| blob | 084308ec9fa502077144884eca117c86480fc3ea |
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. */
59 HOST_WIDE_INT));
62 HOST_WIDE_INT *));
85 HOST_WIDE_INT *,
105 static tree fold_binary_op_with_conditional_arg
109 #if defined(HOST_EBCDIC)
110 /* bit 8 is significant in EBCDIC */
111 #define CHARMASK 0xff
112 #else
113 #define CHARMASK 0x7f
114 #endif
116 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
117 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
118 and SUM1. Then this yields nonzero if overflow occurred during the
119 addition.
121 Overflow occurs if A and B have the same sign, but A and SUM differ in
122 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
123 sign. */
124 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
125 \f
126 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
127 We do that by representing the two-word integer in 4 words, with only
128 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
129 number. The value of the word is LOWPART + HIGHPART * BASE. */
131 #define LOWPART(x) \
132 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
133 #define HIGHPART(x) \
134 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
135 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
137 /* Unpack a two-word integer into 4 words.
138 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
139 WORDS points to the array of HOST_WIDE_INTs. */
141 static void
145 HOST_WIDE_INT hi;
146 {
151 }
153 /* Pack an array of 4 words into a two-word integer.
154 WORDS points to the array of words.
155 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
157 static void
162 {
165 }
166 \f
167 /* Make the integer constant T valid for its type by setting to 0 or 1 all
168 the bits in the constant that don't belong in the type.
170 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
171 nonzero, a signed overflow has already occurred in calculating T, so
172 propagate it.
174 Make the real constant T valid for its type by calling CHECK_FLOAT_VALUE,
175 if it exists. */
177 int
179 tree t;
181 {
183 HOST_WIDE_INT high;
187 {
188 #ifdef CHECK_FLOAT_VALUE
190 overflow);
191 #endif
193 }
203 else
206 /* First clear all bits that are beyond the type's precision. */
209 ;
213 else
214 {
218 }
220 /* Unsigned types do not suffer sign extension or overflow unless they
221 are a sizetype. */
227 /* If the value's sign bit is set, extend the sign. */
235 {
236 /* Value is negative:
237 set to 1 all the bits that are outside this type's precision. */
241 else
242 {
246 }
247 }
249 /* Return nonzero if signed overflow occurred. */
250 return
253 }
254 \f
255 /* Add two doubleword integers with doubleword result.
256 Each argument is given as two `HOST_WIDE_INT' pieces.
257 One argument is L1 and H1; the other, L2 and H2.
258 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
260 int
266 {
268 HOST_WIDE_INT h;
276 }
278 /* Negate a doubleword integer with doubleword result.
279 Return nonzero if the operation overflows, assuming it's signed.
280 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
281 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
283 int
286 HOST_WIDE_INT h1;
289 {
291 {
295 }
296 else
297 {
301 }
302 }
303 \f
304 /* Multiply two doubleword integers with doubleword result.
305 Return nonzero if the operation overflows, assuming it's signed.
306 Each argument is given as two `HOST_WIDE_INT' pieces.
307 One argument is L1 and H1; the other, L2 and H2.
308 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
310 int
316 {
331 {
334 {
336 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
338 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
342 }
344 }
348 /* Check for overflow by calculating the top half of the answer in full;
349 it should agree with the low half's sign bit. */
352 {
355 }
357 {
360 }
362 }
363 \f
364 /* Shift the doubleword integer in L1, H1 left by COUNT places
365 keeping only PREC bits of result.
366 Shift right if COUNT is negative.
367 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
368 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
370 void
378 {
382 {
385 }
387 #ifdef SHIFT_COUNT_TRUNCATED
390 #endif
393 {
394 /* Shifting by the host word size is undefined according to the
395 ANSI standard, so we must handle this as a special case. */
398 }
400 {
403 }
404 else
405 {
409 }
411 /* Sign extend all bits that are beyond the precision. */
418 ;
420 {
423 }
424 else
425 {
429 }
430 }
432 /* Shift the doubleword integer in L1, H1 right by COUNT places
433 keeping only PREC bits of result. COUNT must be positive.
434 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
435 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
437 void
445 {
448 signmask = (arith
452 #ifdef SHIFT_COUNT_TRUNCATED
455 #endif
458 {
459 /* Shifting by the host word size is undefined according to the
460 ANSI standard, so we must handle this as a special case. */
463 }
465 {
468 }
469 else
470 {
474 }
476 /* Zero / sign extend all bits that are beyond the precision. */
479 {
482 }
484 ;
486 {
489 }
490 else
491 {
495 }
496 }
497 \f
498 /* Rotate the doubleword integer in L1, H1 left by COUNT places
499 keeping only PREC bits of result.
500 Rotate right if COUNT is negative.
501 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
503 void
510 {
522 }
524 /* Rotate the doubleword integer in L1, H1 left by COUNT places
525 keeping only PREC bits of result. COUNT must be positive.
526 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
528 void
535 {
547 }
548 \f
549 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
550 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
551 CODE is a tree code for a kind of division, one of
552 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
553 or EXACT_DIV_EXPR
554 It controls how the quotient is rounded to an integer.
555 Return nonzero if the operation overflows.
556 UNS nonzero says do unsigned division. */
558 int
565 HOST_WIDE_INT hnum_orig;
567 HOST_WIDE_INT hden_orig;
570 {
586 /* calculate quotient sign and convert operands to unsigned. */
588 {
590 {
592 /* (minimum integer) / (-1) is the only overflow case. */
596 }
598 {
601 }
602 }
607 /* This unsigned division rounds toward zero. */
610 }
614 /* hden != 0 already checked. */
619 }
629 /* Special code for when the divisor < BASE. */
631 {
632 /* hnum != 0 already checked. */
634 {
638 }
639 }
640 else
641 {
642 /* Full double precision division,
643 with thanks to Don Knuth's "Seminumerical Algorithms". */
647 /* Find the highest non-zero divisor digit. */
650 {
653 }
655 /* Insure that the first digit of the divisor is at least BASE/2.
656 This is required by the quotient digit estimation algorithm. */
663 {
667 }
672 {
677 }
678 }
682 /* Main loop */
684 {
685 /* Guess the next quotient digit, quo_est, by dividing the first
686 two remaining dividend digits by the high order quotient digit.
687 quo_est is never low and is at most 2 high. */
694 else
697 /* Refine quo_est so it's usually correct, and at most one high. */
702 quo_est--;
704 /* Try QUO_EST as the quotient digit, by multiplying the
705 divisor by QUO_EST and subtracting from the remaining dividend.
706 Keep in mind that QUO_EST is the I - 1st digit. */
710 {
716 }
718 /* If quo_est was high by one, then num[i] went negative and
719 we need to correct things. */
721 {
722 quo_est--;
725 {
729 }
732 }
734 /* Store the quotient digit. */
736 }
737 }
741 finish_up:
742 /* if result is negative, make it so. */
746 /* compute trial remainder: rem = num - (quo * den) */
752 {
761 {
762 /* quo = quo - 1; */
765 }
766 else
773 {
776 }
777 else
783 {
789 /* Get absolute values */
795 /* If (2 * abs (lrem) >= abs (lden)) */
804 {
806 /* quo = quo - 1; */
809 else
810 /* quo = quo + 1; */
813 }
814 else
816 }
821 }
823 /* compute true remainder: rem = num - (quo * den) */
828 }
829 \f
830 /* Given T, an expression, return the negation of T. Allow for T to be
831 null, in which case return null. */
833 static tree
835 tree t;
836 {
837 tree type;
838 tree tem;
847 {
860 /* - (A - B) -> B - A */
870 }
873 }
874 \f
875 /* Split a tree IN into a constant, literal and variable parts that could be
876 combined with CODE to make IN. "constant" means an expression with
877 TREE_CONSTANT but that isn't an actual constant. CODE must be a
878 commutative arithmetic operation. Store the constant part into *CONP,
879 the literal in *LITP and return the variable part. If a part isn't
880 present, set it to null. If the tree does not decompose in this way,
881 return the entire tree as the variable part and the other parts as null.
883 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
884 case, we negate an operand that was subtracted. Except if it is a
885 literal for which we use *MINUS_LITP instead.
887 If NEGATE_P is true, we are negating all of IN, again except a literal
888 for which we use *MINUS_LITP instead.
890 If IN is itself a literal or constant, return it as appropriate.
892 Note that we do not guarantee that any of the three values will be the
893 same type as IN, but they will have the same signedness and mode. */
895 static tree
897 tree in;
901 {
908 /* Strip any conversions that don't change the machine mode or signedness. */
915 /* We can associate addition and subtraction together (even
916 though the C standard doesn't say so) for integers because
917 the value is not affected. For reals, the value might be
918 affected, so we can't. */
921 {
927 /* First see if either of the operands is a literal, then a constant. */
938 /* If we haven't dealt with either operand, this is not a case we can
939 decompose. Otherwise, VAR is either of the ones remaining, if any. */
944 else
947 /* Now do any needed negations. */
954 }
957 else
961 {
968 }
971 }
973 /* Re-associate trees split by the above function. T1 and T2 are either
974 expressions to associate or null. Return the new expression, if any. If
975 we build an operation, do it in TYPE and with CODE. */
977 static tree
981 tree type;
982 {
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 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1202 constant. We assume ARG1 and ARG2 have the same data type, or at least
1203 are the same kind of constant and the same machine mode.
1205 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1207 static tree
1212 {
1220 {
1221 REAL_VALUE_TYPE d1;
1222 REAL_VALUE_TYPE d2;
1223 REAL_VALUE_TYPE value;
1224 tree t;
1229 /* If either operand is a NaN, just return it. Otherwise, set up
1230 for floating-point trap; we return an overflow. */
1240 value));
1250 }
1252 {
1258 tree t;
1261 {
1281 notrunc),
1287 notrunc));
1291 {
1292 tree magsquared
1296 notrunc);
1299 const_binop
1304 notrunc),
1306 notrunc),
1307 notrunc),
1309 const_binop
1314 notrunc),
1316 notrunc),
1317 notrunc),
1319 }
1324 }
1326 }
1328 }
1330 /* These are the hash table functions for the hash table of INTEGER_CST
1331 nodes of a sizetype. */
1333 /* Return the hash code code X, an INTEGER_CST. */
1335 static hashval_t
1338 {
1344 }
1346 /* Return non-zero if the value represented by *X (an INTEGER_CST tree node)
1347 is the same as that given by *Y, which is the same. */
1349 static int
1353 {
1361 }
1362 \f
1363 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1364 bits are given by NUMBER and of the sizetype represented by KIND. */
1366 tree
1368 HOST_WIDE_INT number;
1370 {
1372 }
1374 /* Likewise, but the desired type is specified explicitly. */
1376 tree
1378 HOST_WIDE_INT number;
1379 tree type;
1380 {
1386 {
1391 }
1393 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1394 hash table, we return the value from the hash table. Otherwise, we
1395 place that in the hash table and make a new node for the next time. */
1404 {
1410 }
1411 else
1413 }
1415 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1416 is a tree code. The type of the result is taken from the operands.
1417 Both must be the same type integer type and it must be a size type.
1418 If the operands are constant, so is the result. */
1420 tree
1424 {
1431 /* Handle the special case of two integer constants faster. */
1433 {
1434 /* And some specific cases even faster than that. */
1443 /* Handle general case of two integer constants. */
1445 }
1451 }
1453 /* Given two values, either both of sizetype or both of bitsizetype,
1454 compute the difference between the two values. Return the value
1455 in signed type corresponding to the type of the operands. */
1457 tree
1460 {
1462 tree ctype;
1468 /* If the type is already signed, just do the simple thing. */
1475 /* If either operand is not a constant, do the conversions to the signed
1476 type and subtract. The hardware will do the right thing with any
1477 overflow in the subtraction. */
1482 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1483 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1484 overflow) and negate (which can't either). Special-case a result
1485 of zero while we're here. */
1490 else
1493 }
1494 \f
1496 /* Given T, a tree representing type conversion of ARG1, a constant,
1497 return a constant tree representing the result of conversion. */
1499 static tree
1501 tree t;
1502 tree arg1;
1503 {
1508 {
1510 {
1511 /* If we would build a constant wider than GCC supports,
1512 leave the conversion unfolded. */
1516 /* If we are trying to make a sizetype for a small integer, use
1517 size_int to pick up cached types to reduce duplicate nodes. */
1523 /* Given an integer constant, make new constant with new type,
1524 appropriately sign-extended or truncated. */
1528 /* Indicate an overflow if (1) ARG1 already overflowed,
1529 or (2) force_fit_type indicates an overflow.
1530 Tell force_fit_type that an overflow has already occurred
1531 if ARG1 is a too-large unsigned value and T is signed.
1532 But don't indicate an overflow if converting a pointer. */
1542 }
1544 {
1545 /* Don't initialize these, use assignments.
1546 Initialized local aggregates don't work on old compilers. */
1547 REAL_VALUE_TYPE x;
1548 REAL_VALUE_TYPE l;
1549 REAL_VALUE_TYPE u;
1560 /* See if X will be in range after truncation towards 0.
1561 To compensate for truncation, move the bounds away from 0,
1562 but reject if X exactly equals the adjusted bounds. */
1566 /* If X is a NaN, use zero instead and show we have an overflow.
1567 Otherwise, range check. */
1571 && !no_upper_bound
1575 {
1579 }
1585 }
1587 }
1589 {
1593 {
1595 {
1599 }
1610 }
1611 }
1614 }
1615 \f
1616 /* Return an expr equal to X but certainly not valid as an lvalue. */
1618 tree
1620 tree x;
1621 {
1622 tree result;
1624 /* These things are certainly not lvalues. */
1635 }
1637 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1638 Zero means allow extended lvalues. */
1642 /* When pedantic, return an expr equal to X but certainly not valid as a
1643 pedantic lvalue. Otherwise, return X. */
1645 tree
1647 tree x;
1648 {
1651 else
1653 }
1654 \f
1655 /* Given a tree comparison code, return the code that is the logical inverse
1656 of the given code. It is not safe to do this for floating-point
1657 comparisons, except for NE_EXPR and EQ_EXPR. */
1659 static enum tree_code
1662 {
1664 {
1679 }
1680 }
1682 /* Similar, but return the comparison that results if the operands are
1683 swapped. This is safe for floating-point. */
1685 static enum tree_code
1688 {
1690 {
1704 }
1705 }
1707 /* Return nonzero if CODE is a tree code that represents a truth value. */
1709 static int
1712 {
1717 }
1718 \f
1719 /* Return nonzero if two operands are necessarily equal.
1720 If ONLY_CONST is non-zero, only return non-zero for constants.
1721 This function tests whether the operands are indistinguishable;
1722 it does not test whether they are equal using C's == operation.
1723 The distinction is important for IEEE floating point, because
1724 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1725 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1727 int
1731 {
1732 /* If both types don't have the same signedness, then we can't consider
1733 them equal. We must check this before the STRIP_NOPS calls
1734 because they may change the signedness of the arguments. */
1742 /* This is needed for conversions and for COMPONENT_REF.
1743 Might as well play it safe and always test this. */
1749 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1750 We don't care about side effects in that case because the SAVE_EXPR
1751 takes care of that for us. In all other cases, two expressions are
1752 equal if they have no side effects. If we have two identical
1753 expressions with side effects that should be treated the same due
1754 to the only side effects being identical SAVE_EXPR's, that will
1755 be detected in the recursive calls below. */
1761 /* Next handle constant cases, those for which we can return 1 even
1762 if ONLY_CONST is set. */
1765 {
1778 {
1788 {
1793 }
1796 }
1800 only_const)
1802 only_const));
1815 }
1821 {
1823 /* Two conversions are equal only if signedness and modes match. */
1839 /* For commutative ops, allow the other order. */
1852 /* If either of the pointer (or reference) expressions we are dereferencing
1853 contain a side effect, these cannot be equal. */
1859 {
1881 }
1890 }
1891 }
1892 \f
1893 /* Similar to operand_equal_p, but see if ARG0 might have been made by
1894 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
1896 When in doubt, return 0. */
1898 static int
1901 tree other;
1902 {
1914 /* Discard any conversions that don't change the modes of ARG0 and ARG1
1915 and see if the inner values are the same. This removes any
1916 signedness comparison, which doesn't matter here. */
1923 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
1924 actual comparison operand, ARG0.
1926 First throw away any conversions to wider types
1927 already present in the operands. */
1936 {
1939 /* Make sure shorter operand is extended the right way
1940 to match the longer operand. */
1946 }
1949 }
1950 \f
1951 /* See if ARG is an expression that is either a comparison or is performing
1952 arithmetic on comparisons. The comparisons must only be comparing
1953 two different values, which will be stored in *CVAL1 and *CVAL2; if
1954 they are non-zero it means that some operands have already been found.
1955 No variables may be used anywhere else in the expression except in the
1956 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
1957 the expression and save_expr needs to be called with CVAL1 and CVAL2.
1959 If this is true, return 1. Otherwise, return zero. */
1961 static int
1963 tree arg;
1966 {
1970 /* We can handle some of the 'e' cases here. */
1980 {
1981 /* If we've already found a CVAL1 or CVAL2, this expression is
1982 two complex to handle. */
1988 }
1991 {
2014 /* First see if we can handle the first operand, then the second. For
2015 the second operand, we know *CVAL1 can't be zero. It must be that
2016 one side of the comparison is each of the values; test for the
2017 case where this isn't true by failing if the two operands
2018 are the same. */
2027 ;
2031 ;
2032 else
2036 ;
2040 ;
2041 else
2048 }
2049 }
2050 \f
2051 /* ARG is a tree that is known to contain just arithmetic operations and
2052 comparisons. Evaluate the operations in the tree substituting NEW0 for
2053 any occurrence of OLD0 as an operand of a comparison and likewise for
2054 NEW1 and OLD1. */
2056 static tree
2058 tree arg;
2060 {
2065 /* We can handle some of the 'e' cases here. */
2073 {
2088 {
2105 }
2106 /* fall through - ??? */
2109 {
2113 /* We need to check both for exact equality and tree equality. The
2114 former will be true if the operand has a side-effect. In that
2115 case, we know the operand occurred exactly once. */
2128 }
2132 }
2133 }
2134 \f
2135 /* Return a tree for the case when the result of an expression is RESULT
2136 converted to TYPE and OMITTED was previously an operand of the expression
2137 but is now not needed (e.g., we folded OMITTED * 0).
2139 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2140 the conversion of RESULT to TYPE. */
2142 static tree
2145 {
2152 }
2154 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2156 static tree
2159 {
2166 }
2167 \f
2168 /* Return a simplified tree node for the truth-negation of ARG. This
2169 never alters ARG itself. We assume that ARG is an operation that
2170 returns a truth value (0 or 1). */
2172 tree
2174 tree arg;
2175 {
2182 /* If this is a comparison, we can simply invert it, except for
2183 floating-point non-equality comparisons, in which case we just
2184 enclose a TRUTH_NOT_EXPR around what we have. */
2187 {
2189 && !flag_unsafe_math_optimizations
2193 else
2196 }
2199 {
2214 /* Here we can invert either operand. We invert the first operand
2215 unless the second operand is a TRUTH_NOT_EXPR in which case our
2216 result is the XOR of the first operand with the inside of the
2217 negation of the second operand. */
2222 else
2277 }
2281 }
2283 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2284 operands are another bit-wise operation with a common input. If so,
2285 distribute the bit operations to save an operation and possibly two if
2286 constants are involved. For example, convert
2287 (A | B) & (A | C) into A | (B & C)
2288 Further simplification will occur if B and C are constants.
2290 If this optimization cannot be done, 0 will be returned. */
2292 static tree
2295 tree type;
2297 {
2298 tree common;
2308 {
2312 }
2314 {
2318 }
2320 {
2324 }
2326 {
2330 }
2331 else
2336 }
2337 \f
2338 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2339 starting at BITPOS. The field is unsigned if UNSIGNEDP is non-zero. */
2341 static tree
2343 tree inner;
2344 tree type;
2347 {
2354 }
2356 /* Optimize a bit-field compare.
2358 There are two cases: First is a compare against a constant and the
2359 second is a comparison of two items where the fields are at the same
2360 bit position relative to the start of a chunk (byte, halfword, word)
2361 large enough to contain it. In these cases we can avoid the shift
2362 implicit in bitfield extractions.
2364 For constants, we emit a compare of the shifted constant with the
2365 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2366 compared. For two fields at the same position, we do the ANDs with the
2367 similar mask and compare the result of the ANDs.
2369 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2370 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2371 are the left and right operands of the comparison, respectively.
2373 If the optimization described above can be done, we return the resulting
2374 tree. Otherwise we return zero. */
2376 static tree
2379 tree compare_type;
2381 {
2390 tree mask;
2391 tree offset;
2393 /* Get all the information about the extractions being done. If the bit size
2394 if the same as the size of the underlying object, we aren't doing an
2395 extraction at all and so can do nothing. We also don't want to
2396 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2397 then will no longer be able to replace it. */
2405 {
2406 /* If this is not a constant, we can only do something if bit positions,
2407 sizes, and signedness are the same. */
2415 }
2417 /* See if we can find a mode to refer to this field. We should be able to,
2418 but fail if we can't. */
2427 /* Set signed and unsigned types of the precision of this mode for the
2428 shifts below. */
2432 /* Compute the bit position and size for the new reference and our offset
2433 within it. If the new reference is the same size as the original, we
2434 won't optimize anything, so return zero. */
2444 /* Make the mask to be used against the extracted field. */
2454 /* If not comparing with constant, just rework the comparison
2455 and return. */
2460 mask),
2464 mask));
2466 /* Otherwise, we are handling the constant case. See if the constant is too
2467 big for the field. Warn and return a tree of for 0 (false) if so. We do
2468 this not only for its own sake, but to avoid having to test for this
2469 error case below. If we didn't, we might generate wrong code.
2471 For unsigned fields, the constant shifted right by the field length should
2472 be all zero. For signed fields, the high-order bits should agree with
2473 the sign bit. */
2476 {
2480 {
2486 }
2487 }
2488 else
2489 {
2493 {
2499 }
2500 }
2502 /* Single-bit compares should always be against zero. */
2504 {
2507 }
2509 /* Make a new bitfield reference, shift the constant over the
2510 appropriate number of bits and mask it with the computed mask
2511 (in case this was a signed field). If we changed it, make a new one. */
2514 {
2517 }
2527 rhs);
2528 }
2529 \f
2530 /* Subroutine for fold_truthop: decode a field reference.
2532 If EXP is a comparison reference, we return the innermost reference.
2534 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2535 set to the starting bit number.
2537 If the innermost field can be completely contained in a mode-sized
2538 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2540 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2541 otherwise it is not changed.
2543 *PUNSIGNEDP is set to the signedness of the field.
2545 *PMASK is set to the mask used. This is either contained in a
2546 BIT_AND_EXPR or derived from the width of the field.
2548 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2550 Return 0 if this is not a component reference or is one that we can't
2551 do anything with. */
2553 static tree
2556 tree exp;
2562 {
2565 tree unsigned_type;
2568 /* All the optimizations using this function assume integer fields.
2569 There are problems with FP fields since the type_for_size call
2570 below can fail for, e.g., XFmode. */
2577 {
2583 }
2592 /* Compute the mask to access the bitfield. */
2602 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2610 }
2612 /* Return non-zero if MASK represents a mask of SIZE ones in the low-order
2613 bit positions. */
2615 static int
2617 tree mask;
2619 {
2622 tree tmask;
2627 return
2634 }
2636 /* Subroutine for fold_truthop: determine if an operand is simple enough
2637 to be evaluated unconditionally. */
2639 static int
2641 tree exp;
2642 {
2643 /* Strip any conversions that don't change the machine mode. */
2655 /* Don't regard global variables as simple. They may be
2656 allocated in ways unknown to the compiler (shared memory,
2657 #pragma weak, etc). */
2660 /* Loading a static variable is unduly expensive, but global
2661 registers aren't expensive. */
2663 }
2664 \f
2665 /* The following functions are subroutines to fold_range_test and allow it to
2666 try to change a logical combination of comparisons into a range test.
2668 For example, both
2669 X == 2 || X == 3 || X == 4 || X == 5
2670 and
2671 X >= 2 && X <= 5
2672 are converted to
2673 (unsigned) (X - 2) <= 3
2675 We describe each set of comparisons as being either inside or outside
2676 a range, using a variable named like IN_P, and then describe the
2677 range with a lower and upper bound. If one of the bounds is omitted,
2678 it represents either the highest or lowest value of the type.
2680 In the comments below, we represent a range by two numbers in brackets
2681 preceded by a "+" to designate being inside that range, or a "-" to
2682 designate being outside that range, so the condition can be inverted by
2683 flipping the prefix. An omitted bound is represented by a "-". For
2684 example, "- [-, 10]" means being outside the range starting at the lowest
2685 possible value and ending at 10, in other words, being greater than 10.
2686 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2687 always false.
2689 We set up things so that the missing bounds are handled in a consistent
2690 manner so neither a missing bound nor "true" and "false" need to be
2691 handled using a special case. */
2693 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2694 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2695 and UPPER1_P are nonzero if the respective argument is an upper bound
2696 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2697 must be specified for a comparison. ARG1 will be converted to ARG0's
2698 type if both are specified. */
2700 static tree
2703 tree type;
2706 {
2707 tree tem;
2711 /* If neither arg represents infinity, do the normal operation.
2712 Else, if not a comparison, return infinity. Else handle the special
2713 comparison rules. Note that most of the cases below won't occur, but
2714 are handled for consistency. */
2717 {
2722 }
2727 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2728 for neither. In real maths, we cannot assume open ended ranges are
2729 the same. But, this is computer arithmetic, where numbers are finite.
2730 We can therefore make the transformation of any unbounded range with
2731 the value Z, Z being greater than any representable number. This permits
2732 us to treat unbounded ranges as equal. */
2736 {
2757 }
2760 }
2761 \f
2762 /* Given EXP, a logical expression, set the range it is testing into
2763 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2764 actually being tested. *PLOW and *PHIGH will be made of the same type
2765 as the returned expression. If EXP is not a comparison, we will most
2766 likely not be returning a useful value and range. */
2768 static tree
2770 tree exp;
2773 {
2780 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2781 and see if we can refine the range. Some of the cases below may not
2782 happen, but it doesn't seem worth worrying about this. We "continue"
2783 the outer loop when we've changed something; otherwise we "break"
2784 the switch, which will "break" the while. */
2789 {
2793 {
2804 }
2806 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2807 lose a cast by accident. */
2812 {
2819 /* We can only do something if the range is testing for zero
2820 and if the second operand is an integer constant. Note that
2821 saying something is "in" the range we make is done by
2822 complementing IN_P since it will set in the initial case of
2823 being not equal to zero; "out" is leaving it alone. */
2830 {
2851 }
2855 /* If this is an unsigned comparison, we also know that EXP is
2856 greater than or equal to zero. We base the range tests we make
2857 on that fact, so we record it here so we can parse existing
2858 range tests. */
2860 {
2863 NULL_TREE))
2868 /* If the high bound is missing, but we
2869 have a low bound, reverse the range so
2870 it goes from zero to the low bound minus 1. */
2872 {
2877 }
2878 }
2882 /* (-x) IN [a,b] -> x in [-b, -a] */
2892 /* ~ X -> -X - 1 */
2901 /* If EXP is signed, any overflow in the computation is undefined,
2902 so we don't worry about it so long as our computations on
2903 the bounds don't overflow. For unsigned, overflow is defined
2904 and this is exactly the right thing. */
2913 /* Check for an unsigned range which has wrapped around the maximum
2914 value thus making n_high < n_low, and normalize it. */
2916 {
2922 /* If the range is of the form +/- [ x+1, x ], we won't
2923 be able to normalize it. But then, it represents the
2924 whole range or the empty set, so make it
2925 +/- [ -, - ]. */
2929 else
2931 }
2932 else
2955 /* If we're converting from an unsigned to a signed type,
2956 we will be doing the comparison as unsigned. The tests above
2957 have already verified that LOW and HIGH are both positive.
2959 So we have to make sure that the original unsigned value will
2960 be interpreted as positive. */
2962 {
2965 tree high_positive;
2967 /* A range without an upper bound is, naturally, unbounded.
2968 Since convert would have cropped a very large value, use
2969 the max value for the destination type. */
2970 high_positive
2978 /* If the low bound is specified, "and" the range with the
2979 range for which the original unsigned value will be
2980 positive. */
2982 {
2986 high_positive))
2990 }
2991 else
2992 {
2993 /* Otherwise, "or" the range with the range of the input
2994 that will be interpreted as negative. */
2998 high_positive))
3002 }
3003 }
3011 }
3014 }
3016 /* If EXP is a constant, we can evaluate whether this is true or false. */
3018 {
3025 }
3029 }
3030 \f
3031 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3032 type, TYPE, return an expression to test if EXP is in (or out of, depending
3033 on IN_P) the range. */
3035 static tree
3037 tree type;
3038 tree exp;
3041 {
3065 {
3069 }
3076 else
3078 }
3079 \f
3080 /* Given two ranges, see if we can merge them into one. Return 1 if we
3081 can, 0 if we can't. Set the output range into the specified parameters. */
3083 static int
3089 {
3093 tree tem;
3103 /* Make range 0 be the range that starts first, or ends last if they
3104 start at the same value. Swap them if it isn't. */
3107 || (lowequal
3110 {
3114 }
3116 /* Now flag two cases, whether the ranges are disjoint or whether the
3117 second range is totally subsumed in the first. Note that the tests
3118 below are simplified by the ones above. */
3124 /* We now have four cases, depending on whether we are including or
3125 excluding the two ranges. */
3127 {
3128 /* If they don't overlap, the result is false. If the second range
3129 is a subset it is the result. Otherwise, the range is from the start
3130 of the second to the end of the first. */
3135 else
3137 }
3140 {
3141 /* If they don't overlap, the result is the first range. If they are
3142 equal, the result is false. If the second range is a subset of the
3143 first, and the ranges begin at the same place, we go from just after
3144 the end of the first range to the end of the second. If the second
3145 range is not a subset of the first, or if it is a subset and both
3146 ranges end at the same place, the range starts at the start of the
3147 first range and ends just before the second range.
3148 Otherwise, we can't describe this as a single range. */
3154 {
3158 }
3160 {
3164 }
3165 else
3167 }
3170 {
3171 /* If they don't overlap, the result is the second range. If the second
3172 is a subset of the first, the result is false. Otherwise,
3173 the range starts just after the first range and ends at the
3174 end of the second. */
3179 else
3180 {
3184 }
3185 }
3187 else
3188 {
3189 /* The case where we are excluding both ranges. Here the complex case
3190 is if they don't overlap. In that case, the only time we have a
3191 range is if they are adjacent. If the second is a subset of the
3192 first, the result is the first. Otherwise, the range to exclude
3193 starts at the beginning of the first range and ends at the end of the
3194 second. */
3196 {
3203 else
3205 }
3208 else
3210 }
3214 }
3215 \f
3216 /* EXP is some logical combination of boolean tests. See if we can
3217 merge it into some range test. Return the new tree if so. */
3219 static tree
3221 tree exp;
3222 {
3229 tree tem;
3231 /* If this is an OR operation, invert both sides; we will invert
3232 again at the end. */
3236 /* If both expressions are the same, if we can merge the ranges, and we
3237 can build the range test, return it or it inverted. If one of the
3238 ranges is always true or always false, consider it to be the same
3239 expression as the other. */
3249 /* On machines where the branch cost is expensive, if this is a
3250 short-circuited branch and the underlying object on both sides
3251 is the same, make a non-short-circuit operation. */
3257 {
3258 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3259 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3260 which cases we can't do this. */
3269 {
3281 }
3282 }
3285 }
3286 \f
3287 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3288 bit value. Arrange things so the extra bits will be set to zero if and
3289 only if C is signed-extended to its full width. If MASK is nonzero,
3290 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3292 static tree
3294 tree c;
3297 tree mask;
3298 {
3301 tree temp;
3306 /* We work by getting just the sign bit into the low-order bit, then
3307 into the high-order bit, then sign-extend. We then XOR that value
3308 with C. */
3312 /* We must use a signed type in order to get an arithmetic right shift.
3313 However, we must also avoid introducing accidental overflows, so that
3314 a subsequent call to integer_zerop will work. Hence we must
3315 do the type conversion here. At this point, the constant is either
3316 zero or one, and the conversion to a signed type can never overflow.
3317 We could get an overflow if this conversion is done anywhere else. */
3325 /* If necessary, convert the type back to match the type of C. */
3330 }
3331 \f
3332 /* Find ways of folding logical expressions of LHS and RHS:
3333 Try to merge two comparisons to the same innermost item.
3334 Look for range tests like "ch >= '0' && ch <= '9'".
3335 Look for combinations of simple terms on machines with expensive branches
3336 and evaluate the RHS unconditionally.
3338 For example, if we have p->a == 2 && p->b == 4 and we can make an
3339 object large enough to span both A and B, we can do this with a comparison
3340 against the object ANDed with the a mask.
3342 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3343 operations to do this with one comparison.
3345 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3346 function and the one above.
3348 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3349 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3351 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3352 two operands.
3354 We return the simplified tree or 0 if no optimization is possible. */
3356 static tree
3360 {
3361 /* If this is the "or" of two comparisons, we can do something if
3362 the comparisons are NE_EXPR. If this is the "and", we can do something
3363 if the comparisons are EQ_EXPR. I.e.,
3364 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3366 WANTED_CODE is this operation code. For single bit fields, we can
3367 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3368 comparison for one-bit fields. */
3388 /* Start by getting the comparison codes. Fail if anything is volatile.
3389 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3390 it were surrounded with a NE_EXPR. */
3415 /* If the RHS can be evaluated unconditionally and its operands are
3416 simple, it wins to evaluate the RHS unconditionally on machines
3417 with expensive branches. In this case, this isn't a comparison
3418 that can be merged. Avoid doing this if the RHS is a floating-point
3419 comparison since those can trap. */
3427 /* See if the comparisons can be merged. Then get all the parameters for
3428 each side. */
3452 /* It must be true that the inner operation on the lhs of each
3453 comparison must be the same if we are to be able to do anything.
3454 Then see if we have constants. If not, the same must be true for
3455 the rhs's. */
3466 else
3469 /* If either comparison code is not correct for our logical operation,
3470 fail. However, we can convert a one-bit comparison against zero into
3471 the opposite comparison against that bit being set in the field. */
3475 {
3477 {
3478 /* Make the left operand unsigned, since we are only interested
3479 in the value of one bit. Otherwise we are doing the wrong
3480 thing below. */
3483 }
3484 else
3486 }
3488 /* This is analogous to the code for l_const above. */
3490 {
3492 {
3495 }
3496 else
3498 }
3500 /* See if we can find a mode that contains both fields being compared on
3501 the left. If we can't, fail. Otherwise, update all constants and masks
3502 to be relative to a field of that size. */
3507 volatilep);
3517 {
3520 }
3528 {
3536 {
3540 wanted_code == NE_EXPR
3542 }
3543 }
3545 {
3553 {
3557 wanted_code == NE_EXPR
3559 }
3560 }
3562 /* If the right sides are not constant, do the same for it. Also,
3563 disallow this optimization if a size or signedness mismatch occurs
3564 between the left and right sides. */
3566 {
3569 /* Make sure the two fields on the right
3570 correspond to the left without being swapped. */
3578 volatilep);
3588 {
3591 }
3598 /* Make a mask that corresponds to both fields being compared.
3599 Do this for both items being compared. If the operands are the
3600 same size and the bits being compared are in the same position
3601 then we can do this by masking both and comparing the masked
3602 results. */
3606 {
3618 }
3620 /* There is still another way we can do something: If both pairs of
3621 fields being compared are adjacent, we may be able to make a wider
3622 field containing them both.
3624 Note that we still must mask the lhs/rhs expressions. Furthermore,
3625 the mask must be shifted to account for the shift done by
3626 make_bit_field_ref. */
3631 {
3632 tree type;
3644 /* Convert to the smaller type before masking out unwanted bits. */
3647 {
3649 {
3653 }
3655 {
3659 }
3660 }
3669 }
3672 }
3674 /* Handle the case of comparisons with constants. If there is something in
3675 common between the masks, those bits of the constants must be the same.
3676 If not, the condition is always false. Test for this to avoid generating
3677 incorrect code below. */
3682 {
3684 {
3687 }
3688 else
3689 {
3692 }
3693 }
3695 /* Construct the expression we will return. First get the component
3696 reference we will make. Unless the mask is all ones the width of
3697 that field, perform the mask operation. Then compare with the
3698 merged constant. */
3708 }
3709 \f
3710 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3711 constant. */
3713 static tree
3715 tree t;
3716 {
3721 tree minmax_const;
3723 tree inner;
3733 /* If something does not permit us to optimize, return the original tree. */
3741 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3742 and GT_EXPR, doing the rest with recursive calls using logical
3743 simplifications. */
3745 {
3747 return
3751 return
3753 optimize_minmax_comparison
3755 optimize_minmax_comparison
3760 /* MAX (X, 0) == 0 -> X <= 0 */
3764 /* MAX (X, 0) == 5 -> X == 5 */
3768 /* MAX (X, 0) == -1 -> false */
3772 /* MIN (X, 0) == 0 -> X >= 0 */
3776 /* MIN (X, 0) == 5 -> false */
3779 else
3780 /* MIN (X, 0) == -1 -> X == -1 */
3785 /* MAX (X, 0) > 0 -> X > 0
3786 MAX (X, 0) > 5 -> X > 5 */
3790 /* MAX (X, 0) > -1 -> true */
3794 /* MIN (X, 0) > 0 -> false
3795 MIN (X, 0) > 5 -> false */
3798 else
3799 /* MIN (X, 0) > -1 -> X > -1 */
3804 }
3805 }
3806 \f
3807 /* T is an integer expression that is being multiplied, divided, or taken a
3808 modulus (CODE says which and what kind of divide or modulus) by a
3809 constant C. See if we can eliminate that operation by folding it with
3810 other operations already in T. WIDE_TYPE, if non-null, is a type that
3811 should be used for the computation if wider than our type.
3813 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
3814 (X * 2) + (Y * 4). We must, however, be assured that either the original
3815 expression would not overflow or that overflow is undefined for the type
3816 in the language in question.
3818 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
3819 the machine has a multiply-accumulate insn or that this is part of an
3820 addressing calculation.
3822 If we return a non-null expression, it is an equivalent form of the
3823 original computation, but need not be in the original type. */
3825 static tree
3827 tree t;
3828 tree c;
3830 tree wide_type;
3831 {
3841 /* Don't deal with constants of zero here; they confuse the code below. */
3851 /* Note that we need not handle conditional operations here since fold
3852 already handles those cases. So just do arithmetic here. */
3854 {
3856 /* For a constant, we can always simplify if we are a multiply
3857 or (for divide and modulus) if it is a multiple of our constant. */
3864 /* If op0 is an expression, and is unsigned, and the type is
3865 smaller than ctype, then we cannot widen the expression. */
3877 /* Pass the constant down and see if we can make a simplification. If
3878 we can, replace this expression with the inner simplification for
3879 possible later conversion to our or some other type. */
3891 /* If widening the type changes the signedness, then we can't perform
3892 this optimization as that changes the result. */
3896 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
3899 {
3905 }
3915 /* If this has not been evaluated and the operand has no side effects,
3916 we can see if we can do something inside it and make a new one.
3917 Note that this test is overly conservative since we can do this
3918 if the only reason it had side effects is that it was another
3919 similar SAVE_EXPR, but that isn't worth bothering with. */
3922 wide_type)))
3923 {
3930 }
3934 /* If the second operand is constant, this is a multiplication
3935 or floor division, by a power of two, so we can treat it that
3936 way unless the multiplier or divisor overflows. */
3938 /* const_binop may not detect overflow correctly,
3939 so check for it explicitly here. */
3953 /* See if we can eliminate the operation on both sides. If we can, we
3954 can return a new PLUS or MINUS. If we can't, the only remaining
3955 cases where we can do anything are if the second operand is a
3956 constant. */
3961 /* If not multiplication, we can only do this if either operand
3962 is divisible by c. */
3968 /* If this was a subtraction, negate OP1 and set it to be an addition.
3969 This simplifies the logic below. */
3976 /* If either OP1 or C are negative, this optimization is not safe for
3977 some of the division and remainder types while for others we need
3978 to change the code. */
3980 {
3988 }
3990 /* If it's a multiply or a division/modulus operation of a multiple
3991 of our constant, do the operation and verify it doesn't overflow. */
3994 {
3998 }
3999 else
4002 /* If we have an unsigned type is not a sizetype, we cannot widen
4003 the operation since it will change the result if the original
4004 computation overflowed. */
4010 /* If we were able to eliminate our operation from the first side,
4011 apply our operation to the second side and reform the PLUS. */
4015 /* The last case is if we are a multiply. In that case, we can
4016 apply the distributive law to commute the multiply and addition
4017 if the multiplication of the constants doesn't overflow. */
4022 op1));
4027 /* We have a special case here if we are doing something like
4028 (C * 8) % 4 since we know that's zero. */
4035 /* ... fall through ... */
4039 /* If we can extract our operation from the LHS, do so and return a
4040 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4041 do something only if the second operand is a constant. */
4053 /* If these are the same operation types, we can associate them
4054 assuming no overflow. */
4061 /* If these operations "cancel" each other, we have the main
4062 optimizations of this pass, which occur when either constant is a
4063 multiple of the other, in which case we replace this with either an
4064 operation or CODE or TCODE.
4066 If we have an unsigned type that is not a sizetype, we cannot do
4067 this since it will change the result if the original computation
4068 overflowed. */
4075 {
4086 }
4091 }
4094 }
4095 \f
4096 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4097 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4098 that we may sometimes modify the tree. */
4100 static tree
4102 tree t;
4103 tree s;
4104 {
4107 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4112 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4113 don't bother handling any other types. */
4115 {
4119 }
4124 {
4127 }
4130 }
4131 \f
4132 /* Return a node which has the indicated constant VALUE (either 0 or
4133 1), and is of the indicated TYPE. */
4135 static tree
4138 tree type;
4139 {
4144 integer_zero_node);
4145 else
4146 {
4151 }
4152 }
4154 /* Utility function for the following routine, to see how complex a nesting of
4155 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4156 we don't care (to avoid spending too much time on complex expressions.). */
4158 static int
4160 tree expr;
4162 {
4173 }
4175 /* Transform `a + (b ? x : y)' into `x ? (a + b) : (a + y)'.
4176 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4177 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4178 expression, and ARG to `a'. If COND_FIRST_P is non-zero, then the
4179 COND is the first argument to CODE; otherwise (as in the example
4180 given here), it is the second argument. TYPE is the type of the
4181 original expression. */
4183 static tree
4186 tree type;
4187 tree cond;
4188 tree arg;
4190 {
4194 /* In the end, we'll produce a COND_EXPR. Both arms of the
4195 conditional expression will be binary operations. The left-hand
4196 side of the expression to be executed if the condition is true
4197 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4198 of the expression to be executed if the condition is true will be
4199 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4200 but apply to the expression to be executed if the conditional is
4201 false. */
4206 /* These are the codes to use for the left-hand side and right-hand
4207 side of the COND_EXPR. Normally, they are the same as CODE. */
4210 /* And these are the types of the expressions. */
4215 {
4219 }
4220 else
4221 {
4225 }
4228 {
4232 /* If this operand throws an expression, then it does not make
4233 sense to try to perform a logical or arithmetic operation
4234 involving it. Instead of building `a + throw 3' for example,
4235 we simply build `a, throw 3'. */
4237 {
4241 }
4243 {
4247 }
4248 }
4249 else
4250 {
4255 }
4257 /* If ARG is complex we want to make sure we only evaluate
4258 it once. Though this is only required if it is volatile, it
4259 might be more efficient even if it is not. However, if we
4260 succeed in folding one part to a constant, we do not need
4261 to make this SAVE_EXPR. Since we do this optimization
4262 primarily to see if we do end up with constant and this
4263 SAVE_EXPR interferes with later optimizations, suppressing
4264 it when we can is important.
4266 If we are not in a function, we can't make a SAVE_EXPR, so don't
4267 try to do so. Don't try to see if the result is a constant
4268 if an arm is a COND_EXPR since we get exponential behavior
4269 in that case. */
4276 {
4286 }
4299 else
4301 }
4303 \f
4304 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4306 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4307 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4308 ADDEND is the same as X.
4310 X + 0 and X - 0 both give X when X is NaN, infinite, or non-zero
4311 and finite. The problematic cases are when X is zero, and its mode
4312 has signed zeros. In the case of rounding towards -infinity,
4313 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4314 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4316 static bool
4320 {
4324 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4328 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4333 /* The mode has signed zeros, and we have to honor their sign.
4334 In this situation, there is only one case we can return true for.
4335 X - 0 is the same as X unless rounding towards -infinity is
4336 supported. */
4338 }
4341 /* Perform constant folding and related simplification of EXPR.
4342 The related simplifications include x*1 => x, x*0 => 0, etc.,
4343 and application of the associative law.
4344 NOP_EXPR conversions may be removed freely (as long as we
4345 are careful not to change the C type of the overall expression)
4346 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4347 but we can constant-fold them if they have constant operands. */
4349 tree
4351 tree expr;
4352 {
4355 tree tem;
4361 /* WINS will be nonzero when the switch is done
4362 if all operands are constant. */
4365 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4366 Likewise for a SAVE_EXPR that's already been evaluated. */
4370 /* Return right away if a constant. */
4374 #ifdef MAX_INTEGER_COMPUTATION_MODE
4376 #endif
4379 {
4380 tree subop;
4382 /* Special case for conversion ops that can have fixed point args. */
4385 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4391 else
4396 )
4397 /* Note that TREE_CONSTANT isn't enough:
4398 static var addresses are constant but we can't
4399 do arithmetic on them. */
4401 }
4403 {
4407 {
4409 tree subop;
4415 {
4416 /* Signedness matters here. Perhaps we can refine this
4417 later. */
4419 }
4420 else
4421 /* Strip any conversions that don't change the mode. */
4426 else
4431 /* Note that TREE_CONSTANT isn't enough:
4432 static var addresses are constant but we can't
4433 do arithmetic on them. */
4440 }
4441 }
4443 /* If this is a commutative operation, and ARG0 is a constant, move it
4444 to ARG1 to reduce the number of tests below. */
4449 {
4454 }
4456 /* Now WINS is set as described above,
4457 ARG0 is the first operand of EXPR,
4458 and ARG1 is the second operand (if it has more than one operand).
4460 First check for cases where an arithmetic operation is applied to a
4461 compound, conditional, or comparison operation. Push the arithmetic
4462 operation inside the compound or conditional to see if any folding
4463 can then be done. Convert comparison to conditional for this purpose.
4464 The also optimizes non-constant cases that used to be done in
4465 expand_expr.
4467 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4468 one of the operands is a comparison and the other is a comparison, a
4469 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4470 code below would make the expression more complex. Change it to a
4471 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4472 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4484 {
4494 }
4497 {
4502 {
4507 /* If this was a conversion, and all we did was to move into
4508 inside the COND_EXPR, bring it back out. But leave it if
4509 it is a conversion from integer to integer and the
4510 result precision is no wider than a word since such a
4511 conversion is cheap and may be optimized away by combine,
4512 while it couldn't if it were outside the COND_EXPR. Then return
4513 so we don't get into an infinite recursion loop taking the
4514 conversion out and then back in. */
4524 && (INTEGRAL_TYPE_P
4535 }
4540 }
4543 {
4556 return
4570 return
4573 }
4584 {
4600 /* Other kinds of FIX are not handled properly by fold_convert. */
4605 /* Handle cases of two conversions in a row. */
4608 {
4628 /* In addition to the cases of two conversions in a row
4629 handled below, if we are converting something to its own
4630 type via an object of identical or wider precision, neither
4631 conversion is needed. */
4637 /* Likewise, if the intermediate and final types are either both
4638 float or both integer, we don't need the middle conversion if
4639 it is wider than the final type and doesn't change the signedness
4640 (for integers). Avoid this if the final type is a pointer
4641 since then we sometimes need the inner conversion. Likewise if
4642 the outer has a precision not equal to the size of its mode. */
4652 /* If we have a sign-extension of a zero-extended value, we can
4653 replace that by a single zero-extension. */
4659 /* Two conversions in a row are not needed unless:
4660 - some conversion is floating-point (overstrict for now), or
4661 - the intermediate type is narrower than both initial and
4662 final, or
4663 - the intermediate type and innermost type differ in signedness,
4664 and the outermost type is wider than the intermediate, or
4665 - the initial type is a pointer type and the precisions of the
4666 intermediate and final types differ, or
4667 - the final type is a pointer type and the precisions of the
4668 initial and intermediate types differ. */
4682 }
4686 /* Detect assigning a bitfield. */
4689 {
4690 /* Don't leave an assignment inside a conversion
4691 unless assigning a bitfield. */
4694 /* First do the assignment, then return converted constant. */
4698 }
4700 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
4701 constants (if x has signed type, the sign bit cannot be set
4702 in c). This folds extension into the BIT_AND_EXPR. */
4706 {
4716 <= HOST_BITS_PER_WIDE_INT
4718 {
4725 #ifdef LOAD_EXTEND_OP
4729 {
4733 }
4734 #endif
4735 }
4740 }
4743 {
4746 }
4757 {
4761 }
4770 {
4772 {
4774 HOST_WIDE_INT high;
4785 }
4788 }
4792 /* Convert - (a - b) to (b - a) for non-floating-point. */
4802 {
4804 {
4805 /* If the value is unsigned, then the absolute value is
4806 the same as the ordinary value. */
4809 /* Similarly, if the value is non-negative. */
4812 /* If the value is negative, then the absolute value is
4813 its negation. */
4814 else
4815 {
4817 HOST_WIDE_INT high;
4828 }
4829 }
4831 {
4835 }
4836 }
4863 {
4870 }
4876 /* A + (-B) -> A - B */
4879 /* (-A) + B -> B - A */
4883 {
4887 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
4888 with a constant, and the two constants have no bits in common,
4889 we should treat this as a BIT_IOR_EXPR since this may produce more
4890 simplifications. */
4898 {
4901 }
4903 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
4904 (plus (plus (mult) (mult)) (foo)) so that we can
4905 take advantage of the factoring cases below. */
4910 {
4915 else
4926 parg1));
4931 parg0));
4932 }
4935 {
4939 /* (A * C) + (B * C) -> (A+B) * C.
4940 We are most concerned about the case where C is a constant,
4941 but other combinations show up during loop reduction. Since
4942 it is not difficult, try all four possibilities. */
4959 /* No identical multiplicands; see if we can find a common
4960 power-of-two factor in non-power-of-two multiplies. This
4961 can help in multi-dimensional array access. */
4966 {
4971 /* Move min of absolute values to int11. */
4974 {
4978 }
4981 {
4986 }
4987 }
4992 same));
4993 }
4994 }
4996 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5000 /* Likewise if the operands are reversed. */
5004 bit_rotate:
5005 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5006 is a rotate of A by C1 bits. */
5007 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5008 is a rotate of A by B bits. */
5009 {
5018 {
5037 {
5045 TYPE_PRECISION
5050 ? LROTATE_EXPR
5053 }
5055 {
5063 TYPE_PRECISION
5068 ? LROTATE_EXPR
5071 }
5072 }
5073 }
5075 associate:
5076 /* In most languages, can't associate operations on floats through
5077 parentheses. Rather than remember where the parentheses were, we
5078 don't associate floats at all. It shouldn't matter much. However,
5079 associating multiplications is only very slightly inaccurate, so do
5080 that if -funsafe-math-optimizations is specified. */
5085 {
5089 /* Split both trees into variables, constants, and literals. Then
5090 associate each group together, the constants with literals,
5091 then the result with variables. This increases the chances of
5092 literals being recombined later and of generating relocatable
5093 expressions for the sum of a constant and literal. */
5098 /* Only do something if we found more than two objects. Otherwise,
5099 nothing has changed and we risk infinite recursion. */
5104 {
5105 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5114 /* Preserve the MINUS_EXPR if the negative part of the literal is
5115 greater than the positive part. Otherwise, the multiplicative
5116 folding code (i.e extract_muldiv) may be fooled in case
5117 unsigned constants are substracted, like in the following
5118 example: ((X*2 + 4) - 8U)/2. */
5120 {
5122 {
5126 }
5127 else
5128 {
5132 }
5133 }
5135 {
5139 else
5140 {
5145 }
5146 }
5150 }
5151 }
5153 binary:
5157 {
5158 /* The return value should always have
5159 the same type as the original expression. */
5164 }
5168 /* A - (-B) -> A + B */
5171 /* (-A) - CST -> (-CST) - A for floating point (what about ints ?) */
5173 return
5180 {
5186 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5187 about the case where C is a constant, just try one of the
5188 four possibilities. */
5198 }
5200 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5204 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5205 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5206 (-ARG1 + ARG0) reduces to -ARG1. */
5210 /* Fold &x - &x. This can happen from &x.foo - &x.
5211 This is unsafe for certain floats even in non-IEEE formats.
5212 In IEEE, it is unsafe because it does wrong for NaNs.
5213 Also note that operand_equal_p is always false if an operand
5214 is volatile. */
5223 /* (-A) * (-B) -> A * B */
5229 {
5235 /* (a * (1 << b)) is (a << b) */
5250 }
5251 else
5252 {
5253 /* Maybe fold x * 0 to 0. The expressions aren't the same
5254 when x is NaN, since x * 0 is also NaN. Nor are they the
5255 same in modes with signed zeros, since multiplying a
5256 negative value by 0 gives -0, not +0. */
5261 /* In IEEE floating point, x*1 is not equivalent to x for snans.
5262 However, ANSI says we can drop signals,
5263 so we can do this anyway. */
5266 /* x*2 is x+x */
5270 {
5273 }
5274 }
5278 bit_ior:
5287 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5289 This results in more efficient code for machines without a NAND
5290 instruction. Combine will canonicalize to the first form
5291 which will allow use of NAND instructions provided by the
5292 backend if they exist. */
5295 {
5300 }
5302 /* See if this can be simplified into a rotate first. If that
5303 is unsuccessful continue in the association code. */
5312 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5313 with a constant, and the two constants have no bits in common,
5314 we should treat this as a BIT_IOR_EXPR since this may produce more
5315 simplifications. */
5323 {
5326 }
5328 /* See if this can be simplified into a rotate first. If that
5329 is unsuccessful continue in the association code. */
5333 bit_and:
5341 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5344 {
5345 unsigned int prec
5352 }
5355 {
5356 unsigned int prec
5363 }
5365 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5367 This results in more efficient code for machines without a NOR
5368 instruction. Combine will canonicalize to the first form
5369 which will allow use of NOR instructions provided by the
5370 backend if they exist. */
5373 {
5378 }
5388 {
5392 }
5396 /* Don't touch a floating-point divide by zero unless the mode
5397 of the constant can represent infinity. */
5403 /* (-A) / (-B) -> A / B */
5408 /* In IEEE floating point, x/1 is not equivalent to x for snans.
5409 However, ANSI says we can drop signals, so we can do this anyway. */
5413 /* If ARG1 is a constant, we can convert this to a multiply by the
5414 reciprocal. This does not have the same rounding properties,
5415 so only do this if -funsafe-math-optimizations. We can actually
5416 always safely do it if ARG1 is a power of two, but it's hard to
5417 tell if it is or not in a portable manner. */
5419 {
5424 /* Find the reciprocal if optimizing and the result is exact. */
5426 {
5427 REAL_VALUE_TYPE r;
5430 {
5433 }
5434 }
5435 }
5436 /* Convert A/B/C to A/(B*C). */
5439 {
5442 arg1)));
5443 }
5444 /* Convert A/(B/C) to (A/B)*C. */
5447 {
5452 }
5465 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5466 operation, EXACT_DIV_EXPR.
5468 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5469 At one time others generated faster code, it's not clear if they do
5470 after the last round to changes to the DIV code in expmed.c. */
5504 /* Since negative shift count is not well-defined,
5505 don't try to compute it in the compiler. */
5508 /* Rewrite an LROTATE_EXPR by a constant into an
5509 RROTATE_EXPR by a new constant. */
5511 {
5515 = const_binop
5522 }
5524 /* If we have a rotate of a bit operation with the rotate count and
5525 the second operand of the bit operation both constant,
5526 permute the two operations. */
5539 /* Two consecutive rotates adding up to the width of the mode can
5540 be ignored. */
5571 /* Note that the operand of this must be an int
5572 and its values must be 0 or 1.
5573 ("true" is a fixed value perhaps depending on the language,
5574 but we don't handle values other than 1 correctly yet.) */
5576 /* Avoid infinite recursion. */
5582 /* Note that the operands of this must be ints
5583 and their values must be 0 or 1.
5584 ("true" is a fixed value perhaps depending on the language.) */
5585 /* If first arg is constant zero, return it. */
5589 /* If either arg is constant true, drop it. */
5593 /* Preserve sequence points. */
5596 /* If second arg is constant zero, result is zero, but first arg
5597 must be evaluated. */
5600 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
5601 case will be handled here. */
5605 truth_andor:
5606 /* We only do these simplifications if we are optimizing. */
5610 /* Check for things like (A || B) && (A || C). We can convert this
5611 to A || (B && C). Note that either operator can be any of the four
5612 truth and/or operations and the transformation will still be
5613 valid. Also note that we only care about order for the
5614 ANDIF and ORIF operators. If B contains side effects, this
5615 might change the truth-value of A. */
5622 {
5642 /* This case if tricky because we must either have commutative
5643 operators or else A10 must not have side-effects. */
5649 a01));
5650 }
5652 /* See if we can build a range comparison. */
5656 /* Check for the possibility of merging component references. If our
5657 lhs is another similar operation, try to merge its rhs with our
5658 rhs. Then try to merge our lhs and rhs. */
5670 /* Note that the operands of this must be ints
5671 and their values must be 0 or true.
5672 ("true" is a fixed value perhaps depending on the language.) */
5673 /* If first arg is constant true, return it. */
5677 /* If either arg is constant zero, drop it. */
5681 /* Preserve sequence points. */
5684 /* If second arg is constant true, result is true, but we must
5685 evaluate first arg. */
5688 /* Likewise for first arg, but note this only occurs here for
5689 TRUTH_OR_EXPR. */
5695 /* If either arg is constant zero, drop it. */
5700 /* If either arg is constant true, this is a logical inversion. */
5714 {
5715 /* (-a) CMP (-b) -> b CMP a */
5720 /* (-a) CMP CST -> a swap(CMP) (-CST) */
5722 return
5728 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
5729 /* a CMP (-0) -> a CMP 0 */
5734 }
5736 /* If one arg is a constant integer, put it last. */
5739 {
5746 }
5748 /* Convert foo++ == CONST into ++foo == CONST + INCR.
5749 First, see if one arg is constant; find the constant arg
5750 and the other one. */
5751 {
5761 {
5762 /* This optimization is invalid for ordered comparisons
5763 if CONST+INCR overflows or if foo+incr might overflow.
5764 This optimization is invalid for floating point due to rounding.
5765 For pointer types we assume overflow doesn't happen. */
5769 {
5770 tree newconst
5774 /* Do not overwrite the current varop to be a preincrement,
5775 create a new node so that we won't confuse our caller who
5776 might create trees and throw them away, reusing the
5777 arguments that they passed to build. This shows up in
5778 the THEN or ELSE parts of ?: being postincrements. */
5783 /* If VAROP is a reference to a bitfield, we must mask
5784 the constant by the width of the field. */
5788 {
5789 int size
5791 (TREE_OPERAND
5795 tree folded_compare;
5797 /* First check whether the comparison would come out
5798 always the same. If we don't do that we would
5799 change the meaning with the masking. */
5803 else
5806 constop));
5821 mask)));
5822 }
5828 }
5829 }
5831 {
5835 {
5836 tree newconst
5840 /* Do not overwrite the current varop to be a predecrement,
5841 create a new node so that we won't confuse our caller who
5842 might create trees and throw them away, reusing the
5843 arguments that they passed to build. This shows up in
5844 the THEN or ELSE parts of ?: being postdecrements. */
5852 {
5853 int size
5855 (TREE_OPERAND
5859 tree folded_compare;
5864 else
5867 constop));
5882 mask)));
5883 }
5889 }
5890 }
5891 }
5893 /* Change X >= CST to X > (CST - 1) if CST is positive. */
5897 {
5899 {
5914 }
5915 }
5917 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
5918 a MINUS_EXPR of a constant, we can convert it into a comparison with
5919 a revised constant as long as no overflow occurs. */
5931 /* Similarly for a NEGATE_EXPR. */
5940 /* If we have X - Y == 0, we can convert that to X == Y and similarly
5941 for !=. Don't do this for ordered comparisons due to overflow. */
5947 /* If we are widening one operand of an integer comparison,
5948 see if the other operand is similarly being widened. Perhaps we
5949 can do the comparison in the narrower type. */
5959 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
5960 constant, we can simplify it. */
5967 /* If we are comparing an ABS_EXPR with a constant, we can
5968 convert all the cases into explicit comparisons, but they may
5969 well not be faster than doing the ABS and one comparison.
5970 But ABS (X) <= C is a range comparison, which becomes a subtraction
5971 and a comparison, and is probably faster. */
5983 /* If this is an EQ or NE comparison with zero and ARG0 is
5984 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
5985 two operations, but the latter can be done in one less insn
5986 on machines that have only two-operand insns or on which a
5987 constant cannot be the first operand. */
5990 {
5993 return
6001 integer_one_node)),
6002 arg1));
6005 return
6013 integer_one_node)),
6014 arg1));
6015 }
6017 /* If this is an NE or EQ comparison of zero against the result of a
6018 signed MOD operation whose second operand is a power of 2, make
6019 the MOD operation unsigned since it is simpler and equivalent. */
6028 {
6035 }
6037 /* If this is an NE comparison of zero with an AND of one, remove the
6038 comparison since the AND will give the correct value. */
6044 /* If we have (A & C) == C where C is a power of 2, convert this into
6045 (A & C) != 0. Similarly for NE_EXPR. */
6053 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6054 and similarly for >= into !=. */
6070 return
6077 /* Simplify comparison of something with itself. (For IEEE
6078 floating-point, we can only do some of these simplifications.) */
6080 {
6082 {
6093 /* For NE, we can only do this simplification if integer. */
6096 /* ... fall through ... */
6102 }
6103 }
6105 /* An unsigned comparison against 0 can be simplified. */
6110 {
6112 {
6124 arg0);
6128 arg0);
6131 }
6132 }
6134 /* Comparisons with the highest or lowest possible integer of
6135 the specified size will have known values and an unsigned
6136 <= 0x7fffffff can be simplified. */
6137 {
6145 {
6151 {
6155 arg0);
6163 arg0);
6170 }
6177 {
6181 arg0);
6189 arg0);
6196 }
6202 /* signed_type does not work on pointer types. */
6204 {
6206 {
6210 return fold
6214 }
6215 }
6221 {
6225 arg0);
6233 arg0);
6240 }
6241 }
6242 }
6244 /* If we are comparing an expression that just has comparisons
6245 of two integer values, arithmetic expressions of those comparisons,
6246 and constants, we can simplify it. There are only three cases
6247 to check: the two values can either be equal, the first can be
6248 greater, or the second can be greater. Fold the expression for
6249 those three values. Since each value must be 0 or 1, we have
6250 eight possibilities, each of which corresponds to the constant 0
6251 or 1 or one of the six possible comparisons.
6253 This handles common cases like (a > b) == 0 but also handles
6254 expressions like ((x > y) - (y > x)) > 0, which supposedly
6255 occur in macroized code. */
6258 {
6263 /* Don't handle degenerate cases here; they should already
6264 have been handled anyway. */
6273 {
6277 /* We can't just pass T to eval_subst in case cval1 or cval2
6278 was the same as ARG1. */
6280 tree high_result
6283 arg1));
6284 tree equal_result
6287 arg1));
6288 tree low_result
6291 arg1));
6293 /* All three of these results should be 0 or 1. Confirm they
6294 are. Then use those values to select the proper code
6295 to use. */
6303 {
6304 /* Make a 3-bit mask with the high-order bit being the
6305 value for `>', the next for '=', and the low for '<'. */
6309 {
6311 /* Always false. */
6332 /* Always true. */
6334 }
6339 else
6341 }
6342 }
6343 }
6345 /* If this is a comparison of a field, we may be able to simplify it. */
6349 /* Handle the constant case even without -O
6350 to make sure the warnings are given. */
6352 {
6355 }
6357 /* If this is a comparison of complex values and either or both sides
6358 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6359 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6360 This may prevent needless evaluations. */
6367 {
6380 type,
6383 }
6385 /* Optimize comparisons of strlen vs zero to a compare of the
6386 first character of the string vs zero. To wit,
6387 strlen(ptr) == 0 => *ptr == 0
6388 strlen(ptr) != 0 => *ptr != 0
6389 Other cases should reduce to one of these two (or a constant)
6390 due to the return value of strlen being unsigned. */
6395 {
6397 tree arglist;
6409 integer_zero_node));
6410 }
6412 /* From here on, the only cases we handle are when the result is
6413 known to be a constant.
6415 To compute GT, swap the arguments and do LT.
6416 To compute GE, do LT and invert the result.
6417 To compute LE, swap the arguments, do LT and invert the result.
6418 To compute NE, do EQ and invert the result.
6420 Therefore, the code below must handle only EQ and LT. */
6423 {
6426 }
6428 /* Note that it is safe to invert for real values here because we
6429 will check below in the one case that it matters. */
6434 {
6437 }
6439 /* Compute a result for LT or EQ if args permit;
6440 otherwise return T. */
6442 {
6445 else
6450 }
6453 /* Assume a nonexplicit constant cannot equal an explicit one,
6454 since such code would be undefined anyway.
6455 Exception: on sysvr4, using #pragma weak,
6456 a label can come out as 0. */
6463 #endif
6464 /* Two real constants can be compared explicitly. */
6466 {
6467 /* If either operand is a NaN, the result is false with two
6468 exceptions: First, an NE_EXPR is true on NaNs, but that case
6469 is already handled correctly since we will be inverting the
6470 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6471 or a GE_EXPR into a LT_EXPR, we must return true so that it
6472 will be inverted into false. */
6482 else
6486 }
6500 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6501 so all simple results must be passed through pedantic_non_lvalue. */
6503 return pedantic_non_lvalue
6508 /* If the second operand is zero, invert the comparison and swap
6509 the second and third operands. Likewise if the second operand
6510 is constant and the third is not or if the third operand is
6511 equivalent to the first operand of the comparison. */
6519 {
6520 /* See if this can be inverted. If it can't, possibly because
6521 it was a floating-point inequality comparison, don't do
6522 anything. */
6526 {
6530 /* arg1 should be the first argument of the new T. */
6533 }
6534 }
6536 /* If we have A op B ? A : C, we may be able to convert this to a
6537 simpler expression, depending on the operation and the values
6538 of B and C. Signed zeros prevent all of these transformations,
6539 for reasons given above each one. */
6545 {
6551 /* If we have A op 0 ? A : -A, consider applying the following
6552 transformations:
6554 A == 0? A : -A same as -A
6555 A != 0? A : -A same as A
6556 A >= 0? A : -A same as abs (A)
6557 A > 0? A : -A same as abs (A)
6558 A <= 0? A : -A same as -abs (A)
6559 A < 0? A : -A same as -abs (A)
6561 None of these transformations work for modes with signed
6562 zeros. If A is +/-0, the first two transformations will
6563 change the sign of the result (from +0 to -0, or vice
6564 versa). The last four will fix the sign of the result,
6565 even though the original expressions could be positive or
6566 negative, depending on the sign of A.
6568 Note that all these transformations are correct if A is
6569 NaN, since the two alternatives (A and -A) are also NaNs. */
6576 {
6578 return
6579 pedantic_non_lvalue
6581 negate_expr
6583 arg1))));
6591 return pedantic_non_lvalue
6599 return pedantic_non_lvalue
6603 arg1)))));
6606 }
6608 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
6609 A == 0 ? A : 0 is always 0 unless A is -0. Note that
6610 both transformations are correct when A is NaN: A != 0
6611 is then true, and A == 0 is false. */
6614 {
6619 }
6621 /* Try some transformations of A op B ? A : B.
6623 A == B? A : B same as B
6624 A != B? A : B same as A
6625 A >= B? A : B same as max (A, B)
6626 A > B? A : B same as max (B, A)
6627 A <= B? A : B same as min (A, B)
6628 A < B? A : B same as min (B, A)
6630 As above, these transformations don't work in the presence
6631 of signed zeros. For example, if A and B are zeros of
6632 opposite sign, the first two transformations will change
6633 the sign of the result. In the last four, the original
6634 expressions give different results for (A=+0, B=-0) and
6635 (A=-0, B=+0), but the transformed expressions do not.
6637 The first two transformations are correct if either A or B
6638 is a NaN. In the first transformation, the condition will
6639 be false, and B will indeed be chosen. In the case of the
6640 second transformation, the condition A != B will be true,
6641 and A will be chosen.
6643 The conversions to max() and min() are not correct if B is
6644 a number and A is not. The conditions in the original
6645 expressions will be false, so all four give B. The min()
6646 and max() versions would give a NaN instead. */
6649 {
6654 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
6659 {
6666 /* In C++ a ?: expression can be an lvalue, so put the
6667 operand which will be used if they are equal first
6668 so that we can convert this back to the
6669 corresponding COND_EXPR. */
6671 return pedantic_non_lvalue
6681 return pedantic_non_lvalue
6690 }
6691 }
6693 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
6694 we might still be able to simplify this. For example,
6695 if C1 is one less or one more than C2, this might have started
6696 out as a MIN or MAX and been transformed by this function.
6697 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
6703 {
6705 /* We can replace A with C1 in this case. */
6712 /* If C1 is C2 + 1, this is min(A, C2). */
6717 return pedantic_non_lvalue
6722 /* If C1 is C2 - 1, this is min(A, C2). */
6727 return pedantic_non_lvalue
6732 /* If C1 is C2 - 1, this is max(A, C2). */
6737 return pedantic_non_lvalue
6742 /* If C1 is C2 + 1, this is max(A, C2). */
6747 return pedantic_non_lvalue
6754 }
6755 }
6757 /* If the second operand is simpler than the third, swap them
6758 since that produces better jump optimization results. */
6764 {
6765 /* See if this can be inverted. If it can't, possibly because
6766 it was a floating-point inequality comparison, don't do
6767 anything. */
6771 {
6775 /* arg1 should be the first argument of the new T. */
6778 }
6779 }
6781 /* Convert A ? 1 : 0 to simply A. */
6784 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
6785 call to fold will try to move the conversion inside
6786 a COND, which will recurse. In that case, the COND_EXPR
6787 is probably the best choice, so leave it alone. */
6791 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
6792 operation is simply A & 2. */
6806 /* When pedantic, a compound expression can be neither an lvalue
6807 nor an integer constant expression. */
6810 /* Don't let (0, 0) be null pointer constant. */
6852 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
6853 appropriate. */
6858 {
6862 tree arg01;
6873 {
6887 arg01));
6888 }
6891 }
6894 /* Check for a built-in function. */
6899 {
6903 }
6909 }
6911 /* Determine if first argument is a multiple of second argument. Return 0 if
6912 it is not, or we cannot easily determined it to be.
6914 An example of the sort of thing we care about (at this point; this routine
6915 could surely be made more general, and expanded to do what the *_DIV_EXPR's
6916 fold cases do now) is discovering that
6918 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
6920 is a multiple of
6922 SAVE_EXPR (J * 8)
6924 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
6926 This code also handles discovering that
6928 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
6930 is a multiple of 8 so we don't have to worry about dealing with a
6931 possible remainder.
6933 Note that we *look* inside a SAVE_EXPR only to determine how it was
6934 calculated; it is not safe for fold to do much of anything else with the
6935 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
6936 at run time. For example, the latter example above *cannot* be implemented
6937 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
6938 evaluation time of the original SAVE_EXPR is not necessarily the same at
6939 the time the new expression is evaluated. The only optimization of this
6940 sort that would be valid is changing
6942 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
6944 divided by 8 to
6946 SAVE_EXPR (I) * SAVE_EXPR (J)
6948 (where the same SAVE_EXPR (J) is used in the original and the
6949 transformed version). */
6951 static int
6953 tree type;
6954 tree top;
6955 tree bottom;
6956 {
6964 {
6976 {
6980 /* const_binop may not detect overflow correctly,
6981 so check for it explicitly here. */
6990 }
6994 /* Can't handle conversions from non-integral or wider integral type. */
7000 /* .. fall through ... */
7016 }
7017 }
7019 /* Return true if `t' is known to be non-negative. */
7021 int
7023 tree t;
7024 {
7026 {
7067 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7069 else
7070 /* We don't know sign of `t', so be conservative and return false. */
7072 }
7073 }
7075 /* Return true if `r' is known to be non-negative.
7076 Only handles constants at the moment. */
7078 int
7080 rtx r;
7081 {
7083 {
7093 {
7095 rtx elt;
7100 {
7104 }
7107 }
7111 /* These are always nonnegative. */
7116 }
7117 }