can't
4757 do arithmetic on them. */
4764 }
4765 }
4767 /* If this is a commutative operation, and ARG0 is a constant, move it
4768 to ARG1 to reduce the number of tests below. */
4773 {
4778 }
4780 /* Now WINS is set as described above,
4781 ARG0 is the first operand of EXPR,
4782 and ARG1 is the second operand (if it has more than one operand).
4784 First check for cases where an arithmetic operation is applied to a
4785 compound, conditional, or comparison operation. Push the arithmetic
4786 operation inside the compound or conditional to see if any folding
4787 can then be done. Convert comparison to conditional for this purpose.
4788 The also optimizes non-constant cases that used to be done in
4789 expand_expr.
4791 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4792 one of the operands is a comparison and the other is a comparison, a
4793 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4794 code below would make the expression more complex. Change it to a
4795 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4796 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4808 {
4818 }
4821 {
4826 {
4836 /* If this was a conversion, and all we did was to move into
4837 inside the COND_EXPR, bring it back out. But leave it if
4838 it is a conversion from integer to integer and the
4839 result precision is no wider than a word since such a
4840 conversion is cheap and may be optimized away by combine,
4841 while it couldn't if it were outside the COND_EXPR. Then return
4842 so we don't get into an infinite recursion loop taking the
4843 conversion out and then back in. */
4855 && (INTEGRAL_TYPE_P
4866 }
4871 }
4874 {
4889 return
4903 return
4906 }
4917 {
4933 /* Other kinds of FIX are not handled properly by fold_convert. */
4938 /* Handle cases of two conversions in a row. */
4941 {
4961 /* In addition to the cases of two conversions in a row
4962 handled below, if we are converting something to its own
4963 type via an object of identical or wider precision, neither
4964 conversion is needed. */
4970 /* Likewise, if the intermediate and final types are either both
4971 float or both integer, we don't need the middle conversion if
4972 it is wider than the final type and doesn't change the signedness
4973 (for integers). Avoid this if the final type is a pointer
4974 since then we sometimes need the inner conversion. Likewise if
4975 the outer has a precision not equal to the size of its mode. */
4985 /* If we have a sign-extension of a zero-extended value, we can
4986 replace that by a single zero-extension. */
4992 /* Two conversions in a row are not needed unless:
4993 - some conversion is floating-point (overstrict for now), or
4994 - the intermediate type is narrower than both initial and
4995 final, or
4996 - the intermediate type and innermost type differ in signedness,
4997 and the outermost type is wider than the intermediate, or
4998 - the initial type is a pointer type and the precisions of the
4999 intermediate and final types differ, or
5000 - the final type is a pointer type and the precisions of the
5001 initial and intermediate types differ. */
5015 }
5019 /* Detect assigning a bitfield. */
5022 {
5023 /* Don't leave an assignment inside a conversion
5024 unless assigning a bitfield. */
5027 /* First do the assignment, then return converted constant. */
5031 }
5033 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
5034 constants (if x has signed type, the sign bit cannot be set
5035 in c). This folds extension into the BIT_AND_EXPR. */
5040 {
5050 <= HOST_BITS_PER_WIDE_INT
5052 {
5059 #ifdef LOAD_EXTEND_OP
5063 {
5067 }
5068 #endif
5069 }
5074 }
5077 {
5080 }
5091 {
5095 }
5104 {
5106 {
5108 HOST_WIDE_INT high;
5119 }
5122 }
5125 /* Convert -((double)float) into (double)(-float). */
5128 {
5133 }
5135 /* Convert - (a - b) to (b - a) for non-floating-point. */
5145 {
5147 {
5148 /* If the value is unsigned, then the absolute value is
5149 the same as the ordinary value. */
5152 /* Similarly, if the value is non-negative. */
5155 /* If the value is negative, then the absolute value is
5156 its negation. */
5157 else
5158 {
5160 HOST_WIDE_INT high;
5171 }
5172 }
5174 {
5178 }
5179 }
5182 /* Convert fabs((double)float) into (double)fabsf(float). */
5185 {
5190 }
5191 else
5192 {
5193 /* fabs(sqrt(x)) = sqrt(x) and fabs(exp(x)) = exp(x). */
5202 }
5227 {
5234 }
5240 /* A + (-B) -> A - B */
5243 /* (-A) + B -> B - A */
5247 {
5251 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
5252 with a constant, and the two constants have no bits in common,
5253 we should treat this as a BIT_IOR_EXPR since this may produce more
5254 simplifications. */
5262 {
5265 }
5267 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
5268 (plus (plus (mult) (mult)) (foo)) so that we can
5269 take advantage of the factoring cases below. */
5274 {
5279 else
5290 parg1));
5295 parg0));
5296 }
5299 {
5303 /* (A * C) + (B * C) -> (A+B) * C.
5304 We are most concerned about the case where C is a constant,
5305 but other combinations show up during loop reduction. Since
5306 it is not difficult, try all four possibilities. */
5323 /* No identical multiplicands; see if we can find a common
5324 power-of-two factor in non-power-of-two multiplies. This
5325 can help in multi-dimensional array access. */
5330 {
5335 /* Move min of absolute values to int11. */
5338 {
5342 }
5345 {
5350 }
5351 }
5356 same));
5357 }
5358 }
5360 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5364 /* Likewise if the operands are reversed. */
5368 bit_rotate:
5369 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5370 is a rotate of A by C1 bits. */
5371 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5372 is a rotate of A by B bits. */
5373 {
5382 {
5401 {
5409 TYPE_PRECISION
5414 ? LROTATE_EXPR
5417 }
5419 {
5427 TYPE_PRECISION
5432 ? LROTATE_EXPR
5435 }
5436 }
5437 }
5439 associate:
5440 /* In most languages, can't associate operations on floats through
5441 parentheses. Rather than remember where the parentheses were, we
5442 don't associate floats at all. It shouldn't matter much. However,
5443 associating multiplications is only very slightly inaccurate, so do
5444 that if -funsafe-math-optimizations is specified. */
5449 {
5453 /* Split both trees into variables, constants, and literals. Then
5454 associate each group together, the constants with literals,
5455 then the result with variables. This increases the chances of
5456 literals being recombined later and of generating relocatable
5457 expressions for the sum of a constant and literal. */
5462 /* Only do something if we found more than two objects. Otherwise,
5463 nothing has changed and we risk infinite recursion. */
5468 {
5469 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5478 /* Preserve the MINUS_EXPR if the negative part of the literal is
5479 greater than the positive part. Otherwise, the multiplicative
5480 folding code (i.e extract_muldiv) may be fooled in case
5481 unsigned constants are substracted, like in the following
5482 example: ((X*2 + 4) - 8U)/2. */
5484 {
5486 {
5490 }
5491 else
5492 {
5496 }
5497 }
5499 {
5503 else
5504 {
5509 }
5510 }
5514 }
5515 }
5517 binary:
5521 {
5522 /* The return value should always have
5523 the same type as the original expression. */
5528 }
5532 /* A - (-B) -> A + B */
5535 /* (-A) - B -> (-B) - A where B is easily negated and we can swap. */
5545 {
5551 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5552 about the case where C is a constant, just try one of the
5553 four possibilities. */
5563 }
5565 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5569 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5570 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5571 (-ARG1 + ARG0) reduces to -ARG1. */
5575 /* Fold &x - &x. This can happen from &x.foo - &x.
5576 This is unsafe for certain floats even in non-IEEE formats.
5577 In IEEE, it is unsafe because it does wrong for NaNs.
5578 Also note that operand_equal_p is always false if an operand
5579 is volatile. */
5588 /* (-A) * (-B) -> A * B */
5594 {
5600 /* (a * (1 << b)) is (a << b) */
5615 }
5616 else
5617 {
5618 /* Maybe fold x * 0 to 0. The expressions aren't the same
5619 when x is NaN, since x * 0 is also NaN. Nor are they the
5620 same in modes with signed zeros, since multiplying a
5621 negative value by 0 gives -0, not +0. */
5626 /* In IEEE floating point, x*1 is not equivalent to x for snans. */
5631 /* Transform x * -1.0 into -x. */
5636 /* x*2 is x+x */
5640 {
5643 }
5646 {
5650 /* Optimize sqrt(x)*sqrt(y) as sqrt(x*y). */
5654 {
5661 }
5663 /* Optimize exp(x)*exp(y) as exp(x+y). */
5667 {
5674 }
5675 }
5676 }
5680 bit_ior:
5689 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5691 This results in more efficient code for machines without a NAND
5692 instruction. Combine will canonicalize to the first form
5693 which will allow use of NAND instructions provided by the
5694 backend if they exist. */
5697 {
5702 }
5704 /* See if this can be simplified into a rotate first. If that
5705 is unsuccessful continue in the association code. */
5714 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5715 with a constant, and the two constants have no bits in common,
5716 we should treat this as a BIT_IOR_EXPR since this may produce more
5717 simplifications. */
5725 {
5728 }
5730 /* See if this can be simplified into a rotate first. If that
5731 is unsuccessful continue in the association code. */
5735 bit_and:
5743 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5746 {
5747 unsigned int prec
5754 }
5756 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5758 This results in more efficient code for machines without a NOR
5759 instruction. Combine will canonicalize to the first form
5760 which will allow use of NOR instructions provided by the
5761 backend if they exist. */
5764 {
5769 }
5779 {
5783 }
5787 /* Don't touch a floating-point divide by zero unless the mode
5788 of the constant can represent infinity. */
5794 /* (-A) / (-B) -> A / B */
5799 /* In IEEE floating point, x/1 is not equivalent to x for snans. */
5804 /* If ARG1 is a constant, we can convert this to a multiply by the
5805 reciprocal. This does not have the same rounding properties,
5806 so only do this if -funsafe-math-optimizations. We can actually
5807 always safely do it if ARG1 is a power of two, but it's hard to
5808 tell if it is or not in a portable manner. */
5810 {
5815 /* Find the reciprocal if optimizing and the result is exact. */
5817 {
5818 REAL_VALUE_TYPE r;
5821 {
5824 }
5825 }
5826 }
5827 /* Convert A/B/C to A/(B*C). */
5830 {
5833 arg1)));
5834 }
5835 /* Convert A/(B/C) to (A/B)*C. */
5838 {
5843 }
5845 /* Optimize x/exp(y) into x*exp(-y). */
5847 {
5852 {
5859 }
5860 }
5873 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5874 operation, EXACT_DIV_EXPR.
5876 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5877 At one time others generated faster code, it's not clear if they do
5878 after the last round to changes to the DIV code in expmed.c. */
5913 /* Optimize -1 >> x for arithmetic right shifts. */
5916 /* ... fall through ... */
5919 shift:
5925 /* Since negative shift count is not well-defined,
5926 don't try to compute it in the compiler. */
5929 /* Rewrite an LROTATE_EXPR by a constant into an
5930 RROTATE_EXPR by a new constant. */
5932 {
5936 = const_binop
5943 }
5945 /* If we have a rotate of a bit operation with the rotate count and
5946 the second operand of the bit operation both constant,
5947 permute the two operations. */
5960 /* Two consecutive rotates adding up to the width of the mode can
5961 be ignored. */
5992 /* Note that the operand of this must be an int
5993 and its values must be 0 or 1.
5994 ("true" is a fixed value perhaps depending on the language,
5995 but we don't handle values other than 1 correctly yet.) */
5997 /* Avoid infinite recursion. */
6003 /* Note that the operands of this must be ints
6004 and their values must be 0 or 1.
6005 ("true" is a fixed value perhaps depending on the language.) */
6006 /* If first arg is constant zero, return it. */
6010 /* If either arg is constant true, drop it. */
6014 /* Preserve sequence points. */
6017 /* If second arg is constant zero, result is zero, but first arg
6018 must be evaluated. */
6021 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
6022 case will be handled here. */
6026 truth_andor:
6027 /* We only do these simplifications if we are optimizing. */
6031 /* Check for things like (A || B) && (A || C). We can convert this
6032 to A || (B && C). Note that either operator can be any of the four
6033 truth and/or operations and the transformation will still be
6034 valid. Also note that we only care about order for the
6035 ANDIF and ORIF operators. If B contains side effects, this
6036 might change the truth-value of A. */
6043 {
6063 /* This case if tricky because we must either have commutative
6064 operators or else A10 must not have side-effects. */
6070 a01));
6071 }
6073 /* See if we can build a range comparison. */
6077 /* Check for the possibility of merging component references. If our
6078 lhs is another similar operation, try to merge its rhs with our
6079 rhs. Then try to merge our lhs and rhs. */
6091 /* Note that the operands of this must be ints
6092 and their values must be 0 or true.
6093 ("true" is a fixed value perhaps depending on the language.) */
6094 /* If first arg is constant true, return it. */
6098 /* If either arg is constant zero, drop it. */
6102 /* Preserve sequence points. */
6105 /* If second arg is constant true, result is true, but we must
6106 evaluate first arg. */
6109 /* Likewise for first arg, but note this only occurs here for
6110 TRUTH_OR_EXPR. */
6116 /* If either arg is constant zero, drop it. */
6121 /* If either arg is constant true, this is a logical inversion. */
6134 /* If one arg is a real or integer constant, put it last. */
6139 {
6146 }
6149 {
6157 /* Fold (double)float1 CMP (double)float2 into float1 CMP float2. */
6162 /* (-a) CMP (-b) -> b CMP a */
6167 /* (-a) CMP CST -> a swap(CMP) (-CST) */
6169 return
6175 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
6176 /* a CMP (-0) -> a CMP 0 */
6182 /* If this is a comparison of a real constant with a PLUS_EXPR
6183 or a MINUS_EXPR of a real constant, we can convert it into a
6184 comparison with a revised real constant as long as no overflow
6185 occurs when unsafe_math_optimizations are enabled. */
6196 }
6198 /* Convert foo++ == CONST into ++foo == CONST + INCR.
6199 First, see if one arg is constant; find the constant arg
6200 and the other one. */
6201 {
6211 {
6212 /* This optimization is invalid for ordered comparisons
6213 if CONST+INCR overflows or if foo+incr might overflow.
6214 This optimization is invalid for floating point due to rounding.
6215 For pointer types we assume overflow doesn't happen. */
6219 {
6220 tree newconst
6224 /* Do not overwrite the current varop to be a preincrement,
6225 create a new node so that we won't confuse our caller who
6226 might create trees and throw them away, reusing the
6227 arguments that they passed to build. This shows up in
6228 the THEN or ELSE parts of ?: being postincrements. */
6233 /* If VAROP is a reference to a bitfield, we must mask
6234 the constant by the width of the field. */
6238 {
6239 int size
6241 (TREE_OPERAND
6245 tree folded_compare;
6247 /* First check whether the comparison would come out
6248 always the same. If we don't do that we would
6249 change the meaning with the masking. */
6253 else
6256 constop));
6271 mask)));
6272 }
6278 }
6279 }
6281 {
6285 {
6286 tree newconst
6290 /* Do not overwrite the current varop to be a predecrement,
6291 create a new node so that we won't confuse our caller who
6292 might create trees and throw them away, reusing the
6293 arguments that they passed to build. This shows up in
6294 the THEN or ELSE parts of ?: being postdecrements. */
6302 {
6303 int size
6305 (TREE_OPERAND
6309 tree folded_compare;
6314 else
6317 constop));
6332 mask)));
6333 }
6339 }
6340 }
6341 }
6343 /* Change X >= C to X > (C - 1) and X < C to X <= (C - 1) if C > 0.
6344 This transformation affects the cases which are handled in later
6345 optimizations involving comparisons with non-negative constants. */
6349 {
6351 {
6366 }
6367 }
6369 /* Comparisons with the highest or lowest possible integer of
6370 the specified size will have known values. */
6371 {
6379 {
6386 {
6389 }
6390 else
6391 {
6394 }
6399 {
6403 arg0);
6411 arg0);
6417 /* The GE_EXPR and LT_EXPR cases above are not normally
6418 reached because of previous transformations. */
6422 }
6426 {
6439 }
6443 {
6447 arg0);
6456 arg0);
6464 }
6468 {
6481 }
6486 /* signed_type does not work on pointer types. */
6488 {
6489 /* The following case also applies to X < signed_max+1
6490 and X >= signed_max+1 because previous transformations. */
6492 {
6496 return fold
6500 }
6501 }
6502 }
6503 }
6505 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
6506 a MINUS_EXPR of a constant, we can convert it into a comparison with
6507 a revised constant as long as no overflow occurs. */
6519 /* Similarly for a NEGATE_EXPR. */
6528 /* If we have X - Y == 0, we can convert that to X == Y and similarly
6529 for !=. Don't do this for ordered comparisons due to overflow. */
6535 /* If we are widening one operand of an integer comparison,
6536 see if the other operand is similarly being widened. Perhaps we
6537 can do the comparison in the narrower type. */
6547 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
6548 constant, we can simplify it. */
6555 /* If we are comparing an ABS_EXPR with a constant, we can
6556 convert all the cases into explicit comparisons, but they may
6557 well not be faster than doing the ABS and one comparison.
6558 But ABS (X) <= C is a range comparison, which becomes a subtraction
6559 and a comparison, and is probably faster. */
6571 /* If this is an EQ or NE comparison with zero and ARG0 is
6572 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
6573 two operations, but the latter can be done in one less insn
6574 on machines that have only two-operand insns or on which a
6575 constant cannot be the first operand. */
6578 {
6581 return
6589 integer_one_node)),
6590 arg1));
6593 return
6601 integer_one_node)),
6602 arg1));
6603 }
6605 /* If this is an NE or EQ comparison of zero against the result of a
6606 signed MOD operation whose second operand is a power of 2, make
6607 the MOD operation unsigned since it is simpler and equivalent. */
6616 {
6623 }
6625 /* If this is an NE comparison of zero with an AND of one, remove the
6626 comparison since the AND will give the correct value. */
6632 /* If we have (A & C) == C where C is a power of 2, convert this into
6633 (A & C) != 0. Similarly for NE_EXPR. */
6641 /* If we have (A & C) != 0 where C is the sign bit of A, convert
6642 this into A < 0. Similarly for (A & C) == 0 into A >= 0. */
6646 {
6650 {
6655 }
6656 }
6658 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6659 and similarly for >= into !=. */
6675 return
6682 /* Simplify comparison of something with itself. (For IEEE
6683 floating-point, we can only do some of these simplifications.) */
6685 {
6687 {
6698 /* For NE, we can only do this simplification if integer. */
6701 /* ... fall through ... */
6707 }
6708 }
6710 /* If we are comparing an expression that just has comparisons
6711 of two integer values, arithmetic expressions of those comparisons,
6712 and constants, we can simplify it. There are only three cases
6713 to check: the two values can either be equal, the first can be
6714 greater, or the second can be greater. Fold the expression for
6715 those three values. Since each value must be 0 or 1, we have
6716 eight possibilities, each of which corresponds to the constant 0
6717 or 1 or one of the six possible comparisons.
6719 This handles common cases like (a > b) == 0 but also handles
6720 expressions like ((x > y) - (y > x)) > 0, which supposedly
6721 occur in macroized code. */
6724 {
6729 /* Don't handle degenerate cases here; they should already
6730 have been handled anyway. */
6739 {
6743 /* We can't just pass T to eval_subst in case cval1 or cval2
6744 was the same as ARG1. */
6746 tree high_result
6749 arg1));
6750 tree equal_result
6753 arg1));
6754 tree low_result
6757 arg1));
6759 /* All three of these results should be 0 or 1. Confirm they
6760 are. Then use those values to select the proper code
6761 to use. */
6769 {
6770 /* Make a 3-bit mask with the high-order bit being the
6771 value for `>', the next for '=', and the low for '<'. */
6775 {
6777 /* Always false. */
6798 /* Always true. */
6800 }
6805 else
6807 }
6808 }
6809 }
6811 /* If this is a comparison of a field, we may be able to simplify it. */
6816 /* Handle the constant case even without -O
6817 to make sure the warnings are given. */
6819 {
6822 }
6824 /* If this is a comparison of complex values and either or both sides
6825 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6826 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6827 This may prevent needless evaluations. */
6834 {
6847 type,
6850 }
6852 /* Optimize comparisons of strlen vs zero to a compare of the
6853 first character of the string vs zero. To wit,
6854 strlen(ptr) == 0 => *ptr == 0
6855 strlen(ptr) != 0 => *ptr != 0
6856 Other cases should reduce to one of these two (or a constant)
6857 due to the return value of strlen being unsigned. */
6862 {
6864 tree arglist;
6876 integer_zero_node));
6877 }
6879 /* From here on, the only cases we handle are when the result is
6880 known to be a constant.
6882 To compute GT, swap the arguments and do LT.
6883 To compute GE, do LT and invert the result.
6884 To compute LE, swap the arguments, do LT and invert the result.
6885 To compute NE, do EQ and invert the result.
6887 Therefore, the code below must handle only EQ and LT. */
6890 {
6893 }
6895 /* Note that it is safe to invert for real values here because we
6896 will check below in the one case that it matters. */
6901 {
6904 }
6906 /* Compute a result for LT or EQ if args permit;
6907 otherwise return T. */
6909 {
6912 else
6917 }
6920 /* Assume a nonexplicit constant cannot equal an explicit one,
6921 since such code would be undefined anyway.
6922 Exception: on sysvr4, using #pragma weak,
6923 a label can come out as 0. */
6930 #endif
6931 /* Two real constants can be compared explicitly. */
6933 {
6934 /* If either operand is a NaN, the result is false with two
6935 exceptions: First, an NE_EXPR is true on NaNs, but that case
6936 is already handled correctly since we will be inverting the
6937 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6938 or a GE_EXPR into a LT_EXPR, we must return true so that it
6939 will be inverted into false. */
6949 else
6953 }
6967 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6968 so all simple results must be passed through pedantic_non_lvalue. */
6970 return pedantic_non_lvalue
6975 /* If the second operand is zero, invert the comparison and swap
6976 the second and third operands. Likewise if the second operand
6977 is constant and the third is not or if the third operand is
6978 equivalent to the first operand of the comparison. */
6986 {
6987 /* See if this can be inverted. If it can't, possibly because
6988 it was a floating-point inequality comparison, don't do
6989 anything. */
6993 {
6997 /* arg1 should be the first argument of the new T. */
7000 }
7001 }
7003 /* If we have A op B ? A : C, we may be able to convert this to a
7004 simpler expression, depending on the operation and the values
7005 of B and C. Signed zeros prevent all of these transformations,
7006 for reasons given above each one. */
7012 {
7018 /* If we have A op 0 ? A : -A, consider applying the following
7019 transformations:
7021 A == 0? A : -A same as -A
7022 A != 0? A : -A same as A
7023 A >= 0? A : -A same as abs (A)
7024 A > 0? A : -A same as abs (A)
7025 A <= 0? A : -A same as -abs (A)
7026 A < 0? A : -A same as -abs (A)
7028 None of these transformations work for modes with signed
7029 zeros. If A is +/-0, the first two transformations will
7030 change the sign of the result (from +0 to -0, or vice
7031 versa). The last four will fix the sign of the result,
7032 even though the original expressions could be positive or
7033 negative, depending on the sign of A.
7035 Note that all these transformations are correct if A is
7036 NaN, since the two alternatives (A and -A) are also NaNs. */
7043 {
7045 return
7046 pedantic_non_lvalue
7048 negate_expr
7050 arg1))));
7058 return pedantic_non_lvalue
7066 return pedantic_non_lvalue
7070 arg1)))));
7073 }
7075 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
7076 A == 0 ? A : 0 is always 0 unless A is -0. Note that
7077 both transformations are correct when A is NaN: A != 0
7078 is then true, and A == 0 is false. */
7081 {
7086 }
7088 /* Try some transformations of A op B ? A : B.
7090 A == B? A : B same as B
7091 A != B? A : B same as A
7092 A >= B? A : B same as max (A, B)
7093 A > B? A : B same as max (B, A)
7094 A <= B? A : B same as min (A, B)
7095 A < B? A : B same as min (B, A)
7097 As above, these transformations don't work in the presence
7098 of signed zeros. For example, if A and B are zeros of
7099 opposite sign, the first two transformations will change
7100 the sign of the result. In the last four, the original
7101 expressions give different results for (A=+0, B=-0) and
7102 (A=-0, B=+0), but the transformed expressions do not.
7104 The first two transformations are correct if either A or B
7105 is a NaN. In the first transformation, the condition will
7106 be false, and B will indeed be chosen. In the case of the
7107 second transformation, the condition A != B will be true,
7108 and A will be chosen.
7110 The conversions to max() and min() are not correct if B is
7111 a number and A is not. The conditions in the original
7112 expressions will be false, so all four give B. The min()
7113 and max() versions would give a NaN instead. */
7116 {
7121 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
7123 {
7127 }
7130 {
7137 /* In C++ a ?: expression can be an lvalue, so put the
7138 operand which will be used if they are equal first
7139 so that we can convert this back to the
7140 corresponding COND_EXPR. */
7142 return pedantic_non_lvalue
7152 return pedantic_non_lvalue
7161 }
7162 }
7164 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
7165 we might still be able to simplify this. For example,
7166 if C1 is one less or one more than C2, this might have started
7167 out as a MIN or MAX and been transformed by this function.
7168 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
7174 {
7176 /* We can replace A with C1 in this case. */
7183 /* If C1 is C2 + 1, this is min(A, C2). */
7188 return pedantic_non_lvalue
7193 /* If C1 is C2 - 1, this is min(A, C2). */
7198 return pedantic_non_lvalue
7203 /* If C1 is C2 - 1, this is max(A, C2). */
7208 return pedantic_non_lvalue
7213 /* If C1 is C2 + 1, this is max(A, C2). */
7218 return pedantic_non_lvalue
7225 }
7226 }
7228 /* If the second operand is simpler than the third, swap them
7229 since that produces better jump optimization results. */
7235 {
7236 /* See if this can be inverted. If it can't, possibly because
7237 it was a floating-point inequality comparison, don't do
7238 anything. */
7242 {
7246 /* arg1 should be the first argument of the new T. */
7249 }
7250 }
7252 /* Convert A ? 1 : 0 to simply A. */
7255 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
7256 call to fold will try to move the conversion inside
7257 a COND, which will recurse. In that case, the COND_EXPR
7258 is probably the best choice, so leave it alone. */
7262 /* Convert A ? 0 : 1 to !A. This prefers the use of NOT_EXPR
7263 over COND_EXPR in cases such as floating point comparisons. */
7270 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
7271 operation is simply A & 2. */
7282 /* Convert A ? B : 0 into A && B if A and B are truth values. */
7289 /* Convert A ? B : 1 into !A || B if A and B are truth values. */
7293 {
7294 /* Only perform transformation if ARG0 is easily inverted. */
7299 }
7304 /* When pedantic, a compound expression can be neither an lvalue
7305 nor an integer constant expression. */
7308 /* Don't let (0, 0) be null pointer constant. */
7350 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
7351 appropriate. */
7356 {
7360 tree arg01;
7371 {
7385 arg01));
7386 }
7389 }
7392 /* Check for a built-in function. */
7397 {
7401 }
7407 }
7409 /* Determine if first argument is a multiple of second argument. Return 0 if
7410 it is not, or we cannot easily determined it to be.
7412 An example of the sort of thing we care about (at this point; this routine
7413 could surely be made more general, and expanded to do what the *_DIV_EXPR's
7414 fold cases do now) is discovering that
7416 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7418 is a multiple of
7420 SAVE_EXPR (J * 8)
7422 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
7424 This code also handles discovering that
7426 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7428 is a multiple of 8 so we don't have to worry about dealing with a
7429 possible remainder.
7431 Note that we *look* inside a SAVE_EXPR only to determine how it was
7432 calculated; it is not safe for fold to do much of anything else with the
7433 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
7434 at run time. For example, the latter example above *cannot* be implemented
7435 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
7436 evaluation time of the original SAVE_EXPR is not necessarily the same at
7437 the time the new expression is evaluated. The only optimization of this
7438 sort that would be valid is changing
7440 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
7442 divided by 8 to
7444 SAVE_EXPR (I) * SAVE_EXPR (J)
7446 (where the same SAVE_EXPR (J) is used in the original and the
7447 transformed version). */
7449 static int
7451 tree type;
7452 tree top;
7453 tree bottom;
7454 {
7462 {
7474 {
7478 /* const_binop may not detect overflow correctly,
7479 so check for it explicitly here. */
7488 }
7492 /* Can't handle conversions from non-integral or wider integral type. */
7498 /* .. fall through ... */
7514 }
7515 }
7517 /* Return true if `t' is known to be non-negative. */
7519 int
7521 tree t;
7522 {
7524 {
7533 /* These are undefined at zero. This is true even if
7534 C[LT]Z_DEFINED_VALUE_AT_ZERO is set, since what we're
7535 computing here is a user-visible property. */
7575 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7577 else
7578 /* We don't know sign of `t', so be conservative and return false. */
7580 }
7581 }
7583 /* Return true if `r' is known to be non-negative.
7584 Only handles constants at the moment. */
7586 int
7588 rtx r;
7589 {
7591 {
7601 {
7603 rtx elt;
7608 {
7612 }
7615 }
7619 /* These are always nonnegative. */
7624 }
7625 }