| blob | feb3e5f72fdca6df406510e298eaccfd1a655893 |
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,
3 1999, 2000, 2001, 2002 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. */
58 HOST_WIDE_INT));
61 HOST_WIDE_INT *));
86 HOST_WIDE_INT *,
106 static tree fold_binary_op_with_conditional_arg
110 #ifndef BRANCH_COST
111 #define BRANCH_COST 1
112 #endif
114 #if defined(HOST_EBCDIC)
115 /* bit 8 is significant in EBCDIC */
116 #define CHARMASK 0xff
117 #else
118 #define CHARMASK 0x7f
119 #endif
121 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
122 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
123 and SUM1. Then this yields nonzero if overflow occurred during the
124 addition.
126 Overflow occurs if A and B have the same sign, but A and SUM differ in
127 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
128 sign. */
129 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
130 \f
131 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
132 We do that by representing the two-word integer in 4 words, with only
133 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
134 number. The value of the word is LOWPART + HIGHPART * BASE. */
136 #define LOWPART(x) \
137 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
138 #define HIGHPART(x) \
139 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
140 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
142 /* Unpack a two-word integer into 4 words.
143 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
144 WORDS points to the array of HOST_WIDE_INTs. */
146 static void
150 HOST_WIDE_INT hi;
151 {
156 }
158 /* Pack an array of 4 words into a two-word integer.
159 WORDS points to the array of words.
160 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
162 static void
167 {
170 }
171 \f
172 /* Make the integer constant T valid for its type by setting to 0 or 1 all
173 the bits in the constant that don't belong in the type.
175 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
176 nonzero, a signed overflow has already occurred in calculating T, so
177 propagate it.
179 Make the real constant T valid for its type by calling CHECK_FLOAT_VALUE,
180 if it exists. */
182 int
184 tree t;
186 {
188 HOST_WIDE_INT high;
192 {
193 #ifdef CHECK_FLOAT_VALUE
195 overflow);
196 #endif
198 }
208 else
211 /* First clear all bits that are beyond the type's precision. */
214 ;
218 else
219 {
223 }
225 /* Unsigned types do not suffer sign extension or overflow unless they
226 are a sizetype. */
232 /* If the value's sign bit is set, extend the sign. */
240 {
241 /* Value is negative:
242 set to 1 all the bits that are outside this type's precision. */
246 else
247 {
251 }
252 }
254 /* Return nonzero if signed overflow occurred. */
255 return
258 }
259 \f
260 /* Add two doubleword integers with doubleword result.
261 Each argument is given as two `HOST_WIDE_INT' pieces.
262 One argument is L1 and H1; the other, L2 and H2.
263 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
265 int
271 {
273 HOST_WIDE_INT h;
281 }
283 /* Negate a doubleword integer with doubleword result.
284 Return nonzero if the operation overflows, assuming it's signed.
285 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
286 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
288 int
291 HOST_WIDE_INT h1;
294 {
296 {
300 }
301 else
302 {
306 }
307 }
308 \f
309 /* Multiply two doubleword integers with doubleword result.
310 Return nonzero if the operation overflows, assuming it's signed.
311 Each argument is given as two `HOST_WIDE_INT' pieces.
312 One argument is L1 and H1; the other, L2 and H2.
313 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
315 int
321 {
336 {
339 {
341 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
343 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
347 }
349 }
353 /* Check for overflow by calculating the top half of the answer in full;
354 it should agree with the low half's sign bit. */
357 {
360 }
362 {
365 }
367 }
368 \f
369 /* Shift the doubleword integer in L1, H1 left by COUNT places
370 keeping only PREC bits of result.
371 Shift right if COUNT is negative.
372 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
373 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
375 void
383 {
387 {
390 }
392 #ifdef SHIFT_COUNT_TRUNCATED
395 #endif
398 {
399 /* Shifting by the host word size is undefined according to the
400 ANSI standard, so we must handle this as a special case. */
403 }
405 {
408 }
409 else
410 {
414 }
416 /* Sign extend all bits that are beyond the precision. */
423 ;
425 {
428 }
429 else
430 {
434 }
435 }
437 /* Shift the doubleword integer in L1, H1 right by COUNT places
438 keeping only PREC bits of result. COUNT must be positive.
439 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
440 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
442 void
450 {
453 signmask = (arith
457 #ifdef SHIFT_COUNT_TRUNCATED
460 #endif
463 {
464 /* Shifting by the host word size is undefined according to the
465 ANSI standard, so we must handle this as a special case. */
468 }
470 {
473 }
474 else
475 {
479 }
481 /* Zero / sign extend all bits that are beyond the precision. */
484 {
487 }
489 ;
491 {
494 }
495 else
496 {
500 }
501 }
502 \f
503 /* Rotate the doubleword integer in L1, H1 left by COUNT places
504 keeping only PREC bits of result.
505 Rotate right if COUNT is negative.
506 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
508 void
515 {
527 }
529 /* Rotate the doubleword integer in L1, H1 left by COUNT places
530 keeping only PREC bits of result. COUNT must be positive.
531 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
533 void
540 {
552 }
553 \f
554 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
555 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
556 CODE is a tree code for a kind of division, one of
557 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
558 or EXACT_DIV_EXPR
559 It controls how the quotient is rounded to an integer.
560 Return nonzero if the operation overflows.
561 UNS nonzero says do unsigned division. */
563 int
570 HOST_WIDE_INT hnum_orig;
572 HOST_WIDE_INT hden_orig;
575 {
591 /* calculate quotient sign and convert operands to unsigned. */
593 {
595 {
597 /* (minimum integer) / (-1) is the only overflow case. */
601 }
603 {
606 }
607 }
612 /* This unsigned division rounds toward zero. */
615 }
619 /* hden != 0 already checked. */
624 }
634 /* Special code for when the divisor < BASE. */
636 {
637 /* hnum != 0 already checked. */
639 {
643 }
644 }
645 else
646 {
647 /* Full double precision division,
648 with thanks to Don Knuth's "Seminumerical Algorithms". */
652 /* Find the highest non-zero divisor digit. */
655 {
658 }
660 /* Insure that the first digit of the divisor is at least BASE/2.
661 This is required by the quotient digit estimation algorithm. */
668 {
672 }
677 {
682 }
683 }
687 /* Main loop */
689 {
690 /* Guess the next quotient digit, quo_est, by dividing the first
691 two remaining dividend digits by the high order quotient digit.
692 quo_est is never low and is at most 2 high. */
699 else
702 /* Refine quo_est so it's usually correct, and at most one high. */
707 quo_est--;
709 /* Try QUO_EST as the quotient digit, by multiplying the
710 divisor by QUO_EST and subtracting from the remaining dividend.
711 Keep in mind that QUO_EST is the I - 1st digit. */
715 {
721 }
723 /* If quo_est was high by one, then num[i] went negative and
724 we need to correct things. */
726 {
727 quo_est--;
730 {
734 }
737 }
739 /* Store the quotient digit. */
741 }
742 }
746 finish_up:
747 /* if result is negative, make it so. */
751 /* compute trial remainder: rem = num - (quo * den) */
757 {
766 {
767 /* quo = quo - 1; */
770 }
771 else
778 {
781 }
782 else
788 {
794 /* Get absolute values */
800 /* If (2 * abs (lrem) >= abs (lden)) */
809 {
811 /* quo = quo - 1; */
814 else
815 /* quo = quo + 1; */
818 }
819 else
821 }
826 }
828 /* compute true remainder: rem = num - (quo * den) */
833 }
834 \f
835 /* Given T, an expression, return the negation of T. Allow for T to be
836 null, in which case return null. */
838 static tree
840 tree t;
841 {
842 tree type;
843 tree tem;
852 {
865 /* - (A - B) -> B - A */
875 }
878 }
879 \f
880 /* Split a tree IN into a constant, literal and variable parts that could be
881 combined with CODE to make IN. "constant" means an expression with
882 TREE_CONSTANT but that isn't an actual constant. CODE must be a
883 commutative arithmetic operation. Store the constant part into *CONP,
884 the literal in &LITP and return the variable part. If a part isn't
885 present, set it to null. If the tree does not decompose in this way,
886 return the entire tree as the variable part and the other parts as null.
888 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
889 case, we negate an operand that was subtracted. If NEGATE_P is true, we
890 are negating all of IN.
892 If IN is itself a literal or constant, return it as appropriate.
894 Note that we do not guarantee that any of the three values will be the
895 same type as IN, but they will have the same signedness and mode. */
897 static tree
899 tree in;
903 {
909 /* Strip any conversions that don't change the machine mode or signedness. */
916 /* We can associate addition and subtraction together (even
917 though the C standard doesn't say so) for integers because
918 the value is not affected. For reals, the value might be
919 affected, so we can't. */
922 {
928 /* First see if either of the operands is a literal, then a constant. */
939 /* If we haven't dealt with either operand, this is not a case we can
940 decompose. Otherwise, VAR is either of the ones remaining, if any. */
945 else
948 /* Now do any needed negations. */
952 }
955 else
959 {
963 }
966 }
968 /* Re-associate trees split by the above function. T1 and T2 are either
969 expressions to associate or null. Return the new expression, if any. If
970 we build an operation, do it in TYPE and with CODE, except if CODE is a
971 MINUS_EXPR, in which case we use PLUS_EXPR since split_tree will already
972 have taken care of the negations. */
974 static tree
978 tree type;
979 {
988 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
989 try to fold this since we will have infinite recursion. But do
990 deal with any NEGATE_EXPRs. */
993 {
1000 else
1002 }
1005 }
1006 \f
1007 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1008 to produce a new constant.
1010 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1012 static tree
1017 {
1021 HOST_WIDE_INT hi;
1023 HOST_WIDE_INT garbageh;
1024 tree t;
1027 int is_sizetype
1038 {
1058 /* It's unclear from the C standard whether shifts can overflow.
1059 The following code ignores overflow; perhaps a C standard
1060 interpretation ruling is needed. */
1090 /* This is a shortcut for a common special case. */
1095 {
1101 }
1103 /* ... fall through ... */
1107 {
1110 }
1113 {
1116 }
1123 /* This is a shortcut for a common special case. */
1128 {
1133 }
1135 /* ... fall through ... */
1151 else
1157 else
1163 }
1165 /* If this is for a sizetype, can be represented as one (signed)
1166 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1167 constants. */
1173 else
1174 {
1177 }
1180 = ((notrunc
1187 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1188 So check if force_fit_type truncated the value. */
1199 }
1201 /* Define input and output argument for const_binop_1. */
1202 struct cb_args
1203 {
1208 };
1210 /* Do the real arithmetic for const_binop while protected by a
1211 float overflow handler. */
1213 static void
1215 PTR data;
1216 {
1218 REAL_VALUE_TYPE value;
1222 args->t
1225 }
1227 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1228 constant. We assume ARG1 and ARG2 have the same data type, or at least
1229 are the same kind of constant and the same machine mode.
1231 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1233 static tree
1238 {
1246 {
1247 REAL_VALUE_TYPE d1;
1248 REAL_VALUE_TYPE d2;
1250 tree t;
1256 /* If either operand is a NaN, just return it. Otherwise, set up
1257 for floating-point trap; we return an overflow. */
1263 /* Setup input for const_binop_1() */
1270 /* Receive output from const_binop_1. */
1272 else
1273 {
1274 /* We got an exception from const_binop_1. */
1277 }
1287 }
1289 {
1295 tree t;
1298 {
1318 notrunc),
1324 notrunc));
1328 {
1329 tree magsquared
1333 notrunc);
1336 const_binop
1341 notrunc),
1343 notrunc),
1344 notrunc),
1346 const_binop
1351 notrunc),
1353 notrunc),
1354 notrunc),
1356 }
1361 }
1363 }
1365 }
1367 /* These are the hash table functions for the hash table of INTEGER_CST
1368 nodes of a sizetype. */
1370 /* Return the hash code code X, an INTEGER_CST. */
1372 static hashval_t
1375 {
1381 }
1383 /* Return non-zero if the value represented by *X (an INTEGER_CST tree node)
1384 is the same as that given by *Y, which is the same. */
1386 static int
1390 {
1398 }
1399 \f
1400 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1401 bits are given by NUMBER and of the sizetype represented by KIND. */
1403 tree
1405 HOST_WIDE_INT number;
1407 {
1409 }
1411 /* Likewise, but the desired type is specified explicitly. */
1413 tree
1415 HOST_WIDE_INT number;
1416 tree type;
1417 {
1423 {
1428 }
1430 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1431 hash table, we return the value from the hash table. Otherwise, we
1432 place that in the hash table and make a new node for the next time. */
1441 {
1447 }
1448 else
1450 }
1452 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1453 is a tree code. The type of the result is taken from the operands.
1454 Both must be the same type integer type and it must be a size type.
1455 If the operands are constant, so is the result. */
1457 tree
1461 {
1468 /* Handle the special case of two integer constants faster. */
1470 {
1471 /* And some specific cases even faster than that. */
1480 /* Handle general case of two integer constants. */
1482 }
1488 }
1490 /* Given two values, either both of sizetype or both of bitsizetype,
1491 compute the difference between the two values. Return the value
1492 in signed type corresponding to the type of the operands. */
1494 tree
1497 {
1499 tree ctype;
1505 /* If the type is already signed, just do the simple thing. */
1512 /* If either operand is not a constant, do the conversions to the signed
1513 type and subtract. The hardware will do the right thing with any
1514 overflow in the subtraction. */
1519 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1520 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1521 overflow) and negate (which can't either). Special-case a result
1522 of zero while we're here. */
1527 else
1530 }
1531 \f
1532 /* This structure is used to communicate arguments to fold_convert_1. */
1533 struct fc_args
1534 {
1538 };
1540 /* Function to convert floating-point constants, protected by floating
1541 point exception handler. */
1543 static void
1545 PTR data;
1546 {
1552 }
1554 /* Given T, a tree representing type conversion of ARG1, a constant,
1555 return a constant tree representing the result of conversion. */
1557 static tree
1559 tree t;
1560 tree arg1;
1561 {
1566 {
1568 {
1569 /* If we would build a constant wider than GCC supports,
1570 leave the conversion unfolded. */
1574 /* If we are trying to make a sizetype for a small integer, use
1575 size_int to pick up cached types to reduce duplicate nodes. */
1581 /* Given an integer constant, make new constant with new type,
1582 appropriately sign-extended or truncated. */
1586 /* Indicate an overflow if (1) ARG1 already overflowed,
1587 or (2) force_fit_type indicates an overflow.
1588 Tell force_fit_type that an overflow has already occurred
1589 if ARG1 is a too-large unsigned value and T is signed.
1590 But don't indicate an overflow if converting a pointer. */
1600 }
1602 {
1603 /* Don't initialize these, use assignments.
1604 Initialized local aggregates don't work on old compilers. */
1605 REAL_VALUE_TYPE x;
1606 REAL_VALUE_TYPE l;
1607 REAL_VALUE_TYPE u;
1618 /* See if X will be in range after truncation towards 0.
1619 To compensate for truncation, move the bounds away from 0,
1620 but reject if X exactly equals the adjusted bounds. */
1624 /* If X is a NaN, use zero instead and show we have an overflow.
1625 Otherwise, range check. */
1629 && !no_upper_bound
1633 {
1637 }
1643 }
1645 }
1647 {
1651 {
1655 {
1659 }
1661 /* Setup input for fold_convert_1() */
1666 {
1667 /* Receive output from fold_convert_1() */
1669 }
1670 else
1671 {
1672 /* We got an exception from fold_convert_1() */
1675 }
1682 }
1683 }
1686 }
1687 \f
1688 /* Return an expr equal to X but certainly not valid as an lvalue. */
1690 tree
1692 tree x;
1693 {
1694 tree result;
1696 /* These things are certainly not lvalues. */
1707 }
1709 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1710 Zero means allow extended lvalues. */
1714 /* When pedantic, return an expr equal to X but certainly not valid as a
1715 pedantic lvalue. Otherwise, return X. */
1717 tree
1719 tree x;
1720 {
1723 else
1725 }
1726 \f
1727 /* Given a tree comparison code, return the code that is the logical inverse
1728 of the given code. It is not safe to do this for floating-point
1729 comparisons, except for NE_EXPR and EQ_EXPR. */
1731 static enum tree_code
1734 {
1736 {
1751 }
1752 }
1754 /* Similar, but return the comparison that results if the operands are
1755 swapped. This is safe for floating-point. */
1757 static enum tree_code
1760 {
1762 {
1776 }
1777 }
1779 /* Return nonzero if CODE is a tree code that represents a truth value. */
1781 static int
1784 {
1789 }
1790 \f
1791 /* Return nonzero if two operands are necessarily equal.
1792 If ONLY_CONST is non-zero, only return non-zero for constants.
1793 This function tests whether the operands are indistinguishable;
1794 it does not test whether they are equal using C's == operation.
1795 The distinction is important for IEEE floating point, because
1796 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1797 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1799 int
1803 {
1804 /* If both types don't have the same signedness, then we can't consider
1805 them equal. We must check this before the STRIP_NOPS calls
1806 because they may change the signedness of the arguments. */
1814 /* This is needed for conversions and for COMPONENT_REF.
1815 Might as well play it safe and always test this. */
1821 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1822 We don't care about side effects in that case because the SAVE_EXPR
1823 takes care of that for us. In all other cases, two expressions are
1824 equal if they have no side effects. If we have two identical
1825 expressions with side effects that should be treated the same due
1826 to the only side effects being identical SAVE_EXPR's, that will
1827 be detected in the recursive calls below. */
1833 /* Next handle constant cases, those for which we can return 1 even
1834 if ONLY_CONST is set. */
1837 {
1850 {
1860 {
1865 }
1868 }
1872 only_const)
1874 only_const));
1887 }
1893 {
1895 /* Two conversions are equal only if signedness and modes match. */
1911 /* For commutative ops, allow the other order. */
1924 /* If either of the pointer (or reference) expressions we are dereferencing
1925 contain a side effect, these cannot be equal. */
1931 {
1953 }
1962 }
1963 }
1964 \f
1965 /* Similar to operand_equal_p, but see if ARG0 might have been made by
1966 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
1968 When in doubt, return 0. */
1970 static int
1973 tree other;
1974 {
1986 /* Discard any conversions that don't change the modes of ARG0 and ARG1
1987 and see if the inner values are the same. This removes any
1988 signedness comparison, which doesn't matter here. */
1995 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
1996 actual comparison operand, ARG0.
1998 First throw away any conversions to wider types
1999 already present in the operands. */
2008 {
2011 /* Make sure shorter operand is extended the right way
2012 to match the longer operand. */
2015 primarg1);
2019 }
2022 }
2023 \f
2024 /* See if ARG is an expression that is either a comparison or is performing
2025 arithmetic on comparisons. The comparisons must only be comparing
2026 two different values, which will be stored in *CVAL1 and *CVAL2; if
2027 they are non-zero it means that some operands have already been found.
2028 No variables may be used anywhere else in the expression except in the
2029 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
2030 the expression and save_expr needs to be called with CVAL1 and CVAL2.
2032 If this is true, return 1. Otherwise, return zero. */
2034 static int
2036 tree arg;
2039 {
2043 /* We can handle some of the 'e' cases here. */
2053 {
2054 /* If we've already found a CVAL1 or CVAL2, this expression is
2055 two complex to handle. */
2061 }
2064 {
2087 /* First see if we can handle the first operand, then the second. For
2088 the second operand, we know *CVAL1 can't be zero. It must be that
2089 one side of the comparison is each of the values; test for the
2090 case where this isn't true by failing if the two operands
2091 are the same. */
2100 ;
2104 ;
2105 else
2109 ;
2113 ;
2114 else
2121 }
2122 }
2123 \f
2124 /* ARG is a tree that is known to contain just arithmetic operations and
2125 comparisons. Evaluate the operations in the tree substituting NEW0 for
2126 any occurrence of OLD0 as an operand of a comparison and likewise for
2127 NEW1 and OLD1. */
2129 static tree
2131 tree arg;
2133 {
2138 /* We can handle some of the 'e' cases here. */
2146 {
2161 {
2178 }
2179 /* fall through - ??? */
2182 {
2186 /* We need to check both for exact equality and tree equality. The
2187 former will be true if the operand has a side-effect. In that
2188 case, we know the operand occurred exactly once. */
2201 }
2205 }
2206 }
2207 \f
2208 /* Return a tree for the case when the result of an expression is RESULT
2209 converted to TYPE and OMITTED was previously an operand of the expression
2210 but is now not needed (e.g., we folded OMITTED * 0).
2212 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2213 the conversion of RESULT to TYPE. */
2215 static tree
2218 {
2225 }
2227 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2229 static tree
2232 {
2239 }
2240 \f
2241 /* Return a simplified tree node for the truth-negation of ARG. This
2242 never alters ARG itself. We assume that ARG is an operation that
2243 returns a truth value (0 or 1). */
2245 tree
2247 tree arg;
2248 {
2255 /* If this is a comparison, we can simply invert it, except for
2256 floating-point non-equality comparisons, in which case we just
2257 enclose a TRUTH_NOT_EXPR around what we have. */
2260 {
2262 && !flag_unsafe_math_optimizations
2266 else
2269 }
2272 {
2287 /* Here we can invert either operand. We invert the first operand
2288 unless the second operand is a TRUTH_NOT_EXPR in which case our
2289 result is the XOR of the first operand with the inside of the
2290 negation of the second operand. */
2295 else
2350 }
2354 }
2356 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2357 operands are another bit-wise operation with a common input. If so,
2358 distribute the bit operations to save an operation and possibly two if
2359 constants are involved. For example, convert
2360 (A | B) & (A | C) into A | (B & C)
2361 Further simplification will occur if B and C are constants.
2363 If this optimization cannot be done, 0 will be returned. */
2365 static tree
2368 tree type;
2370 {
2371 tree common;
2381 {
2385 }
2387 {
2391 }
2393 {
2397 }
2399 {
2403 }
2404 else
2409 }
2410 \f
2411 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2412 starting at BITPOS. The field is unsigned if UNSIGNEDP is non-zero. */
2414 static tree
2416 tree inner;
2417 tree type;
2420 {
2427 }
2429 /* Optimize a bit-field compare.
2431 There are two cases: First is a compare against a constant and the
2432 second is a comparison of two items where the fields are at the same
2433 bit position relative to the start of a chunk (byte, halfword, word)
2434 large enough to contain it. In these cases we can avoid the shift
2435 implicit in bitfield extractions.
2437 For constants, we emit a compare of the shifted constant with the
2438 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2439 compared. For two fields at the same position, we do the ANDs with the
2440 similar mask and compare the result of the ANDs.
2442 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2443 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2444 are the left and right operands of the comparison, respectively.
2446 If the optimization described above can be done, we return the resulting
2447 tree. Otherwise we return zero. */
2449 static tree
2452 tree compare_type;
2454 {
2463 tree mask;
2464 tree offset;
2466 /* Get all the information about the extractions being done. If the bit size
2467 if the same as the size of the underlying object, we aren't doing an
2468 extraction at all and so can do nothing. We also don't want to
2469 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2470 then will no longer be able to replace it. */
2478 {
2479 /* If this is not a constant, we can only do something if bit positions,
2480 sizes, and signedness are the same. */
2488 }
2490 /* See if we can find a mode to refer to this field. We should be able to,
2491 but fail if we can't. */
2500 /* Set signed and unsigned types of the precision of this mode for the
2501 shifts below. */
2505 /* Compute the bit position and size for the new reference and our offset
2506 within it. If the new reference is the same size as the original, we
2507 won't optimize anything, so return zero. */
2517 /* Make the mask to be used against the extracted field. */
2527 /* If not comparing with constant, just rework the comparison
2528 and return. */
2533 mask),
2537 mask));
2539 /* Otherwise, we are handling the constant case. See if the constant is too
2540 big for the field. Warn and return a tree of for 0 (false) if so. We do
2541 this not only for its own sake, but to avoid having to test for this
2542 error case below. If we didn't, we might generate wrong code.
2544 For unsigned fields, the constant shifted right by the field length should
2545 be all zero. For signed fields, the high-order bits should agree with
2546 the sign bit. */
2549 {
2553 {
2559 }
2560 }
2561 else
2562 {
2566 {
2572 }
2573 }
2575 /* Single-bit compares should always be against zero. */
2577 {
2580 }
2582 /* Make a new bitfield reference, shift the constant over the
2583 appropriate number of bits and mask it with the computed mask
2584 (in case this was a signed field). If we changed it, make a new one. */
2587 {
2590 }
2600 rhs);
2601 }
2602 \f
2603 /* Subroutine for fold_truthop: decode a field reference.
2605 If EXP is a comparison reference, we return the innermost reference.
2607 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2608 set to the starting bit number.
2610 If the innermost field can be completely contained in a mode-sized
2611 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2613 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2614 otherwise it is not changed.
2616 *PUNSIGNEDP is set to the signedness of the field.
2618 *PMASK is set to the mask used. This is either contained in a
2619 BIT_AND_EXPR or derived from the width of the field.
2621 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2623 Return 0 if this is not a component reference or is one that we can't
2624 do anything with. */
2626 static tree
2629 tree exp;
2635 {
2638 tree unsigned_type;
2641 /* All the optimizations using this function assume integer fields.
2642 There are problems with FP fields since the type_for_size call
2643 below can fail for, e.g., XFmode. */
2650 {
2656 }
2665 /* Compute the mask to access the bitfield. */
2675 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2683 }
2685 /* Return non-zero if MASK represents a mask of SIZE ones in the low-order
2686 bit positions. */
2688 static int
2690 tree mask;
2692 {
2695 tree tmask;
2700 return
2707 }
2709 /* Subroutine for fold_truthop: determine if an operand is simple enough
2710 to be evaluated unconditionally. */
2712 static int
2714 tree exp;
2715 {
2716 /* Strip any conversions that don't change the machine mode. */
2728 /* Don't regard global variables as simple. They may be
2729 allocated in ways unknown to the compiler (shared memory,
2730 #pragma weak, etc). */
2733 /* Loading a static variable is unduly expensive, but global
2734 registers aren't expensive. */
2736 }
2737 \f
2738 /* The following functions are subroutines to fold_range_test and allow it to
2739 try to change a logical combination of comparisons into a range test.
2741 For example, both
2742 X == 2 || X == 3 || X == 4 || X == 5
2743 and
2744 X >= 2 && X <= 5
2745 are converted to
2746 (unsigned) (X - 2) <= 3
2748 We describe each set of comparisons as being either inside or outside
2749 a range, using a variable named like IN_P, and then describe the
2750 range with a lower and upper bound. If one of the bounds is omitted,
2751 it represents either the highest or lowest value of the type.
2753 In the comments below, we represent a range by two numbers in brackets
2754 preceded by a "+" to designate being inside that range, or a "-" to
2755 designate being outside that range, so the condition can be inverted by
2756 flipping the prefix. An omitted bound is represented by a "-". For
2757 example, "- [-, 10]" means being outside the range starting at the lowest
2758 possible value and ending at 10, in other words, being greater than 10.
2759 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2760 always false.
2762 We set up things so that the missing bounds are handled in a consistent
2763 manner so neither a missing bound nor "true" and "false" need to be
2764 handled using a special case. */
2766 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2767 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2768 and UPPER1_P are nonzero if the respective argument is an upper bound
2769 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2770 must be specified for a comparison. ARG1 will be converted to ARG0's
2771 type if both are specified. */
2773 static tree
2776 tree type;
2779 {
2780 tree tem;
2784 /* If neither arg represents infinity, do the normal operation.
2785 Else, if not a comparison, return infinity. Else handle the special
2786 comparison rules. Note that most of the cases below won't occur, but
2787 are handled for consistency. */
2790 {
2795 }
2800 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2801 for neither. In real maths, we cannot assume open ended ranges are
2802 the same. But, this is computer arithmetic, where numbers are finite.
2803 We can therefore make the transformation of any unbounded range with
2804 the value Z, Z being greater than any representable number. This permits
2805 us to treat unbounded ranges as equal. */
2809 {
2830 }
2833 }
2834 \f
2835 /* Given EXP, a logical expression, set the range it is testing into
2836 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2837 actually being tested. *PLOW and *PHIGH will be made of the same type
2838 as the returned expression. If EXP is not a comparison, we will most
2839 likely not be returning a useful value and range. */
2841 static tree
2843 tree exp;
2846 {
2853 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2854 and see if we can refine the range. Some of the cases below may not
2855 happen, but it doesn't seem worth worrying about this. We "continue"
2856 the outer loop when we've changed something; otherwise we "break"
2857 the switch, which will "break" the while. */
2862 {
2866 {
2877 }
2879 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2880 lose a cast by accident. */
2885 {
2892 /* We can only do something if the range is testing for zero
2893 and if the second operand is an integer constant. Note that
2894 saying something is "in" the range we make is done by
2895 complementing IN_P since it will set in the initial case of
2896 being not equal to zero; "out" is leaving it alone. */
2903 {
2924 }
2928 /* If this is an unsigned comparison, we also know that EXP is
2929 greater than or equal to zero. We base the range tests we make
2930 on that fact, so we record it here so we can parse existing
2931 range tests. */
2933 {
2936 NULL_TREE))
2941 /* If the high bound is missing, but we
2942 have a low bound, reverse the range so
2943 it goes from zero to the low bound minus 1. */
2945 {
2950 }
2951 }
2955 /* (-x) IN [a,b] -> x in [-b, -a] */
2965 /* ~ X -> -X - 1 */
2974 /* If EXP is signed, any overflow in the computation is undefined,
2975 so we don't worry about it so long as our computations on
2976 the bounds don't overflow. For unsigned, overflow is defined
2977 and this is exactly the right thing. */
2986 /* Check for an unsigned range which has wrapped around the maximum
2987 value thus making n_high < n_low, and normalize it. */
2989 {
2995 /* If the range is of the form +/- [ x+1, x ], we won't
2996 be able to normalize it. But then, it represents the
2997 whole range or the empty set, so make it
2998 +/- [ -, - ]. */
3002 else
3004 }
3005 else
3028 /* If we're converting from an unsigned to a signed type,
3029 we will be doing the comparison as unsigned. The tests above
3030 have already verified that LOW and HIGH are both positive.
3032 So we have to make sure that the original unsigned value will
3033 be interpreted as positive. */
3035 {
3037 tree high_positive;
3039 /* A range without an upper bound is, naturally, unbounded.
3040 Since convert would have cropped a very large value, use
3041 the max value for the destination type. */
3042 high_positive
3050 /* If the low bound is specified, "and" the range with the
3051 range for which the original unsigned value will be
3052 positive. */
3054 {
3058 high_positive))
3062 }
3063 else
3064 {
3065 /* Otherwise, "or" the range with the range of the input
3066 that will be interpreted as negative. */
3070 high_positive))
3074 }
3075 }
3083 }
3086 }
3088 /* If EXP is a constant, we can evaluate whether this is true or false. */
3090 {
3097 }
3101 }
3102 \f
3103 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3104 type, TYPE, return an expression to test if EXP is in (or out of, depending
3105 on IN_P) the range. */
3107 static tree
3109 tree type;
3110 tree exp;
3113 {
3137 {
3141 }
3148 else
3150 }
3151 \f
3152 /* Given two ranges, see if we can merge them into one. Return 1 if we
3153 can, 0 if we can't. Set the output range into the specified parameters. */
3155 static int
3161 {
3165 tree tem;
3175 /* Make range 0 be the range that starts first, or ends last if they
3176 start at the same value. Swap them if it isn't. */
3179 || (lowequal
3182 {
3186 }
3188 /* Now flag two cases, whether the ranges are disjoint or whether the
3189 second range is totally subsumed in the first. Note that the tests
3190 below are simplified by the ones above. */
3196 /* We now have four cases, depending on whether we are including or
3197 excluding the two ranges. */
3199 {
3200 /* If they don't overlap, the result is false. If the second range
3201 is a subset it is the result. Otherwise, the range is from the start
3202 of the second to the end of the first. */
3207 else
3209 }
3212 {
3213 /* If they don't overlap, the result is the first range. If they are
3214 equal, the result is false. If the second range is a subset of the
3215 first, and the ranges begin at the same place, we go from just after
3216 the end of the first range to the end of the second. If the second
3217 range is not a subset of the first, or if it is a subset and both
3218 ranges end at the same place, the range starts at the start of the
3219 first range and ends just before the second range.
3220 Otherwise, we can't describe this as a single range. */
3226 {
3230 }
3232 {
3236 }
3237 else
3239 }
3242 {
3243 /* If they don't overlap, the result is the second range. If the second
3244 is a subset of the first, the result is false. Otherwise,
3245 the range starts just after the first range and ends at the
3246 end of the second. */
3251 else
3252 {
3256 }
3257 }
3259 else
3260 {
3261 /* The case where we are excluding both ranges. Here the complex case
3262 is if they don't overlap. In that case, the only time we have a
3263 range is if they are adjacent. If the second is a subset of the
3264 first, the result is the first. Otherwise, the range to exclude
3265 starts at the beginning of the first range and ends at the end of the
3266 second. */
3268 {
3275 else
3277 }
3280 else
3282 }
3286 }
3287 \f
3288 /* EXP is some logical combination of boolean tests. See if we can
3289 merge it into some range test. Return the new tree if so. */
3291 static tree
3293 tree exp;
3294 {
3301 tree tem;
3303 /* If this is an OR operation, invert both sides; we will invert
3304 again at the end. */
3308 /* If both expressions are the same, if we can merge the ranges, and we
3309 can build the range test, return it or it inverted. If one of the
3310 ranges is always true or always false, consider it to be the same
3311 expression as the other. */
3321 /* On machines where the branch cost is expensive, if this is a
3322 short-circuited branch and the underlying object on both sides
3323 is the same, make a non-short-circuit operation. */
3329 {
3330 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3331 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3332 which cases we can't do this. */
3341 {
3353 }
3354 }
3357 }
3358 \f
3359 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3360 bit value. Arrange things so the extra bits will be set to zero if and
3361 only if C is signed-extended to its full width. If MASK is nonzero,
3362 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3364 static tree
3366 tree c;
3369 tree mask;
3370 {
3373 tree temp;
3378 /* We work by getting just the sign bit into the low-order bit, then
3379 into the high-order bit, then sign-extend. We then XOR that value
3380 with C. */
3384 /* We must use a signed type in order to get an arithmetic right shift.
3385 However, we must also avoid introducing accidental overflows, so that
3386 a subsequent call to integer_zerop will work. Hence we must
3387 do the type conversion here. At this point, the constant is either
3388 zero or one, and the conversion to a signed type can never overflow.
3389 We could get an overflow if this conversion is done anywhere else. */
3397 /* If necessary, convert the type back to match the type of C. */
3402 }
3403 \f
3404 /* Find ways of folding logical expressions of LHS and RHS:
3405 Try to merge two comparisons to the same innermost item.
3406 Look for range tests like "ch >= '0' && ch <= '9'".
3407 Look for combinations of simple terms on machines with expensive branches
3408 and evaluate the RHS unconditionally.
3410 For example, if we have p->a == 2 && p->b == 4 and we can make an
3411 object large enough to span both A and B, we can do this with a comparison
3412 against the object ANDed with the a mask.
3414 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3415 operations to do this with one comparison.
3417 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3418 function and the one above.
3420 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3421 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3423 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3424 two operands.
3426 We return the simplified tree or 0 if no optimization is possible. */
3428 static tree
3432 {
3433 /* If this is the "or" of two comparisons, we can do something if
3434 the comparisons are NE_EXPR. If this is the "and", we can do something
3435 if the comparisons are EQ_EXPR. I.e.,
3436 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3438 WANTED_CODE is this operation code. For single bit fields, we can
3439 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3440 comparison for one-bit fields. */
3460 /* Start by getting the comparison codes. Fail if anything is volatile.
3461 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3462 it were surrounded with a NE_EXPR. */
3487 /* If the RHS can be evaluated unconditionally and its operands are
3488 simple, it wins to evaluate the RHS unconditionally on machines
3489 with expensive branches. In this case, this isn't a comparison
3490 that can be merged. Avoid doing this if the RHS is a floating-point
3491 comparison since those can trap. */
3499 /* See if the comparisons can be merged. Then get all the parameters for
3500 each side. */
3524 /* It must be true that the inner operation on the lhs of each
3525 comparison must be the same if we are to be able to do anything.
3526 Then see if we have constants. If not, the same must be true for
3527 the rhs's. */
3538 else
3541 /* If either comparison code is not correct for our logical operation,
3542 fail. However, we can convert a one-bit comparison against zero into
3543 the opposite comparison against that bit being set in the field. */
3547 {
3549 {
3550 /* Make the left operand unsigned, since we are only interested
3551 in the value of one bit. Otherwise we are doing the wrong
3552 thing below. */
3555 }
3556 else
3558 }
3560 /* This is analogous to the code for l_const above. */
3562 {
3564 {
3567 }
3568 else
3570 }
3572 /* See if we can find a mode that contains both fields being compared on
3573 the left. If we can't, fail. Otherwise, update all constants and masks
3574 to be relative to a field of that size. */
3579 volatilep);
3589 {
3592 }
3600 {
3608 {
3612 wanted_code == NE_EXPR
3614 }
3615 }
3617 {
3625 {
3629 wanted_code == NE_EXPR
3631 }
3632 }
3634 /* If the right sides are not constant, do the same for it. Also,
3635 disallow this optimization if a size or signedness mismatch occurs
3636 between the left and right sides. */
3638 {
3641 /* Make sure the two fields on the right
3642 correspond to the left without being swapped. */
3650 volatilep);
3660 {
3663 }
3670 /* Make a mask that corresponds to both fields being compared.
3671 Do this for both items being compared. If the operands are the
3672 same size and the bits being compared are in the same position
3673 then we can do this by masking both and comparing the masked
3674 results. */
3678 {
3690 }
3692 /* There is still another way we can do something: If both pairs of
3693 fields being compared are adjacent, we may be able to make a wider
3694 field containing them both.
3696 Note that we still must mask the lhs/rhs expressions. Furthermore,
3697 the mask must be shifted to account for the shift done by
3698 make_bit_field_ref. */
3703 {
3704 tree type;
3716 /* Convert to the smaller type before masking out unwanted bits. */
3719 {
3721 {
3725 }
3727 {
3731 }
3732 }
3741 }
3744 }
3746 /* Handle the case of comparisons with constants. If there is something in
3747 common between the masks, those bits of the constants must be the same.
3748 If not, the condition is always false. Test for this to avoid generating
3749 incorrect code below. */
3754 {
3756 {
3759 }
3760 else
3761 {
3764 }
3765 }
3767 /* Construct the expression we will return. First get the component
3768 reference we will make. Unless the mask is all ones the width of
3769 that field, perform the mask operation. Then compare with the
3770 merged constant. */
3780 }
3781 \f
3782 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3783 constant. */
3785 static tree
3787 tree t;
3788 {
3793 tree minmax_const;
3795 tree inner;
3805 /* If something does not permit us to optimize, return the original tree. */
3813 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3814 and GT_EXPR, doing the rest with recursive calls using logical
3815 simplifications. */
3817 {
3819 return
3823 return
3825 optimize_minmax_comparison
3827 optimize_minmax_comparison
3832 /* MAX (X, 0) == 0 -> X <= 0 */
3836 /* MAX (X, 0) == 5 -> X == 5 */
3840 /* MAX (X, 0) == -1 -> false */
3844 /* MIN (X, 0) == 0 -> X >= 0 */
3848 /* MIN (X, 0) == 5 -> false */
3851 else
3852 /* MIN (X, 0) == -1 -> X == -1 */
3857 /* MAX (X, 0) > 0 -> X > 0
3858 MAX (X, 0) > 5 -> X > 5 */
3862 /* MAX (X, 0) > -1 -> true */
3866 /* MIN (X, 0) > 0 -> false
3867 MIN (X, 0) > 5 -> false */
3870 else
3871 /* MIN (X, 0) > -1 -> X > -1 */
3876 }
3877 }
3878 \f
3879 /* T is an integer expression that is being multiplied, divided, or taken a
3880 modulus (CODE says which and what kind of divide or modulus) by a
3881 constant C. See if we can eliminate that operation by folding it with
3882 other operations already in T. WIDE_TYPE, if non-null, is a type that
3883 should be used for the computation if wider than our type.
3885 For example, if we are dividing (X * 8) + (Y + 16) by 4, we can return
3886 (X * 2) + (Y + 4). We must, however, be assured that either the original
3887 expression would not overflow or that overflow is undefined for the type
3888 in the language in question.
3890 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
3891 the machine has a multiply-accumulate insn or that this is part of an
3892 addressing calculation.
3894 If we return a non-null expression, it is an equivalent form of the
3895 original computation, but need not be in the original type. */
3897 static tree
3899 tree t;
3900 tree c;
3902 tree wide_type;
3903 {
3913 /* Don't deal with constants of zero here; they confuse the code below. */
3923 /* Note that we need not handle conditional operations here since fold
3924 already handles those cases. So just do arithmetic here. */
3926 {
3928 /* For a constant, we can always simplify if we are a multiply
3929 or (for divide and modulus) if it is a multiple of our constant. */
3936 /* If op0 is an expression, and is unsigned, and the type is
3937 smaller than ctype, then we cannot widen the expression. */
3949 /* Pass the constant down and see if we can make a simplification. If
3950 we can, replace this expression with the inner simplification for
3951 possible later conversion to our or some other type. */
3963 /* If widening the type changes the signedness, then we can't perform
3964 this optimization as that changes the result. */
3968 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
3971 {
3977 }
3987 /* If this has not been evaluated and the operand has no side effects,
3988 we can see if we can do something inside it and make a new one.
3989 Note that this test is overly conservative since we can do this
3990 if the only reason it had side effects is that it was another
3991 similar SAVE_EXPR, but that isn't worth bothering with. */
3994 wide_type)))
3995 {
4002 }
4006 /* If the second operand is constant, this is a multiplication
4007 or floor division, by a power of two, so we can treat it that
4008 way unless the multiplier or divisor overflows. */
4010 /* const_binop may not detect overflow correctly,
4011 so check for it explicitly here. */
4025 /* See if we can eliminate the operation on both sides. If we can, we
4026 can return a new PLUS or MINUS. If we can't, the only remaining
4027 cases where we can do anything are if the second operand is a
4028 constant. */
4033 /* If not multiplication, we can only do this if either operand
4034 is divisible by c. */
4040 /* If this was a subtraction, negate OP1 and set it to be an addition.
4041 This simplifies the logic below. */
4048 /* If either OP1 or C are negative, this optimization is not safe for
4049 some of the division and remainder types while for others we need
4050 to change the code. */
4052 {
4060 }
4062 /* If it's a multiply or a division/modulus operation of a multiple
4063 of our constant, do the operation and verify it doesn't overflow. */
4066 {
4070 }
4071 else
4074 /* If we have an unsigned type is not a sizetype, we cannot widen
4075 the operation since it will change the result if the original
4076 computation overflowed. */
4082 /* If we were able to eliminate our operation from the first side,
4083 apply our operation to the second side and reform the PLUS. */
4087 /* The last case is if we are a multiply. In that case, we can
4088 apply the distributive law to commute the multiply and addition
4089 if the multiplication of the constants doesn't overflow. */
4094 op1));
4099 /* We have a special case here if we are doing something like
4100 (C * 8) % 4 since we know that's zero. */
4107 /* ... fall through ... */
4111 /* If we can extract our operation from the LHS, do so and return a
4112 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4113 do something only if the second operand is a constant. */
4125 /* If these are the same operation types, we can associate them
4126 assuming no overflow. */
4133 /* If these operations "cancel" each other, we have the main
4134 optimizations of this pass, which occur when either constant is a
4135 multiple of the other, in which case we replace this with either an
4136 operation or CODE or TCODE.
4138 If we have an unsigned type that is not a sizetype, we cannot do
4139 this since it will change the result if the original computation
4140 overflowed. */
4147 {
4158 }
4163 }
4166 }
4167 \f
4168 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4169 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4170 that we may sometimes modify the tree. */
4172 static tree
4174 tree t;
4175 tree s;
4176 {
4179 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4184 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4185 don't bother handling any other types. */
4187 {
4191 }
4196 {
4199 }
4202 }
4203 \f
4204 /* Return a node which has the indicated constant VALUE (either 0 or
4205 1), and is of the indicated TYPE. */
4207 static tree
4210 tree type;
4211 {
4216 integer_zero_node);
4217 else
4218 {
4223 }
4224 }
4226 /* Utility function for the following routine, to see how complex a nesting of
4227 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4228 we don't care (to avoid spending too much time on complex expressions.). */
4230 static int
4232 tree expr;
4234 {
4245 }
4247 /* Transform `a + (b ? x : y)' into `x ? (a + b) : (a + y)'.
4248 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4249 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4250 expression, and ARG to `a'. If COND_FIRST_P is non-zero, then the
4251 COND is the first argument to CODE; otherwise (as in the example
4252 given here), it is the second argument. TYPE is the type of the
4253 original expression. */
4255 static tree
4258 tree type;
4259 tree cond;
4260 tree arg;
4262 {
4266 /* In the end, we'll produce a COND_EXPR. Both arms of the
4267 conditional expression will be binary operations. The left-hand
4268 side of the expression to be executed if the condition is true
4269 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4270 of the expression to be executed if the condition is true will be
4271 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4272 but apply to the expression to be executed if the conditional is
4273 false. */
4278 /* These are the codes to use for the left-hand side and right-hand
4279 side of the COND_EXPR. Normally, they are the same as CODE. */
4282 /* And these are the types of the expressions. */
4287 {
4291 }
4292 else
4293 {
4297 }
4300 {
4304 /* If this operand throws an expression, then it does not make
4305 sense to try to perform a logical or arithmetic operation
4306 involving it. Instead of building `a + throw 3' for example,
4307 we simply build `a, throw 3'. */
4309 {
4313 }
4315 {
4319 }
4320 }
4321 else
4322 {
4327 }
4329 /* If ARG is complex we want to make sure we only evaluate
4330 it once. Though this is only required if it is volatile, it
4331 might be more efficient even if it is not. However, if we
4332 succeed in folding one part to a constant, we do not need
4333 to make this SAVE_EXPR. Since we do this optimization
4334 primarily to see if we do end up with constant and this
4335 SAVE_EXPR interferes with later optimizations, suppressing
4336 it when we can is important.
4338 If we are not in a function, we can't make a SAVE_EXPR, so don't
4339 try to do so. Don't try to see if the result is a constant
4340 if an arm is a COND_EXPR since we get exponential behavior
4341 in that case. */
4348 {
4358 }
4371 else
4373 }
4375 \f
4376 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4378 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4379 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4380 ADDEND is the same as X.
4382 X + 0 and X - 0 both give X when X is NaN, infinite, or non-zero
4383 and finite. The problematic cases are when X is zero, and its mode
4384 has signed zeros. In the case of rounding towards -infinity,
4385 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4386 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4388 static bool
4392 {
4396 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4400 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4405 /* The mode has signed zeros, and we have to honor their sign.
4406 In this situation, there is only one case we can return true for.
4407 X - 0 is the same as X unless rounding towards -infinity is
4408 supported. */
4410 }
4413 /* Perform constant folding and related simplification of EXPR.
4414 The related simplifications include x*1 => x, x*0 => 0, etc.,
4415 and application of the associative law.
4416 NOP_EXPR conversions may be removed freely (as long as we
4417 are careful not to change the C type of the overall expression)
4418 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4419 but we can constant-fold them if they have constant operands. */
4421 tree
4423 tree expr;
4424 {
4427 tree tem;
4433 /* WINS will be nonzero when the switch is done
4434 if all operands are constant. */
4437 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4438 Likewise for a SAVE_EXPR that's already been evaluated. */
4442 /* Return right away if a constant. */
4446 #ifdef MAX_INTEGER_COMPUTATION_MODE
4448 #endif
4451 {
4452 tree subop;
4454 /* Special case for conversion ops that can have fixed point args. */
4457 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4463 else
4468 )
4469 /* Note that TREE_CONSTANT isn't enough:
4470 static var addresses are constant but we can't
4471 do arithmetic on them. */
4473 }
4475 {
4479 {
4481 tree subop;
4487 {
4488 /* Signedness matters here. Perhaps we can refine this
4489 later. */
4491 }
4492 else
4493 /* Strip any conversions that don't change the mode. */
4498 else
4503 /* Note that TREE_CONSTANT isn't enough:
4504 static var addresses are constant but we can't
4505 do arithmetic on them. */
4512 }
4513 }
4515 /* If this is a commutative operation, and ARG0 is a constant, move it
4516 to ARG1 to reduce the number of tests below. */
4521 {
4526 }
4528 /* Now WINS is set as described above,
4529 ARG0 is the first operand of EXPR,
4530 and ARG1 is the second operand (if it has more than one operand).
4532 First check for cases where an arithmetic operation is applied to a
4533 compound, conditional, or comparison operation. Push the arithmetic
4534 operation inside the compound or conditional to see if any folding
4535 can then be done. Convert comparison to conditional for this purpose.
4536 The also optimizes non-constant cases that used to be done in
4537 expand_expr.
4539 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4540 one of the operands is a comparison and the other is a comparison, a
4541 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4542 code below would make the expression more complex. Change it to a
4543 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4544 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4556 {
4566 }
4569 {
4574 {
4579 /* If this was a conversion, and all we did was to move into
4580 inside the COND_EXPR, bring it back out. But leave it if
4581 it is a conversion from integer to integer and the
4582 result precision is no wider than a word since such a
4583 conversion is cheap and may be optimized away by combine,
4584 while it couldn't if it were outside the COND_EXPR. Then return
4585 so we don't get into an infinite recursion loop taking the
4586 conversion out and then back in. */
4596 && (INTEGRAL_TYPE_P
4607 }
4612 }
4615 {
4628 return
4642 return
4645 }
4656 {
4672 /* Other kinds of FIX are not handled properly by fold_convert. */
4677 /* Handle cases of two conversions in a row. */
4680 {
4700 /* In addition to the cases of two conversions in a row
4701 handled below, if we are converting something to its own
4702 type via an object of identical or wider precision, neither
4703 conversion is needed. */
4709 /* Likewise, if the intermediate and final types are either both
4710 float or both integer, we don't need the middle conversion if
4711 it is wider than the final type and doesn't change the signedness
4712 (for integers). Avoid this if the final type is a pointer
4713 since then we sometimes need the inner conversion. Likewise if
4714 the outer has a precision not equal to the size of its mode. */
4724 /* If we have a sign-extension of a zero-extended value, we can
4725 replace that by a single zero-extension. */
4731 /* Two conversions in a row are not needed unless:
4732 - some conversion is floating-point (overstrict for now), or
4733 - the intermediate type is narrower than both initial and
4734 final, or
4735 - the intermediate type and innermost type differ in signedness,
4736 and the outermost type is wider than the intermediate, or
4737 - the initial type is a pointer type and the precisions of the
4738 intermediate and final types differ, or
4739 - the final type is a pointer type and the precisions of the
4740 initial and intermediate types differ. */
4754 }
4758 /* Detect assigning a bitfield. */
4761 {
4762 /* Don't leave an assignment inside a conversion
4763 unless assigning a bitfield. */
4766 /* First do the assignment, then return converted constant. */
4770 }
4772 {
4775 }
4786 {
4789 /* Fold an expression like: "foo"[2] */
4793 {
4797 }
4798 }
4804 {
4808 }
4817 {
4819 {
4821 HOST_WIDE_INT high;
4832 }
4835 }
4839 /* Convert - (a - b) to (b - a) for non-floating-point. */
4849 {
4851 {
4852 /* If the value is unsigned, then the absolute value is
4853 the same as the ordinary value. */
4856 /* Similarly, if the value is non-negative. */
4859 /* If the value is negative, then the absolute value is
4860 its negation. */
4861 else
4862 {
4864 HOST_WIDE_INT high;
4875 }
4876 }
4878 {
4882 }
4883 }
4910 {
4917 }
4923 /* A + (-B) -> A - B */
4926 /* (-A) + B -> B - A */
4930 {
4934 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
4935 with a constant, and the two constants have no bits in common,
4936 we should treat this as a BIT_IOR_EXPR since this may produce more
4937 simplifications. */
4945 {
4948 }
4950 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
4951 (plus (plus (mult) (mult)) (foo)) so that we can
4952 take advantage of the factoring cases below. */
4957 {
4962 else
4973 parg1));
4978 parg0));
4979 }
4982 {
4986 /* (A * C) + (B * C) -> (A+B) * C.
4987 We are most concerned about the case where C is a constant,
4988 but other combinations show up during loop reduction. Since
4989 it is not difficult, try all four possibilities. */
5006 /* No identical multiplicands; see if we can find a common
5007 power-of-two factor in non-power-of-two multiplies. This
5008 can help in multi-dimensional array access. */
5013 {
5018 /* Move min of absolute values to int11. */
5021 {
5025 }
5028 {
5033 }
5034 }
5039 same));
5040 }
5041 }
5043 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5047 /* Likewise if the operands are reversed. */
5051 bit_rotate:
5052 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5053 is a rotate of A by C1 bits. */
5054 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5055 is a rotate of A by B bits. */
5056 {
5065 {
5084 {
5092 TYPE_PRECISION
5097 ? LROTATE_EXPR
5100 }
5102 {
5110 TYPE_PRECISION
5115 ? LROTATE_EXPR
5118 }
5119 }
5120 }
5122 associate:
5123 /* In most languages, can't associate operations on floats through
5124 parentheses. Rather than remember where the parentheses were, we
5125 don't associate floats at all. It shouldn't matter much. However,
5126 associating multiplications is only very slightly inaccurate, so do
5127 that if -funsafe-math-optimizations is specified. */
5132 {
5135 /* Split both trees into variables, constants, and literals. Then
5136 associate each group together, the constants with literals,
5137 then the result with variables. This increases the chances of
5138 literals being recombined later and of generating relocatable
5139 expressions for the sum of a constant and literal. */
5143 /* Only do something if we found more than two objects. Otherwise,
5144 nothing has changed and we risk infinite recursion. */
5147 {
5153 }
5154 }
5156 binary:
5160 {
5161 /* The return value should always have
5162 the same type as the original expression. */
5167 }
5171 /* A - (-B) -> A + B */
5174 /* (-A) - CST -> (-CST) - A for floating point (what about ints ?) */
5176 return
5183 {
5189 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5190 about the case where C is a constant, just try one of the
5191 four possibilities. */
5201 }
5203 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5207 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5208 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5209 (-ARG1 + ARG0) reduces to -ARG1. */
5213 /* Fold &x - &x. This can happen from &x.foo - &x.
5214 This is unsafe for certain floats even in non-IEEE formats.
5215 In IEEE, it is unsafe because it does wrong for NaNs.
5216 Also note that operand_equal_p is always false if an operand
5217 is volatile. */
5226 /* (-A) * (-B) -> A * B */
5232 {
5238 /* (a * (1 << b)) is (a << b) */
5253 }
5254 else
5255 {
5256 /* Maybe fold x * 0 to 0. The expressions aren't the same
5257 when x is NaN, since x * 0 is also NaN. Nor are they the
5258 same in modes with signed zeros, since multiplying a
5259 negative value by 0 gives -0, not +0. */
5264 /* In IEEE floating point, x*1 is not equivalent to x for snans.
5265 However, ANSI says we can drop signals,
5266 so we can do this anyway. */
5269 /* x*2 is x+x */
5272 {
5275 }
5276 }
5280 bit_ior:
5289 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5291 This results in more efficient code for machines without a NAND
5292 instruction. Combine will canonicalize to the first form
5293 which will allow use of NAND instructions provided by the
5294 backend if they exist. */
5297 {
5302 }
5304 /* See if this can be simplified into a rotate first. If that
5305 is unsuccessful continue in the association code. */
5314 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5315 with a constant, and the two constants have no bits in common,
5316 we should treat this as a BIT_IOR_EXPR since this may produce more
5317 simplifications. */
5325 {
5328 }
5330 /* See if this can be simplified into a rotate first. If that
5331 is unsuccessful continue in the association code. */
5335 bit_and:
5343 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5346 {
5347 unsigned int prec
5354 }
5357 {
5358 unsigned int prec
5365 }
5367 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5369 This results in more efficient code for machines without a NOR
5370 instruction. Combine will canonicalize to the first form
5371 which will allow use of NOR instructions provided by the
5372 backend if they exist. */
5375 {
5380 }
5390 {
5394 }
5398 /* In most cases, do nothing with a divide by zero. */
5399 #ifndef REAL_INFINITY
5402 #endif
5404 /* (-A) / (-B) -> A / B */
5409 /* In IEEE floating point, x/1 is not equivalent to x for snans.
5410 However, ANSI says we can drop signals, so we can do this anyway. */
5414 /* If ARG1 is a constant, we can convert this to a multiply by the
5415 reciprocal. This does not have the same rounding properties,
5416 so only do this if -funsafe-math-optimizations. We can actually
5417 always safely do it if ARG1 is a power of two, but it's hard to
5418 tell if it is or not in a portable manner. */
5420 {
5425 /* Find the reciprocal if optimizing and the result is exact. */
5427 {
5428 REAL_VALUE_TYPE r;
5431 {
5434 }
5435 }
5436 }
5437 /* Convert A/B/C to A/(B*C). */
5440 {
5443 arg1)));
5444 }
5445 /* Convert A/(B/C) to (A/B)*C. */
5448 {
5453 }
5466 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5467 operation, EXACT_DIV_EXPR.
5469 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5470 At one time others generated faster code, it's not clear if they do
5471 after the last round to changes to the DIV code in expmed.c. */
5505 /* Since negative shift count is not well-defined,
5506 don't try to compute it in the compiler. */
5509 /* Rewrite an LROTATE_EXPR by a constant into an
5510 RROTATE_EXPR by a new constant. */
5512 {
5516 = const_binop
5523 }
5525 /* If we have a rotate of a bit operation with the rotate count and
5526 the second operand of the bit operation both constant,
5527 permute the two operations. */
5540 /* Two consecutive rotates adding up to the width of the mode can
5541 be ignored. */
5572 /* Note that the operand of this must be an int
5573 and its values must be 0 or 1.
5574 ("true" is a fixed value perhaps depending on the language,
5575 but we don't handle values other than 1 correctly yet.) */
5577 /* Avoid infinite recursion. */
5583 /* Note that the operands of this must be ints
5584 and their values must be 0 or 1.
5585 ("true" is a fixed value perhaps depending on the language.) */
5586 /* If first arg is constant zero, return it. */
5590 /* If either arg is constant true, drop it. */
5594 /* Preserve sequence points. */
5597 /* If second arg is constant zero, result is zero, but first arg
5598 must be evaluated. */
5601 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
5602 case will be handled here. */
5606 truth_andor:
5607 /* We only do these simplifications if we are optimizing. */
5611 /* Check for things like (A || B) && (A || C). We can convert this
5612 to A || (B && C). Note that either operator can be any of the four
5613 truth and/or operations and the transformation will still be
5614 valid. Also note that we only care about order for the
5615 ANDIF and ORIF operators. If B contains side effects, this
5616 might change the truth-value of A. */
5623 {
5643 /* This case if tricky because we must either have commutative
5644 operators or else A10 must not have side-effects. */
5650 a01));
5651 }
5653 /* See if we can build a range comparison. */
5657 /* Check for the possibility of merging component references. If our
5658 lhs is another similar operation, try to merge its rhs with our
5659 rhs. Then try to merge our lhs and rhs. */
5671 /* Note that the operands of this must be ints
5672 and their values must be 0 or true.
5673 ("true" is a fixed value perhaps depending on the language.) */
5674 /* If first arg is constant true, return it. */
5678 /* If either arg is constant zero, drop it. */
5682 /* Preserve sequence points. */
5685 /* If second arg is constant true, result is true, but we must
5686 evaluate first arg. */
5689 /* Likewise for first arg, but note this only occurs here for
5690 TRUTH_OR_EXPR. */
5696 /* If either arg is constant zero, drop it. */
5701 /* If either arg is constant true, this is a logical inversion. */
5715 {
5716 /* (-a) CMP (-b) -> b CMP a */
5721 /* (-a) CMP CST -> a swap(CMP) (-CST) */
5723 return
5729 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
5730 /* a CMP (-0) -> a CMP 0 */
5735 }
5737 /* If one arg is a constant integer, put it last. */
5740 {
5747 }
5749 /* Convert foo++ == CONST into ++foo == CONST + INCR.
5750 First, see if one arg is constant; find the constant arg
5751 and the other one. */
5752 {
5762 {
5763 /* This optimization is invalid for ordered comparisons
5764 if CONST+INCR overflows or if foo+incr might overflow.
5765 This optimization is invalid for floating point due to rounding.
5766 For pointer types we assume overflow doesn't happen. */
5770 {
5771 tree newconst
5775 /* Do not overwrite the current varop to be a preincrement,
5776 create a new node so that we won't confuse our caller who
5777 might create trees and throw them away, reusing the
5778 arguments that they passed to build. This shows up in
5779 the THEN or ELSE parts of ?: being postincrements. */
5784 /* If VAROP is a reference to a bitfield, we must mask
5785 the constant by the width of the field. */
5789 {
5790 int size
5792 (TREE_OPERAND
5796 tree folded_compare;
5798 /* First check whether the comparison would come out
5799 always the same. If we don't do that we would
5800 change the meaning with the masking. */
5804 else
5807 constop));
5822 mask)));
5823 }
5829 }
5830 }
5832 {
5836 {
5837 tree newconst
5841 /* Do not overwrite the current varop to be a predecrement,
5842 create a new node so that we won't confuse our caller who
5843 might create trees and throw them away, reusing the
5844 arguments that they passed to build. This shows up in
5845 the THEN or ELSE parts of ?: being postdecrements. */
5853 {
5854 int size
5856 (TREE_OPERAND
5860 tree folded_compare;
5865 else
5868 constop));
5883 mask)));
5884 }
5890 }
5891 }
5892 }
5894 /* Change X >= CST to X > (CST - 1) if CST is positive. */
5898 {
5900 {
5915 }
5916 }
5918 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
5919 a MINUS_EXPR of a constant, we can convert it into a comparison with
5920 a revised constant as long as no overflow occurs. */
5932 /* Similarly for a NEGATE_EXPR. */
5941 /* If we have X - Y == 0, we can convert that to X == Y and similarly
5942 for !=. Don't do this for ordered comparisons due to overflow. */
5948 /* If we are widening one operand of an integer comparison,
5949 see if the other operand is similarly being widened. Perhaps we
5950 can do the comparison in the narrower type. */
5960 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
5961 constant, we can simplify it. */
5968 /* If we are comparing an ABS_EXPR with a constant, we can
5969 convert all the cases into explicit comparisons, but they may
5970 well not be faster than doing the ABS and one comparison.
5971 But ABS (X) <= C is a range comparison, which becomes a subtraction
5972 and a comparison, and is probably faster. */
5984 /* If this is an EQ or NE comparison with zero and ARG0 is
5985 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
5986 two operations, but the latter can be done in one less insn
5987 on machines that have only two-operand insns or on which a
5988 constant cannot be the first operand. */
5991 {
5994 return
6002 integer_one_node)),
6003 arg1));
6006 return
6014 integer_one_node)),
6015 arg1));
6016 }
6018 /* If this is an NE or EQ comparison of zero against the result of a
6019 signed MOD operation whose second operand is a power of 2, make
6020 the MOD operation unsigned since it is simpler and equivalent. */
6029 {
6036 }
6038 /* If this is an NE comparison of zero with an AND of one, remove the
6039 comparison since the AND will give the correct value. */
6045 /* If we have (A & C) == C where C is a power of 2, convert this into
6046 (A & C) != 0. Similarly for NE_EXPR. */
6054 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6055 and similarly for >= into !=. */
6071 return
6078 /* Simplify comparison of something with itself. (For IEEE
6079 floating-point, we can only do some of these simplifications.) */
6081 {
6083 {
6094 /* For NE, we can only do this simplification if integer. */
6097 /* ... fall through ... */
6103 }
6104 }
6106 /* An unsigned comparison against 0 can be simplified. */
6111 {
6113 {
6125 arg0);
6129 arg0);
6132 }
6133 }
6135 /* Comparisons with the highest or lowest possible integer of
6136 the specified size will have known values and an unsigned
6137 <= 0x7fffffff can be simplified. */
6138 {
6146 {
6152 {
6156 arg0);
6164 arg0);
6171 }
6178 {
6182 arg0);
6190 arg0);
6197 }
6203 /* signed_type does not work on pointer types. */
6206 {
6210 arg0),
6212 integer_zero_node)));
6216 arg0),
6218 integer_zero_node)));
6222 }
6229 {
6233 arg0);
6241 arg0);
6248 }
6249 }
6250 }
6252 /* If we are comparing an expression that just has comparisons
6253 of two integer values, arithmetic expressions of those comparisons,
6254 and constants, we can simplify it. There are only three cases
6255 to check: the two values can either be equal, the first can be
6256 greater, or the second can be greater. Fold the expression for
6257 those three values. Since each value must be 0 or 1, we have
6258 eight possibilities, each of which corresponds to the constant 0
6259 or 1 or one of the six possible comparisons.
6261 This handles common cases like (a > b) == 0 but also handles
6262 expressions like ((x > y) - (y > x)) > 0, which supposedly
6263 occur in macroized code. */
6266 {
6271 /* Don't handle degenerate cases here; they should already
6272 have been handled anyway. */
6281 {
6285 /* We can't just pass T to eval_subst in case cval1 or cval2
6286 was the same as ARG1. */
6288 tree high_result
6291 arg1));
6292 tree equal_result
6295 arg1));
6296 tree low_result
6299 arg1));
6301 /* All three of these results should be 0 or 1. Confirm they
6302 are. Then use those values to select the proper code
6303 to use. */
6311 {
6312 /* Make a 3-bit mask with the high-order bit being the
6313 value for `>', the next for '=', and the low for '<'. */
6317 {
6319 /* Always false. */
6340 /* Always true. */
6342 }
6347 else
6349 }
6350 }
6351 }
6353 /* If this is a comparison of a field, we may be able to simplify it. */
6357 /* Handle the constant case even without -O
6358 to make sure the warnings are given. */
6360 {
6363 }
6365 /* If this is a comparison of complex values and either or both sides
6366 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6367 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6368 This may prevent needless evaluations. */
6375 {
6388 type,
6391 }
6393 /* Optimize comparisons of strlen vs zero to a compare of the
6394 first character of the string vs zero. To wit,
6395 strlen(ptr) == 0 => *ptr == 0
6396 strlen(ptr) != 0 => *ptr != 0
6397 Other cases should reduce to one of these two (or a constant)
6398 due to the return value of strlen being unsigned. */
6403 {
6405 tree arglist;
6417 integer_zero_node));
6418 }
6420 /* From here on, the only cases we handle are when the result is
6421 known to be a constant.
6423 To compute GT, swap the arguments and do LT.
6424 To compute GE, do LT and invert the result.
6425 To compute LE, swap the arguments, do LT and invert the result.
6426 To compute NE, do EQ and invert the result.
6428 Therefore, the code below must handle only EQ and LT. */
6431 {
6434 }
6436 /* Note that it is safe to invert for real values here because we
6437 will check below in the one case that it matters. */
6442 {
6445 }
6447 /* Compute a result for LT or EQ if args permit;
6448 otherwise return T. */
6450 {
6453 else
6458 }
6461 /* Assume a nonexplicit constant cannot equal an explicit one,
6462 since such code would be undefined anyway.
6463 Exception: on sysvr4, using #pragma weak,
6464 a label can come out as 0. */
6471 #endif
6472 /* Two real constants can be compared explicitly. */
6474 {
6475 /* If either operand is a NaN, the result is false with two
6476 exceptions: First, an NE_EXPR is true on NaNs, but that case
6477 is already handled correctly since we will be inverting the
6478 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6479 or a GE_EXPR into a LT_EXPR, we must return true so that it
6480 will be inverted into false. */
6490 else
6494 }
6508 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6509 so all simple results must be passed through pedantic_non_lvalue. */
6511 return pedantic_non_lvalue
6516 /* If the second operand is zero, invert the comparison and swap
6517 the second and third operands. Likewise if the second operand
6518 is constant and the third is not or if the third operand is
6519 equivalent to the first operand of the comparison. */
6527 {
6528 /* See if this can be inverted. If it can't, possibly because
6529 it was a floating-point inequality comparison, don't do
6530 anything. */
6534 {
6538 /* arg1 should be the first argument of the new T. */
6541 }
6542 }
6544 /* If we have A op B ? A : C, we may be able to convert this to a
6545 simpler expression, depending on the operation and the values
6546 of B and C. Signed zeros prevent all of these transformations,
6547 for reasons given above each one. */
6553 {
6559 /* If we have A op 0 ? A : -A, consider applying the following
6560 transformations:
6562 A == 0? A : -A same as -A
6563 A != 0? A : -A same as A
6564 A >= 0? A : -A same as abs (A)
6565 A > 0? A : -A same as abs (A)
6566 A <= 0? A : -A same as -abs (A)
6567 A < 0? A : -A same as -abs (A)
6569 None of these transformations work for modes with signed
6570 zeros. If A is +/-0, the first two transformations will
6571 change the sign of the result (from +0 to -0, or vice
6572 versa). The last four will fix the sign of the result,
6573 even though the original expressions could be positive or
6574 negative, depending on the sign of A.
6576 Note that all these transformations are correct if A is
6577 NaN, since the two alternatives (A and -A) are also NaNs. */
6584 {
6586 return
6587 pedantic_non_lvalue
6589 negate_expr
6591 arg1))));
6598 return pedantic_non_lvalue
6605 return pedantic_non_lvalue
6609 arg1)))));
6612 }
6614 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
6615 A == 0 ? A : 0 is always 0 unless A is -0. Note that
6616 both transformations are correct when A is NaN: A != 0
6617 is then true, and A == 0 is false. */
6620 {
6625 }
6627 /* Try some transformations of A op B ? A : B.
6629 A == B? A : B same as B
6630 A != B? A : B same as A
6631 A >= B? A : B same as max (A, B)
6632 A > B? A : B same as max (B, A)
6633 A <= B? A : B same as min (A, B)
6634 A < B? A : B same as min (B, A)
6636 As above, these transformations don't work in the presence
6637 of signed zeros. For example, if A and B are zeros of
6638 opposite sign, the first two transformations will change
6639 the sign of the result. In the last four, the original
6640 expressions give different results for (A=+0, B=-0) and
6641 (A=-0, B=+0), but the transformed expressions do not.
6643 The first two transformations are correct if either A or B
6644 is a NaN. In the first transformation, the condition will
6645 be false, and B will indeed be chosen. In the case of the
6646 second transformation, the condition A != B will be true,
6647 and A will be chosen.
6649 The conversions to max() and min() are not correct if B is
6650 a number and A is not. The conditions in the original
6651 expressions will be false, so all four give B. The min()
6652 and max() versions would give a NaN instead. */
6655 {
6660 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
6665 {
6672 /* In C++ a ?: expression can be an lvalue, so put the
6673 operand which will be used if they are equal first
6674 so that we can convert this back to the
6675 corresponding COND_EXPR. */
6677 return pedantic_non_lvalue
6687 return pedantic_non_lvalue
6696 }
6697 }
6699 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
6700 we might still be able to simplify this. For example,
6701 if C1 is one less or one more than C2, this might have started
6702 out as a MIN or MAX and been transformed by this function.
6703 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
6709 {
6711 /* We can replace A with C1 in this case. */
6718 /* If C1 is C2 + 1, this is min(A, C2). */
6723 return pedantic_non_lvalue
6728 /* If C1 is C2 - 1, this is min(A, C2). */
6733 return pedantic_non_lvalue
6738 /* If C1 is C2 - 1, this is max(A, C2). */
6743 return pedantic_non_lvalue
6748 /* If C1 is C2 + 1, this is max(A, C2). */
6753 return pedantic_non_lvalue
6760 }
6761 }
6763 /* If the second operand is simpler than the third, swap them
6764 since that produces better jump optimization results. */
6770 {
6771 /* See if this can be inverted. If it can't, possibly because
6772 it was a floating-point inequality comparison, don't do
6773 anything. */
6777 {
6781 /* arg1 should be the first argument of the new T. */
6784 }
6785 }
6787 /* Convert A ? 1 : 0 to simply A. */
6790 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
6791 call to fold will try to move the conversion inside
6792 a COND, which will recurse. In that case, the COND_EXPR
6793 is probably the best choice, so leave it alone. */
6797 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
6798 operation is simply A & 2. */
6812 /* When pedantic, a compound expression can be neither an lvalue
6813 nor an integer constant expression. */
6816 /* Don't let (0, 0) be null pointer constant. */
6858 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
6859 appropriate. */
6864 {
6868 tree arg01;
6879 {
6893 arg01));
6894 }
6897 }
6900 /* Check for a built-in function. */
6905 {
6909 }
6915 }
6917 /* Determine if first argument is a multiple of second argument. Return 0 if
6918 it is not, or we cannot easily determined it to be.
6920 An example of the sort of thing we care about (at this point; this routine
6921 could surely be made more general, and expanded to do what the *_DIV_EXPR's
6922 fold cases do now) is discovering that
6924 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
6926 is a multiple of
6928 SAVE_EXPR (J * 8)
6930 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
6932 This code also handles discovering that
6934 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
6936 is a multiple of 8 so we don't have to worry about dealing with a
6937 possible remainder.
6939 Note that we *look* inside a SAVE_EXPR only to determine how it was
6940 calculated; it is not safe for fold to do much of anything else with the
6941 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
6942 at run time. For example, the latter example above *cannot* be implemented
6943 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
6944 evaluation time of the original SAVE_EXPR is not necessarily the same at
6945 the time the new expression is evaluated. The only optimization of this
6946 sort that would be valid is changing
6948 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
6950 divided by 8 to
6952 SAVE_EXPR (I) * SAVE_EXPR (J)
6954 (where the same SAVE_EXPR (J) is used in the original and the
6955 transformed version). */
6957 static int
6959 tree type;
6960 tree top;
6961 tree bottom;
6962 {
6970 {
6982 {
6986 /* const_binop may not detect overflow correctly,
6987 so check for it explicitly here. */
6996 }
7000 /* Can't handle conversions from non-integral or wider integral type. */
7006 /* .. fall through ... */
7022 }
7023 }
7025 /* Return true if `t' is known to be non-negative. */
7027 int
7029 tree t;
7030 {
7032 {
7073 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7075 else
7076 /* We don't know sign of `t', so be conservative and return false. */
7078 }
7079 }
7081 /* Return true if `r' is known to be non-negative.
7082 Only handles constants at the moment. */
7084 int
7086 rtx r;
7087 {
7089 {
7099 {
7101 rtx elt;
7106 {
7110 }
7113 }
7117 /* These are always nonnegative. */
7122 }
7123 }