| blob | 947d575dd9398442d5faab5fe6eea3f204f7dbf4 |
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. */
60 HOST_WIDE_INT));
63 HOST_WIDE_INT *));
86 HOST_WIDE_INT *,
107 static tree fold_binary_op_with_conditional_arg
111 #if defined(HOST_EBCDIC)
112 /* bit 8 is significant in EBCDIC */
113 #define CHARMASK 0xff
114 #else
115 #define CHARMASK 0x7f
116 #endif
118 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
119 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
120 and SUM1. Then this yields nonzero if overflow occurred during the
121 addition.
123 Overflow occurs if A and B have the same sign, but A and SUM differ in
124 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
125 sign. */
126 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
127 \f
128 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
129 We do that by representing the two-word integer in 4 words, with only
130 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
131 number. The value of the word is LOWPART + HIGHPART * BASE. */
133 #define LOWPART(x) \
134 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
135 #define HIGHPART(x) \
136 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
137 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
139 /* Unpack a two-word integer into 4 words.
140 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
141 WORDS points to the array of HOST_WIDE_INTs. */
143 static void
147 HOST_WIDE_INT hi;
148 {
153 }
155 /* Pack an array of 4 words into a two-word integer.
156 WORDS points to the array of words.
157 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
159 static void
164 {
167 }
168 \f
169 /* Make the integer constant T valid for its type by setting to 0 or 1 all
170 the bits in the constant that don't belong in the type.
172 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
173 nonzero, a signed overflow has already occurred in calculating T, so
174 propagate it.
176 Make the real constant T valid for its type by calling CHECK_FLOAT_VALUE,
177 if it exists. */
179 int
181 tree t;
183 {
185 HOST_WIDE_INT high;
189 {
190 #ifdef CHECK_FLOAT_VALUE
192 overflow);
193 #endif
195 }
205 else
208 /* First clear all bits that are beyond the type's precision. */
211 ;
215 else
216 {
220 }
222 /* Unsigned types do not suffer sign extension or overflow unless they
223 are a sizetype. */
229 /* If the value's sign bit is set, extend the sign. */
237 {
238 /* Value is negative:
239 set to 1 all the bits that are outside this type's precision. */
243 else
244 {
248 }
249 }
251 /* Return nonzero if signed overflow occurred. */
252 return
255 }
256 \f
257 /* Add two doubleword integers with doubleword result.
258 Each argument is given as two `HOST_WIDE_INT' pieces.
259 One argument is L1 and H1; the other, L2 and H2.
260 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
262 int
268 {
270 HOST_WIDE_INT h;
278 }
280 /* Negate a doubleword integer with doubleword result.
281 Return nonzero if the operation overflows, assuming it's signed.
282 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
283 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
285 int
288 HOST_WIDE_INT h1;
291 {
293 {
297 }
298 else
299 {
303 }
304 }
305 \f
306 /* Multiply two doubleword integers with doubleword result.
307 Return nonzero if the operation overflows, assuming it's signed.
308 Each argument is given as two `HOST_WIDE_INT' pieces.
309 One argument is L1 and H1; the other, L2 and H2.
310 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
312 int
318 {
333 {
336 {
338 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
340 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
344 }
346 }
350 /* Check for overflow by calculating the top half of the answer in full;
351 it should agree with the low half's sign bit. */
354 {
357 }
359 {
362 }
364 }
365 \f
366 /* Shift the doubleword integer in L1, H1 left by COUNT places
367 keeping only PREC bits of result.
368 Shift right if COUNT is negative.
369 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
370 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
372 void
380 {
384 {
387 }
389 #ifdef SHIFT_COUNT_TRUNCATED
392 #endif
395 {
396 /* Shifting by the host word size is undefined according to the
397 ANSI standard, so we must handle this as a special case. */
400 }
402 {
405 }
406 else
407 {
411 }
413 /* Sign extend all bits that are beyond the precision. */
421 ;
423 {
426 }
427 else
428 {
432 }
433 }
435 /* Shift the doubleword integer in L1, H1 right by COUNT places
436 keeping only PREC bits of result. COUNT must be positive.
437 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
438 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
440 void
448 {
451 signmask = (arith
455 #ifdef SHIFT_COUNT_TRUNCATED
458 #endif
461 {
462 /* Shifting by the host word size is undefined according to the
463 ANSI standard, so we must handle this as a special case. */
466 }
468 {
471 }
472 else
473 {
477 }
479 /* Zero / sign extend all bits that are beyond the precision. */
482 {
485 }
487 ;
489 {
492 }
493 else
494 {
498 }
499 }
500 \f
501 /* Rotate the doubleword integer in L1, H1 left by COUNT places
502 keeping only PREC bits of result.
503 Rotate right if COUNT is negative.
504 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
506 void
513 {
525 }
527 /* Rotate the doubleword integer in L1, H1 left by COUNT places
528 keeping only PREC bits of result. COUNT must be positive.
529 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
531 void
538 {
550 }
551 \f
552 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
553 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
554 CODE is a tree code for a kind of division, one of
555 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
556 or EXACT_DIV_EXPR
557 It controls how the quotient is rounded to an integer.
558 Return nonzero if the operation overflows.
559 UNS nonzero says do unsigned division. */
561 int
568 HOST_WIDE_INT hnum_orig;
570 HOST_WIDE_INT hden_orig;
573 {
589 /* calculate quotient sign and convert operands to unsigned. */
591 {
593 {
595 /* (minimum integer) / (-1) is the only overflow case. */
599 }
601 {
604 }
605 }
610 /* This unsigned division rounds toward zero. */
613 }
617 /* hden != 0 already checked. */
622 }
632 /* Special code for when the divisor < BASE. */
634 {
635 /* hnum != 0 already checked. */
637 {
641 }
642 }
643 else
644 {
645 /* Full double precision division,
646 with thanks to Don Knuth's "Seminumerical Algorithms". */
650 /* Find the highest non-zero divisor digit. */
653 {
656 }
658 /* Insure that the first digit of the divisor is at least BASE/2.
659 This is required by the quotient digit estimation algorithm. */
666 {
670 }
675 {
680 }
681 }
685 /* Main loop */
687 {
688 /* Guess the next quotient digit, quo_est, by dividing the first
689 two remaining dividend digits by the high order quotient digit.
690 quo_est is never low and is at most 2 high. */
697 else
700 /* Refine quo_est so it's usually correct, and at most one high. */
705 quo_est--;
707 /* Try QUO_EST as the quotient digit, by multiplying the
708 divisor by QUO_EST and subtracting from the remaining dividend.
709 Keep in mind that QUO_EST is the I - 1st digit. */
713 {
719 }
721 /* If quo_est was high by one, then num[i] went negative and
722 we need to correct things. */
724 {
725 quo_est--;
728 {
732 }
735 }
737 /* Store the quotient digit. */
739 }
740 }
744 finish_up:
745 /* if result is negative, make it so. */
749 /* compute trial remainder: rem = num - (quo * den) */
755 {
764 {
765 /* quo = quo - 1; */
768 }
769 else
776 {
779 }
780 else
786 {
792 /* Get absolute values */
798 /* If (2 * abs (lrem) >= abs (lden)) */
807 {
809 /* quo = quo - 1; */
812 else
813 /* quo = quo + 1; */
816 }
817 else
819 }
824 }
826 /* compute true remainder: rem = num - (quo * den) */
831 }
832 \f
833 /* Given T, an expression, return the negation of T. Allow for T to be
834 null, in which case return null. */
836 static tree
838 tree t;
839 {
840 tree type;
841 tree tem;
850 {
863 /* - (A - B) -> B - A */
873 }
876 }
877 \f
878 /* Split a tree IN into a constant, literal and variable parts that could be
879 combined with CODE to make IN. "constant" means an expression with
880 TREE_CONSTANT but that isn't an actual constant. CODE must be a
881 commutative arithmetic operation. Store the constant part into *CONP,
882 the literal in *LITP and return the variable part. If a part isn't
883 present, set it to null. If the tree does not decompose in this way,
884 return the entire tree as the variable part and the other parts as null.
886 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
887 case, we negate an operand that was subtracted. Except if it is a
888 literal for which we use *MINUS_LITP instead.
890 If NEGATE_P is true, we are negating all of IN, again except a literal
891 for which we use *MINUS_LITP instead.
893 If IN is itself a literal or constant, return it as appropriate.
895 Note that we do not guarantee that any of the three values will be the
896 same type as IN, but they will have the same signedness and mode. */
898 static tree
900 tree in;
904 {
911 /* Strip any conversions that don't change the machine mode or signedness. */
918 /* We can associate addition and subtraction together (even
919 though the C standard doesn't say so) for integers because
920 the value is not affected. For reals, the value might be
921 affected, so we can't. */
924 {
930 /* First see if either of the operands is a literal, then a constant. */
941 /* If we haven't dealt with either operand, this is not a case we can
942 decompose. Otherwise, VAR is either of the ones remaining, if any. */
947 else
950 /* Now do any needed negations. */
957 }
960 else
964 {
971 }
974 }
976 /* Re-associate trees split by the above function. T1 and T2 are either
977 expressions to associate or null. Return the new expression, if any. If
978 we build an operation, do it in TYPE and with CODE. */
980 static tree
984 tree type;
985 {
991 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
992 try to fold this since we will have infinite recursion. But do
993 deal with any NEGATE_EXPRs. */
996 {
1003 else
1005 }
1008 }
1009 \f
1010 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1011 to produce a new constant.
1013 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1015 static tree
1020 {
1024 HOST_WIDE_INT hi;
1026 HOST_WIDE_INT garbageh;
1027 tree t;
1030 int is_sizetype
1041 {
1061 /* It's unclear from the C standard whether shifts can overflow.
1062 The following code ignores overflow; perhaps a C standard
1063 interpretation ruling is needed. */
1093 /* This is a shortcut for a common special case. */
1098 {
1104 }
1106 /* ... fall through ... */
1110 {
1113 }
1116 {
1119 }
1126 /* This is a shortcut for a common special case. */
1131 {
1136 }
1138 /* ... fall through ... */
1154 else
1160 else
1166 }
1168 /* If this is for a sizetype, can be represented as one (signed)
1169 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1170 constants. */
1176 else
1177 {
1180 }
1183 = ((notrunc
1190 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1191 So check if force_fit_type truncated the value. */
1202 }
1204 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1205 constant. We assume ARG1 and ARG2 have the same data type, or at least
1206 are the same kind of constant and the same machine mode.
1208 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1210 static tree
1215 {
1223 {
1224 REAL_VALUE_TYPE d1;
1225 REAL_VALUE_TYPE d2;
1226 REAL_VALUE_TYPE value;
1227 tree t;
1232 /* If either operand is a NaN, just return it. Otherwise, set up
1233 for floating-point trap; we return an overflow. */
1243 value));
1253 }
1255 {
1261 tree t;
1264 {
1284 notrunc),
1290 notrunc));
1294 {
1295 tree magsquared
1299 notrunc);
1302 const_binop
1307 notrunc),
1309 notrunc),
1310 notrunc),
1312 const_binop
1317 notrunc),
1319 notrunc),
1320 notrunc),
1322 }
1327 }
1329 }
1331 }
1333 /* These are the hash table functions for the hash table of INTEGER_CST
1334 nodes of a sizetype. */
1336 /* Return the hash code code X, an INTEGER_CST. */
1338 static hashval_t
1341 {
1347 }
1349 /* Return non-zero if the value represented by *X (an INTEGER_CST tree node)
1350 is the same as that given by *Y, which is the same. */
1352 static int
1356 {
1364 }
1365 \f
1366 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1367 bits are given by NUMBER and of the sizetype represented by KIND. */
1369 tree
1371 HOST_WIDE_INT number;
1373 {
1375 }
1377 /* Likewise, but the desired type is specified explicitly. */
1379 tree
1381 HOST_WIDE_INT number;
1382 tree type;
1383 {
1389 {
1394 }
1396 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1397 hash table, we return the value from the hash table. Otherwise, we
1398 place that in the hash table and make a new node for the next time. */
1407 {
1413 }
1414 else
1416 }
1418 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1419 is a tree code. The type of the result is taken from the operands.
1420 Both must be the same type integer type and it must be a size type.
1421 If the operands are constant, so is the result. */
1423 tree
1427 {
1434 /* Handle the special case of two integer constants faster. */
1436 {
1437 /* And some specific cases even faster than that. */
1446 /* Handle general case of two integer constants. */
1448 }
1454 }
1456 /* Given two values, either both of sizetype or both of bitsizetype,
1457 compute the difference between the two values. Return the value
1458 in signed type corresponding to the type of the operands. */
1460 tree
1463 {
1465 tree ctype;
1471 /* If the type is already signed, just do the simple thing. */
1478 /* If either operand is not a constant, do the conversions to the signed
1479 type and subtract. The hardware will do the right thing with any
1480 overflow in the subtraction. */
1485 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1486 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1487 overflow) and negate (which can't either). Special-case a result
1488 of zero while we're here. */
1493 else
1496 }
1497 \f
1499 /* Given T, a tree representing type conversion of ARG1, a constant,
1500 return a constant tree representing the result of conversion. */
1502 static tree
1504 tree t;
1505 tree arg1;
1506 {
1511 {
1513 {
1514 /* If we would build a constant wider than GCC supports,
1515 leave the conversion unfolded. */
1519 /* If we are trying to make a sizetype for a small integer, use
1520 size_int to pick up cached types to reduce duplicate nodes. */
1526 /* Given an integer constant, make new constant with new type,
1527 appropriately sign-extended or truncated. */
1531 /* Indicate an overflow if (1) ARG1 already overflowed,
1532 or (2) force_fit_type indicates an overflow.
1533 Tell force_fit_type that an overflow has already occurred
1534 if ARG1 is a too-large unsigned value and T is signed.
1535 But don't indicate an overflow if converting a pointer. */
1545 }
1547 {
1548 /* Don't initialize these, use assignments.
1549 Initialized local aggregates don't work on old compilers. */
1550 REAL_VALUE_TYPE x;
1551 REAL_VALUE_TYPE l;
1552 REAL_VALUE_TYPE u;
1563 /* See if X will be in range after truncation towards 0.
1564 To compensate for truncation, move the bounds away from 0,
1565 but reject if X exactly equals the adjusted bounds. */
1569 /* If X is a NaN, use zero instead and show we have an overflow.
1570 Otherwise, range check. */
1574 && !no_upper_bound
1578 {
1582 }
1588 }
1590 }
1592 {
1596 {
1598 {
1602 }
1613 }
1614 }
1617 }
1618 \f
1619 /* Return an expr equal to X but certainly not valid as an lvalue. */
1621 tree
1623 tree x;
1624 {
1625 tree result;
1627 /* These things are certainly not lvalues. */
1638 }
1640 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1641 Zero means allow extended lvalues. */
1645 /* When pedantic, return an expr equal to X but certainly not valid as a
1646 pedantic lvalue. Otherwise, return X. */
1648 tree
1650 tree x;
1651 {
1654 else
1656 }
1657 \f
1658 /* Given a tree comparison code, return the code that is the logical inverse
1659 of the given code. It is not safe to do this for floating-point
1660 comparisons, except for NE_EXPR and EQ_EXPR. */
1662 static enum tree_code
1665 {
1667 {
1682 }
1683 }
1685 /* Similar, but return the comparison that results if the operands are
1686 swapped. This is safe for floating-point. */
1688 static enum tree_code
1691 {
1693 {
1707 }
1708 }
1710 /* Return nonzero if CODE is a tree code that represents a truth value. */
1712 static int
1715 {
1720 }
1721 \f
1722 /* Return nonzero if two operands are necessarily equal.
1723 If ONLY_CONST is non-zero, only return non-zero for constants.
1724 This function tests whether the operands are indistinguishable;
1725 it does not test whether they are equal using C's == operation.
1726 The distinction is important for IEEE floating point, because
1727 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1728 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1730 int
1734 {
1735 /* If both types don't have the same signedness, then we can't consider
1736 them equal. We must check this before the STRIP_NOPS calls
1737 because they may change the signedness of the arguments. */
1745 /* This is needed for conversions and for COMPONENT_REF.
1746 Might as well play it safe and always test this. */
1752 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1753 We don't care about side effects in that case because the SAVE_EXPR
1754 takes care of that for us. In all other cases, two expressions are
1755 equal if they have no side effects. If we have two identical
1756 expressions with side effects that should be treated the same due
1757 to the only side effects being identical SAVE_EXPR's, that will
1758 be detected in the recursive calls below. */
1764 /* Next handle constant cases, those for which we can return 1 even
1765 if ONLY_CONST is set. */
1768 {
1781 {
1791 {
1796 }
1799 }
1803 only_const)
1805 only_const));
1818 }
1824 {
1826 /* Two conversions are equal only if signedness and modes match. */
1842 /* For commutative ops, allow the other order. */
1855 /* If either of the pointer (or reference) expressions we are dereferencing
1856 contain a side effect, these cannot be equal. */
1862 {
1884 }
1893 }
1894 }
1895 \f
1896 /* Similar to operand_equal_p, but see if ARG0 might have been made by
1897 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
1899 When in doubt, return 0. */
1901 static int
1904 tree other;
1905 {
1917 /* Discard any conversions that don't change the modes of ARG0 and ARG1
1918 and see if the inner values are the same. This removes any
1919 signedness comparison, which doesn't matter here. */
1926 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
1927 actual comparison operand, ARG0.
1929 First throw away any conversions to wider types
1930 already present in the operands. */
1939 {
1942 /* Make sure shorter operand is extended the right way
1943 to match the longer operand. */
1949 }
1952 }
1953 \f
1954 /* See if ARG is an expression that is either a comparison or is performing
1955 arithmetic on comparisons. The comparisons must only be comparing
1956 two different values, which will be stored in *CVAL1 and *CVAL2; if
1957 they are non-zero it means that some operands have already been found.
1958 No variables may be used anywhere else in the expression except in the
1959 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
1960 the expression and save_expr needs to be called with CVAL1 and CVAL2.
1962 If this is true, return 1. Otherwise, return zero. */
1964 static int
1966 tree arg;
1969 {
1973 /* We can handle some of the 'e' cases here. */
1983 {
1984 /* If we've already found a CVAL1 or CVAL2, this expression is
1985 two complex to handle. */
1991 }
1994 {
2017 /* First see if we can handle the first operand, then the second. For
2018 the second operand, we know *CVAL1 can't be zero. It must be that
2019 one side of the comparison is each of the values; test for the
2020 case where this isn't true by failing if the two operands
2021 are the same. */
2030 ;
2034 ;
2035 else
2039 ;
2043 ;
2044 else
2051 }
2052 }
2053 \f
2054 /* ARG is a tree that is known to contain just arithmetic operations and
2055 comparisons. Evaluate the operations in the tree substituting NEW0 for
2056 any occurrence of OLD0 as an operand of a comparison and likewise for
2057 NEW1 and OLD1. */
2059 static tree
2061 tree arg;
2063 {
2068 /* We can handle some of the 'e' cases here. */
2076 {
2091 {
2108 }
2109 /* fall through - ??? */
2112 {
2116 /* We need to check both for exact equality and tree equality. The
2117 former will be true if the operand has a side-effect. In that
2118 case, we know the operand occurred exactly once. */
2131 }
2135 }
2136 }
2137 \f
2138 /* Return a tree for the case when the result of an expression is RESULT
2139 converted to TYPE and OMITTED was previously an operand of the expression
2140 but is now not needed (e.g., we folded OMITTED * 0).
2142 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2143 the conversion of RESULT to TYPE. */
2145 static tree
2148 {
2155 }
2157 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2159 static tree
2162 {
2169 }
2170 \f
2171 /* Return a simplified tree node for the truth-negation of ARG. This
2172 never alters ARG itself. We assume that ARG is an operation that
2173 returns a truth value (0 or 1). */
2175 tree
2177 tree arg;
2178 {
2185 /* If this is a comparison, we can simply invert it, except for
2186 floating-point non-equality comparisons, in which case we just
2187 enclose a TRUTH_NOT_EXPR around what we have. */
2190 {
2192 && !flag_unsafe_math_optimizations
2196 else
2199 }
2202 {
2217 /* Here we can invert either operand. We invert the first operand
2218 unless the second operand is a TRUTH_NOT_EXPR in which case our
2219 result is the XOR of the first operand with the inside of the
2220 negation of the second operand. */
2225 else
2280 }
2284 }
2286 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2287 operands are another bit-wise operation with a common input. If so,
2288 distribute the bit operations to save an operation and possibly two if
2289 constants are involved. For example, convert
2290 (A | B) & (A | C) into A | (B & C)
2291 Further simplification will occur if B and C are constants.
2293 If this optimization cannot be done, 0 will be returned. */
2295 static tree
2298 tree type;
2300 {
2301 tree common;
2311 {
2315 }
2317 {
2321 }
2323 {
2327 }
2329 {
2333 }
2334 else
2339 }
2340 \f
2341 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2342 starting at BITPOS. The field is unsigned if UNSIGNEDP is non-zero. */
2344 static tree
2346 tree inner;
2347 tree type;
2350 {
2357 }
2359 /* Optimize a bit-field compare.
2361 There are two cases: First is a compare against a constant and the
2362 second is a comparison of two items where the fields are at the same
2363 bit position relative to the start of a chunk (byte, halfword, word)
2364 large enough to contain it. In these cases we can avoid the shift
2365 implicit in bitfield extractions.
2367 For constants, we emit a compare of the shifted constant with the
2368 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2369 compared. For two fields at the same position, we do the ANDs with the
2370 similar mask and compare the result of the ANDs.
2372 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2373 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2374 are the left and right operands of the comparison, respectively.
2376 If the optimization described above can be done, we return the resulting
2377 tree. Otherwise we return zero. */
2379 static tree
2382 tree compare_type;
2384 {
2393 tree mask;
2394 tree offset;
2396 /* Get all the information about the extractions being done. If the bit size
2397 if the same as the size of the underlying object, we aren't doing an
2398 extraction at all and so can do nothing. We also don't want to
2399 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2400 then will no longer be able to replace it. */
2408 {
2409 /* If this is not a constant, we can only do something if bit positions,
2410 sizes, and signedness are the same. */
2418 }
2420 /* See if we can find a mode to refer to this field. We should be able to,
2421 but fail if we can't. */
2430 /* Set signed and unsigned types of the precision of this mode for the
2431 shifts below. */
2435 /* Compute the bit position and size for the new reference and our offset
2436 within it. If the new reference is the same size as the original, we
2437 won't optimize anything, so return zero. */
2447 /* Make the mask to be used against the extracted field. */
2457 /* If not comparing with constant, just rework the comparison
2458 and return. */
2463 mask),
2467 mask));
2469 /* Otherwise, we are handling the constant case. See if the constant is too
2470 big for the field. Warn and return a tree of for 0 (false) if so. We do
2471 this not only for its own sake, but to avoid having to test for this
2472 error case below. If we didn't, we might generate wrong code.
2474 For unsigned fields, the constant shifted right by the field length should
2475 be all zero. For signed fields, the high-order bits should agree with
2476 the sign bit. */
2479 {
2483 {
2489 }
2490 }
2491 else
2492 {
2496 {
2502 }
2503 }
2505 /* Single-bit compares should always be against zero. */
2507 {
2510 }
2512 /* Make a new bitfield reference, shift the constant over the
2513 appropriate number of bits and mask it with the computed mask
2514 (in case this was a signed field). If we changed it, make a new one. */
2517 {
2520 }
2530 rhs);
2531 }
2532 \f
2533 /* Subroutine for fold_truthop: decode a field reference.
2535 If EXP is a comparison reference, we return the innermost reference.
2537 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2538 set to the starting bit number.
2540 If the innermost field can be completely contained in a mode-sized
2541 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2543 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2544 otherwise it is not changed.
2546 *PUNSIGNEDP is set to the signedness of the field.
2548 *PMASK is set to the mask used. This is either contained in a
2549 BIT_AND_EXPR or derived from the width of the field.
2551 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2553 Return 0 if this is not a component reference or is one that we can't
2554 do anything with. */
2556 static tree
2559 tree exp;
2565 {
2568 tree unsigned_type;
2571 /* All the optimizations using this function assume integer fields.
2572 There are problems with FP fields since the type_for_size call
2573 below can fail for, e.g., XFmode. */
2580 {
2586 }
2595 /* Compute the mask to access the bitfield. */
2605 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2613 }
2615 /* Return non-zero if MASK represents a mask of SIZE ones in the low-order
2616 bit positions. */
2618 static int
2620 tree mask;
2622 {
2625 tree tmask;
2630 return
2637 }
2639 /* Subroutine for fold: determine if VAL is the INTEGER_CONST that
2640 represents the sign bit of EXP's type. If EXP represents a sign
2641 or zero extension, also test VAL against the unextended type.
2642 The return value is the (sub)expression whose sign bit is VAL,
2643 or NULL_TREE otherwise. */
2645 static tree
2647 tree exp;
2648 tree val;
2649 {
2651 HOST_WIDE_INT hi;
2653 tree t;
2655 /* Tree EXP must have a integral type. */
2660 /* Tree VAL must be an integer constant. */
2667 {
2670 }
2671 else
2672 {
2675 }
2680 /* Handle extension from a narrower type. */
2686 }
2688 /* Subroutine for fold_truthop: determine if an operand is simple enough
2689 to be evaluated unconditionally. */
2691 static int
2693 tree exp;
2694 {
2695 /* Strip any conversions that don't change the machine mode. */
2707 /* Don't regard global variables as simple. They may be
2708 allocated in ways unknown to the compiler (shared memory,
2709 #pragma weak, etc). */
2712 /* Loading a static variable is unduly expensive, but global
2713 registers aren't expensive. */
2715 }
2716 \f
2717 /* The following functions are subroutines to fold_range_test and allow it to
2718 try to change a logical combination of comparisons into a range test.
2720 For example, both
2721 X == 2 || X == 3 || X == 4 || X == 5
2722 and
2723 X >= 2 && X <= 5
2724 are converted to
2725 (unsigned) (X - 2) <= 3
2727 We describe each set of comparisons as being either inside or outside
2728 a range, using a variable named like IN_P, and then describe the
2729 range with a lower and upper bound. If one of the bounds is omitted,
2730 it represents either the highest or lowest value of the type.
2732 In the comments below, we represent a range by two numbers in brackets
2733 preceded by a "+" to designate being inside that range, or a "-" to
2734 designate being outside that range, so the condition can be inverted by
2735 flipping the prefix. An omitted bound is represented by a "-". For
2736 example, "- [-, 10]" means being outside the range starting at the lowest
2737 possible value and ending at 10, in other words, being greater than 10.
2738 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2739 always false.
2741 We set up things so that the missing bounds are handled in a consistent
2742 manner so neither a missing bound nor "true" and "false" need to be
2743 handled using a special case. */
2745 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2746 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2747 and UPPER1_P are nonzero if the respective argument is an upper bound
2748 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2749 must be specified for a comparison. ARG1 will be converted to ARG0's
2750 type if both are specified. */
2752 static tree
2755 tree type;
2758 {
2759 tree tem;
2763 /* If neither arg represents infinity, do the normal operation.
2764 Else, if not a comparison, return infinity. Else handle the special
2765 comparison rules. Note that most of the cases below won't occur, but
2766 are handled for consistency. */
2769 {
2774 }
2779 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2780 for neither. In real maths, we cannot assume open ended ranges are
2781 the same. But, this is computer arithmetic, where numbers are finite.
2782 We can therefore make the transformation of any unbounded range with
2783 the value Z, Z being greater than any representable number. This permits
2784 us to treat unbounded ranges as equal. */
2788 {
2809 }
2812 }
2813 \f
2814 /* Given EXP, a logical expression, set the range it is testing into
2815 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2816 actually being tested. *PLOW and *PHIGH will be made of the same type
2817 as the returned expression. If EXP is not a comparison, we will most
2818 likely not be returning a useful value and range. */
2820 static tree
2822 tree exp;
2825 {
2832 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2833 and see if we can refine the range. Some of the cases below may not
2834 happen, but it doesn't seem worth worrying about this. We "continue"
2835 the outer loop when we've changed something; otherwise we "break"
2836 the switch, which will "break" the while. */
2841 {
2845 {
2856 }
2858 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2859 lose a cast by accident. */
2864 {
2871 /* We can only do something if the range is testing for zero
2872 and if the second operand is an integer constant. Note that
2873 saying something is "in" the range we make is done by
2874 complementing IN_P since it will set in the initial case of
2875 being not equal to zero; "out" is leaving it alone. */
2882 {
2903 }
2907 /* If this is an unsigned comparison, we also know that EXP is
2908 greater than or equal to zero. We base the range tests we make
2909 on that fact, so we record it here so we can parse existing
2910 range tests. */
2912 {
2915 NULL_TREE))
2920 /* If the high bound is missing, but we
2921 have a low bound, reverse the range so
2922 it goes from zero to the low bound minus 1. */
2924 {
2929 }
2930 }
2934 /* (-x) IN [a,b] -> x in [-b, -a] */
2944 /* ~ X -> -X - 1 */
2953 /* If EXP is signed, any overflow in the computation is undefined,
2954 so we don't worry about it so long as our computations on
2955 the bounds don't overflow. For unsigned, overflow is defined
2956 and this is exactly the right thing. */
2965 /* Check for an unsigned range which has wrapped around the maximum
2966 value thus making n_high < n_low, and normalize it. */
2968 {
2974 /* If the range is of the form +/- [ x+1, x ], we won't
2975 be able to normalize it. But then, it represents the
2976 whole range or the empty set, so make it
2977 +/- [ -, - ]. */
2981 else
2983 }
2984 else
3007 /* If we're converting from an unsigned to a signed type,
3008 we will be doing the comparison as unsigned. The tests above
3009 have already verified that LOW and HIGH are both positive.
3011 So we have to make sure that the original unsigned value will
3012 be interpreted as positive. */
3014 {
3017 tree high_positive;
3019 /* A range without an upper bound is, naturally, unbounded.
3020 Since convert would have cropped a very large value, use
3021 the max value for the destination type. */
3022 high_positive
3030 /* If the low bound is specified, "and" the range with the
3031 range for which the original unsigned value will be
3032 positive. */
3034 {
3038 high_positive))
3042 }
3043 else
3044 {
3045 /* Otherwise, "or" the range with the range of the input
3046 that will be interpreted as negative. */
3050 high_positive))
3054 }
3055 }
3063 }
3066 }
3068 /* If EXP is a constant, we can evaluate whether this is true or false. */
3070 {
3077 }
3081 }
3082 \f
3083 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3084 type, TYPE, return an expression to test if EXP is in (or out of, depending
3085 on IN_P) the range. */
3087 static tree
3089 tree type;
3090 tree exp;
3093 {
3095 tree value;
3114 {
3116 {
3120 }
3122 }
3124 /* Optimize (c>=1) && (c<=127) into (signed char)c > 0. */
3126 {
3128 HOST_WIDE_INT hi;
3133 {
3136 }
3137 else
3138 {
3141 }
3144 {
3146 {
3149 }
3152 }
3153 }
3162 }
3163 \f
3164 /* Given two ranges, see if we can merge them into one. Return 1 if we
3165 can, 0 if we can't. Set the output range into the specified parameters. */
3167 static int
3173 {
3177 tree tem;
3187 /* Make range 0 be the range that starts first, or ends last if they
3188 start at the same value. Swap them if it isn't. */
3191 || (lowequal
3194 {
3198 }
3200 /* Now flag two cases, whether the ranges are disjoint or whether the
3201 second range is totally subsumed in the first. Note that the tests
3202 below are simplified by the ones above. */
3208 /* We now have four cases, depending on whether we are including or
3209 excluding the two ranges. */
3211 {
3212 /* If they don't overlap, the result is false. If the second range
3213 is a subset it is the result. Otherwise, the range is from the start
3214 of the second to the end of the first. */
3219 else
3221 }
3224 {
3225 /* If they don't overlap, the result is the first range. If they are
3226 equal, the result is false. If the second range is a subset of the
3227 first, and the ranges begin at the same place, we go from just after
3228 the end of the first range to the end of the second. If the second
3229 range is not a subset of the first, or if it is a subset and both
3230 ranges end at the same place, the range starts at the start of the
3231 first range and ends just before the second range.
3232 Otherwise, we can't describe this as a single range. */
3238 {
3242 }
3244 {
3248 }
3249 else
3251 }
3254 {
3255 /* If they don't overlap, the result is the second range. If the second
3256 is a subset of the first, the result is false. Otherwise,
3257 the range starts just after the first range and ends at the
3258 end of the second. */
3263 else
3264 {
3268 }
3269 }
3271 else
3272 {
3273 /* The case where we are excluding both ranges. Here the complex case
3274 is if they don't overlap. In that case, the only time we have a
3275 range is if they are adjacent. If the second is a subset of the
3276 first, the result is the first. Otherwise, the range to exclude
3277 starts at the beginning of the first range and ends at the end of the
3278 second. */
3280 {
3287 else
3289 }
3292 else
3294 }
3298 }
3299 \f
3300 /* EXP is some logical combination of boolean tests. See if we can
3301 merge it into some range test. Return the new tree if so. */
3303 static tree
3305 tree exp;
3306 {
3313 tree tem;
3315 /* If this is an OR operation, invert both sides; we will invert
3316 again at the end. */
3320 /* If both expressions are the same, if we can merge the ranges, and we
3321 can build the range test, return it or it inverted. If one of the
3322 ranges is always true or always false, consider it to be the same
3323 expression as the other. */
3333 /* On machines where the branch cost is expensive, if this is a
3334 short-circuited branch and the underlying object on both sides
3335 is the same, make a non-short-circuit operation. */
3341 {
3342 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3343 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3344 which cases we can't do this. */
3353 {
3365 }
3366 }
3369 }
3370 \f
3371 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3372 bit value. Arrange things so the extra bits will be set to zero if and
3373 only if C is signed-extended to its full width. If MASK is nonzero,
3374 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3376 static tree
3378 tree c;
3381 tree mask;
3382 {
3385 tree temp;
3390 /* We work by getting just the sign bit into the low-order bit, then
3391 into the high-order bit, then sign-extend. We then XOR that value
3392 with C. */
3396 /* We must use a signed type in order to get an arithmetic right shift.
3397 However, we must also avoid introducing accidental overflows, so that
3398 a subsequent call to integer_zerop will work. Hence we must
3399 do the type conversion here. At this point, the constant is either
3400 zero or one, and the conversion to a signed type can never overflow.
3401 We could get an overflow if this conversion is done anywhere else. */
3409 /* If necessary, convert the type back to match the type of C. */
3414 }
3415 \f
3416 /* Find ways of folding logical expressions of LHS and RHS:
3417 Try to merge two comparisons to the same innermost item.
3418 Look for range tests like "ch >= '0' && ch <= '9'".
3419 Look for combinations of simple terms on machines with expensive branches
3420 and evaluate the RHS unconditionally.
3422 For example, if we have p->a == 2 && p->b == 4 and we can make an
3423 object large enough to span both A and B, we can do this with a comparison
3424 against the object ANDed with the a mask.
3426 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3427 operations to do this with one comparison.
3429 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3430 function and the one above.
3432 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3433 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3435 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3436 two operands.
3438 We return the simplified tree or 0 if no optimization is possible. */
3440 static tree
3444 {
3445 /* If this is the "or" of two comparisons, we can do something if
3446 the comparisons are NE_EXPR. If this is the "and", we can do something
3447 if the comparisons are EQ_EXPR. I.e.,
3448 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3450 WANTED_CODE is this operation code. For single bit fields, we can
3451 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3452 comparison for one-bit fields. */
3472 /* Start by getting the comparison codes. Fail if anything is volatile.
3473 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3474 it were surrounded with a NE_EXPR. */
3499 /* If the RHS can be evaluated unconditionally and its operands are
3500 simple, it wins to evaluate the RHS unconditionally on machines
3501 with expensive branches. In this case, this isn't a comparison
3502 that can be merged. Avoid doing this if the RHS is a floating-point
3503 comparison since those can trap. */
3509 {
3510 /* Convert (a != 0) || (b != 0) into (a | b) != 0. */
3518 integer_zero_node);
3520 /* Convert (a == 0) && (b == 0) into (a | b) == 0. */
3528 integer_zero_node);
3531 }
3533 /* See if the comparisons can be merged. Then get all the parameters for
3534 each side. */
3558 /* It must be true that the inner operation on the lhs of each
3559 comparison must be the same if we are to be able to do anything.
3560 Then see if we have constants. If not, the same must be true for
3561 the rhs's. */
3572 else
3575 /* If either comparison code is not correct for our logical operation,
3576 fail. However, we can convert a one-bit comparison against zero into
3577 the opposite comparison against that bit being set in the field. */
3581 {
3583 {
3584 /* Make the left operand unsigned, since we are only interested
3585 in the value of one bit. Otherwise we are doing the wrong
3586 thing below. */
3589 }
3590 else
3592 }
3594 /* This is analogous to the code for l_const above. */
3596 {
3598 {
3601 }
3602 else
3604 }
3606 /* See if we can find a mode that contains both fields being compared on
3607 the left. If we can't, fail. Otherwise, update all constants and masks
3608 to be relative to a field of that size. */
3613 volatilep);
3623 {
3626 }
3634 {
3642 {
3646 wanted_code == NE_EXPR
3648 }
3649 }
3651 {
3659 {
3663 wanted_code == NE_EXPR
3665 }
3666 }
3668 /* If the right sides are not constant, do the same for it. Also,
3669 disallow this optimization if a size or signedness mismatch occurs
3670 between the left and right sides. */
3672 {
3675 /* Make sure the two fields on the right
3676 correspond to the left without being swapped. */
3684 volatilep);
3694 {
3697 }
3704 /* Make a mask that corresponds to both fields being compared.
3705 Do this for both items being compared. If the operands are the
3706 same size and the bits being compared are in the same position
3707 then we can do this by masking both and comparing the masked
3708 results. */
3712 {
3724 }
3726 /* There is still another way we can do something: If both pairs of
3727 fields being compared are adjacent, we may be able to make a wider
3728 field containing them both.
3730 Note that we still must mask the lhs/rhs expressions. Furthermore,
3731 the mask must be shifted to account for the shift done by
3732 make_bit_field_ref. */
3737 {
3738 tree type;
3750 /* Convert to the smaller type before masking out unwanted bits. */
3753 {
3755 {
3759 }
3761 {
3765 }
3766 }
3775 }
3778 }
3780 /* Handle the case of comparisons with constants. If there is something in
3781 common between the masks, those bits of the constants must be the same.
3782 If not, the condition is always false. Test for this to avoid generating
3783 incorrect code below. */
3788 {
3790 {
3793 }
3794 else
3795 {
3798 }
3799 }
3801 /* Construct the expression we will return. First get the component
3802 reference we will make. Unless the mask is all ones the width of
3803 that field, perform the mask operation. Then compare with the
3804 merged constant. */
3814 }
3815 \f
3816 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3817 constant. */
3819 static tree
3821 tree t;
3822 {
3827 tree minmax_const;
3829 tree inner;
3839 /* If something does not permit us to optimize, return the original tree. */
3847 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3848 and GT_EXPR, doing the rest with recursive calls using logical
3849 simplifications. */
3851 {
3853 return
3857 return
3859 optimize_minmax_comparison
3861 optimize_minmax_comparison
3866 /* MAX (X, 0) == 0 -> X <= 0 */
3870 /* MAX (X, 0) == 5 -> X == 5 */
3874 /* MAX (X, 0) == -1 -> false */
3878 /* MIN (X, 0) == 0 -> X >= 0 */
3882 /* MIN (X, 0) == 5 -> false */
3885 else
3886 /* MIN (X, 0) == -1 -> X == -1 */
3891 /* MAX (X, 0) > 0 -> X > 0
3892 MAX (X, 0) > 5 -> X > 5 */
3896 /* MAX (X, 0) > -1 -> true */
3900 /* MIN (X, 0) > 0 -> false
3901 MIN (X, 0) > 5 -> false */
3904 else
3905 /* MIN (X, 0) > -1 -> X > -1 */
3910 }
3911 }
3912 \f
3913 /* T is an integer expression that is being multiplied, divided, or taken a
3914 modulus (CODE says which and what kind of divide or modulus) by a
3915 constant C. See if we can eliminate that operation by folding it with
3916 other operations already in T. WIDE_TYPE, if non-null, is a type that
3917 should be used for the computation if wider than our type.
3919 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
3920 (X * 2) + (Y * 4). We must, however, be assured that either the original
3921 expression would not overflow or that overflow is undefined for the type
3922 in the language in question.
3924 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
3925 the machine has a multiply-accumulate insn or that this is part of an
3926 addressing calculation.
3928 If we return a non-null expression, it is an equivalent form of the
3929 original computation, but need not be in the original type. */
3931 static tree
3933 tree t;
3934 tree c;
3936 tree wide_type;
3937 {
3947 /* Don't deal with constants of zero here; they confuse the code below. */
3957 /* Note that we need not handle conditional operations here since fold
3958 already handles those cases. So just do arithmetic here. */
3960 {
3962 /* For a constant, we can always simplify if we are a multiply
3963 or (for divide and modulus) if it is a multiple of our constant. */
3970 /* If op0 is an expression, and is unsigned, and the type is
3971 smaller than ctype, then we cannot widen the expression. */
3983 /* Pass the constant down and see if we can make a simplification. If
3984 we can, replace this expression with the inner simplification for
3985 possible later conversion to our or some other type. */
3997 /* If widening the type changes the signedness, then we can't perform
3998 this optimization as that changes the result. */
4002 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
4005 {
4011 }
4021 /* If this has not been evaluated and the operand has no side effects,
4022 we can see if we can do something inside it and make a new one.
4023 Note that this test is overly conservative since we can do this
4024 if the only reason it had side effects is that it was another
4025 similar SAVE_EXPR, but that isn't worth bothering with. */
4028 wide_type)))
4029 {
4036 }
4040 /* If the second operand is constant, this is a multiplication
4041 or floor division, by a power of two, so we can treat it that
4042 way unless the multiplier or divisor overflows. */
4044 /* const_binop may not detect overflow correctly,
4045 so check for it explicitly here. */
4059 /* See if we can eliminate the operation on both sides. If we can, we
4060 can return a new PLUS or MINUS. If we can't, the only remaining
4061 cases where we can do anything are if the second operand is a
4062 constant. */
4067 /* If not multiplication, we can only do this if either operand
4068 is divisible by c. */
4074 /* If this was a subtraction, negate OP1 and set it to be an addition.
4075 This simplifies the logic below. */
4082 /* If either OP1 or C are negative, this optimization is not safe for
4083 some of the division and remainder types while for others we need
4084 to change the code. */
4086 {
4094 }
4096 /* If it's a multiply or a division/modulus operation of a multiple
4097 of our constant, do the operation and verify it doesn't overflow. */
4100 {
4104 }
4105 else
4108 /* If we have an unsigned type is not a sizetype, we cannot widen
4109 the operation since it will change the result if the original
4110 computation overflowed. */
4116 /* If we were able to eliminate our operation from the first side,
4117 apply our operation to the second side and reform the PLUS. */
4121 /* The last case is if we are a multiply. In that case, we can
4122 apply the distributive law to commute the multiply and addition
4123 if the multiplication of the constants doesn't overflow. */
4128 op1));
4133 /* We have a special case here if we are doing something like
4134 (C * 8) % 4 since we know that's zero. */
4141 /* ... fall through ... */
4145 /* If we can extract our operation from the LHS, do so and return a
4146 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4147 do something only if the second operand is a constant. */
4159 /* If these are the same operation types, we can associate them
4160 assuming no overflow. */
4167 /* If these operations "cancel" each other, we have the main
4168 optimizations of this pass, which occur when either constant is a
4169 multiple of the other, in which case we replace this with either an
4170 operation or CODE or TCODE.
4172 If we have an unsigned type that is not a sizetype, we cannot do
4173 this since it will change the result if the original computation
4174 overflowed. */
4181 {
4192 }
4197 }
4200 }
4201 \f
4202 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4203 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4204 that we may sometimes modify the tree. */
4206 static tree
4208 tree t;
4209 tree s;
4210 {
4213 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4218 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4219 don't bother handling any other types. */
4221 {
4225 }
4230 {
4233 }
4236 }
4237 \f
4238 /* Return a node which has the indicated constant VALUE (either 0 or
4239 1), and is of the indicated TYPE. */
4241 static tree
4244 tree type;
4245 {
4250 integer_zero_node);
4251 else
4252 {
4257 }
4258 }
4260 /* Utility function for the following routine, to see how complex a nesting of
4261 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4262 we don't care (to avoid spending too much time on complex expressions.). */
4264 static int
4266 tree expr;
4268 {
4279 }
4281 /* Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
4282 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4283 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4284 expression, and ARG to `a'. If COND_FIRST_P is non-zero, then the
4285 COND is the first argument to CODE; otherwise (as in the example
4286 given here), it is the second argument. TYPE is the type of the
4287 original expression. */
4289 static tree
4292 tree type;
4293 tree cond;
4294 tree arg;
4296 {
4300 /* In the end, we'll produce a COND_EXPR. Both arms of the
4301 conditional expression will be binary operations. The left-hand
4302 side of the expression to be executed if the condition is true
4303 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4304 of the expression to be executed if the condition is true will be
4305 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4306 but apply to the expression to be executed if the conditional is
4307 false. */
4312 /* These are the codes to use for the left-hand side and right-hand
4313 side of the COND_EXPR. Normally, they are the same as CODE. */
4316 /* And these are the types of the expressions. */
4321 {
4325 }
4326 else
4327 {
4331 }
4334 {
4338 /* If this operand throws an expression, then it does not make
4339 sense to try to perform a logical or arithmetic operation
4340 involving it. Instead of building `a + throw 3' for example,
4341 we simply build `a, throw 3'. */
4343 {
4347 }
4349 {
4353 }
4354 }
4355 else
4356 {
4361 }
4363 /* If ARG is complex we want to make sure we only evaluate
4364 it once. Though this is only required if it is volatile, it
4365 might be more efficient even if it is not. However, if we
4366 succeed in folding one part to a constant, we do not need
4367 to make this SAVE_EXPR. Since we do this optimization
4368 primarily to see if we do end up with constant and this
4369 SAVE_EXPR interferes with later optimizations, suppressing
4370 it when we can is important.
4372 If we are not in a function, we can't make a SAVE_EXPR, so don't
4373 try to do so. Don't try to see if the result is a constant
4374 if an arm is a COND_EXPR since we get exponential behavior
4375 in that case. */
4382 {
4392 }
4405 else
4407 }
4409 \f
4410 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4412 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4413 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4414 ADDEND is the same as X.
4416 X + 0 and X - 0 both give X when X is NaN, infinite, or non-zero
4417 and finite. The problematic cases are when X is zero, and its mode
4418 has signed zeros. In the case of rounding towards -infinity,
4419 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4420 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4422 static bool
4426 {
4430 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4434 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4439 /* The mode has signed zeros, and we have to honor their sign.
4440 In this situation, there is only one case we can return true for.
4441 X - 0 is the same as X unless rounding towards -infinity is
4442 supported. */
4444 }
4447 /* Perform constant folding and related simplification of EXPR.
4448 The related simplifications include x*1 => x, x*0 => 0, etc.,
4449 and application of the associative law.
4450 NOP_EXPR conversions may be removed freely (as long as we
4451 are careful not to change the C type of the overall expression)
4452 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4453 but we can constant-fold them if they have constant operands. */
4455 tree
4457 tree expr;
4458 {
4461 tree tem;
4467 /* WINS will be nonzero when the switch is done
4468 if all operands are constant. */
4471 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4472 Likewise for a SAVE_EXPR that's already been evaluated. */
4476 /* Return right away if a constant. */
4480 #ifdef MAX_INTEGER_COMPUTATION_MODE
4482 #endif
4485 {
4486 tree subop;
4488 /* Special case for conversion ops that can have fixed point args. */
4491 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4497 else
4502 )
4503 /* Note that TREE_CONSTANT isn't enough:
4504 static var addresses are constant but we can't
4505 do arithmetic on them. */
4507 }
4509 {
4513 {
4515 tree subop;
4521 {
4522 /* Signedness matters here. Perhaps we can refine this
4523 later. */
4525 }
4526 else
4527 /* Strip any conversions that don't change the mode. */
4532 else
4537 /* Note that TREE_CONSTANT isn't enough:
4538 static var addresses are constant but we can't
4539 do arithmetic on them. */
4546 }
4547 }
4549 /* If this is a commutative operation, and ARG0 is a constant, move it
4550 to ARG1 to reduce the number of tests below. */
4555 {
4560 }
4562 /* Now WINS is set as described above,
4563 ARG0 is the first operand of EXPR,
4564 and ARG1 is the second operand (if it has more than one operand).
4566 First check for cases where an arithmetic operation is applied to a
4567 compound, conditional, or comparison operation. Push the arithmetic
4568 operation inside the compound or conditional to see if any folding
4569 can then be done. Convert comparison to conditional for this purpose.
4570 The also optimizes non-constant cases that used to be done in
4571 expand_expr.
4573 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4574 one of the operands is a comparison and the other is a comparison, a
4575 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4576 code below would make the expression more complex. Change it to a
4577 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4578 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4590 {
4600 }
4603 {
4608 {
4613 /* If this was a conversion, and all we did was to move into
4614 inside the COND_EXPR, bring it back out. But leave it if
4615 it is a conversion from integer to integer and the
4616 result precision is no wider than a word since such a
4617 conversion is cheap and may be optimized away by combine,
4618 while it couldn't if it were outside the COND_EXPR. Then return
4619 so we don't get into an infinite recursion loop taking the
4620 conversion out and then back in. */
4630 && (INTEGRAL_TYPE_P
4641 }
4646 }
4649 {
4662 return
4676 return
4679 }
4690 {
4706 /* Other kinds of FIX are not handled properly by fold_convert. */
4711 /* Handle cases of two conversions in a row. */
4714 {
4734 /* In addition to the cases of two conversions in a row
4735 handled below, if we are converting something to its own
4736 type via an object of identical or wider precision, neither
4737 conversion is needed. */
4743 /* Likewise, if the intermediate and final types are either both
4744 float or both integer, we don't need the middle conversion if
4745 it is wider than the final type and doesn't change the signedness
4746 (for integers). Avoid this if the final type is a pointer
4747 since then we sometimes need the inner conversion. Likewise if
4748 the outer has a precision not equal to the size of its mode. */
4758 /* If we have a sign-extension of a zero-extended value, we can
4759 replace that by a single zero-extension. */
4765 /* Two conversions in a row are not needed unless:
4766 - some conversion is floating-point (overstrict for now), or
4767 - the intermediate type is narrower than both initial and
4768 final, or
4769 - the intermediate type and innermost type differ in signedness,
4770 and the outermost type is wider than the intermediate, or
4771 - the initial type is a pointer type and the precisions of the
4772 intermediate and final types differ, or
4773 - the final type is a pointer type and the precisions of the
4774 initial and intermediate types differ. */
4788 }
4792 /* Detect assigning a bitfield. */
4795 {
4796 /* Don't leave an assignment inside a conversion
4797 unless assigning a bitfield. */
4800 /* First do the assignment, then return converted constant. */
4804 }
4806 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
4807 constants (if x has signed type, the sign bit cannot be set
4808 in c). This folds extension into the BIT_AND_EXPR. */
4813 {
4823 <= HOST_BITS_PER_WIDE_INT
4825 {
4832 #ifdef LOAD_EXTEND_OP
4836 {
4840 }
4841 #endif
4842 }
4847 }
4850 {
4853 }
4864 {
4868 }
4877 {
4879 {
4881 HOST_WIDE_INT high;
4892 }
4895 }
4899 /* Convert - (a - b) to (b - a) for non-floating-point. */
4909 {
4911 {
4912 /* If the value is unsigned, then the absolute value is
4913 the same as the ordinary value. */
4916 /* Similarly, if the value is non-negative. */
4919 /* If the value is negative, then the absolute value is
4920 its negation. */
4921 else
4922 {
4924 HOST_WIDE_INT high;
4935 }
4936 }
4938 {
4942 }
4943 }
4970 {
4977 }
4983 /* A + (-B) -> A - B */
4986 /* (-A) + B -> B - A */
4990 {
4994 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
4995 with a constant, and the two constants have no bits in common,
4996 we should treat this as a BIT_IOR_EXPR since this may produce more
4997 simplifications. */
5005 {
5008 }
5010 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
5011 (plus (plus (mult) (mult)) (foo)) so that we can
5012 take advantage of the factoring cases below. */
5017 {
5022 else
5033 parg1));
5038 parg0));
5039 }
5042 {
5046 /* (A * C) + (B * C) -> (A+B) * C.
5047 We are most concerned about the case where C is a constant,
5048 but other combinations show up during loop reduction. Since
5049 it is not difficult, try all four possibilities. */
5066 /* No identical multiplicands; see if we can find a common
5067 power-of-two factor in non-power-of-two multiplies. This
5068 can help in multi-dimensional array access. */
5073 {
5078 /* Move min of absolute values to int11. */
5081 {
5085 }
5088 {
5093 }
5094 }
5099 same));
5100 }
5101 }
5103 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5107 /* Likewise if the operands are reversed. */
5111 bit_rotate:
5112 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5113 is a rotate of A by C1 bits. */
5114 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5115 is a rotate of A by B bits. */
5116 {
5125 {
5144 {
5152 TYPE_PRECISION
5157 ? LROTATE_EXPR
5160 }
5162 {
5170 TYPE_PRECISION
5175 ? LROTATE_EXPR
5178 }
5179 }
5180 }
5182 associate:
5183 /* In most languages, can't associate operations on floats through
5184 parentheses. Rather than remember where the parentheses were, we
5185 don't associate floats at all. It shouldn't matter much. However,
5186 associating multiplications is only very slightly inaccurate, so do
5187 that if -funsafe-math-optimizations is specified. */
5192 {
5196 /* Split both trees into variables, constants, and literals. Then
5197 associate each group together, the constants with literals,
5198 then the result with variables. This increases the chances of
5199 literals being recombined later and of generating relocatable
5200 expressions for the sum of a constant and literal. */
5205 /* Only do something if we found more than two objects. Otherwise,
5206 nothing has changed and we risk infinite recursion. */
5211 {
5212 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5221 /* Preserve the MINUS_EXPR if the negative part of the literal is
5222 greater than the positive part. Otherwise, the multiplicative
5223 folding code (i.e extract_muldiv) may be fooled in case
5224 unsigned constants are substracted, like in the following
5225 example: ((X*2 + 4) - 8U)/2. */
5227 {
5229 {
5233 }
5234 else
5235 {
5239 }
5240 }
5242 {
5246 else
5247 {
5252 }
5253 }
5257 }
5258 }
5260 binary:
5264 {
5265 /* The return value should always have
5266 the same type as the original expression. */
5271 }
5275 /* A - (-B) -> A + B */
5278 /* (-A) - CST -> (-CST) - A for floating point (what about ints ?) */
5280 return
5287 {
5293 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5294 about the case where C is a constant, just try one of the
5295 four possibilities. */
5305 }
5307 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5311 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5312 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5313 (-ARG1 + ARG0) reduces to -ARG1. */
5317 /* Fold &x - &x. This can happen from &x.foo - &x.
5318 This is unsafe for certain floats even in non-IEEE formats.
5319 In IEEE, it is unsafe because it does wrong for NaNs.
5320 Also note that operand_equal_p is always false if an operand
5321 is volatile. */
5330 /* (-A) * (-B) -> A * B */
5336 {
5342 /* (a * (1 << b)) is (a << b) */
5357 }
5358 else
5359 {
5360 /* Maybe fold x * 0 to 0. The expressions aren't the same
5361 when x is NaN, since x * 0 is also NaN. Nor are they the
5362 same in modes with signed zeros, since multiplying a
5363 negative value by 0 gives -0, not +0. */
5368 /* In IEEE floating point, x*1 is not equivalent to x for snans.
5369 However, ANSI says we can drop signals,
5370 so we can do this anyway. */
5374 /* Transform x * -1.0 into -x. This should be safe for NaNs,
5375 signed zeros and signed infinities, but is currently
5376 restricted to "unsafe math optimizations" just in case. */
5381 /* x*2 is x+x */
5385 {
5388 }
5389 }
5393 bit_ior:
5402 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5404 This results in more efficient code for machines without a NAND
5405 instruction. Combine will canonicalize to the first form
5406 which will allow use of NAND instructions provided by the
5407 backend if they exist. */
5410 {
5415 }
5417 /* See if this can be simplified into a rotate first. If that
5418 is unsuccessful continue in the association code. */
5427 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5428 with a constant, and the two constants have no bits in common,
5429 we should treat this as a BIT_IOR_EXPR since this may produce more
5430 simplifications. */
5438 {
5441 }
5443 /* See if this can be simplified into a rotate first. If that
5444 is unsuccessful continue in the association code. */
5448 bit_and:
5456 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5459 {
5460 unsigned int prec
5467 }
5469 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5471 This results in more efficient code for machines without a NOR
5472 instruction. Combine will canonicalize to the first form
5473 which will allow use of NOR instructions provided by the
5474 backend if they exist. */
5477 {
5482 }
5492 {
5496 }
5500 /* Don't touch a floating-point divide by zero unless the mode
5501 of the constant can represent infinity. */
5507 /* (-A) / (-B) -> A / B */
5512 /* In IEEE floating point, x/1 is not equivalent to x for snans.
5513 However, ANSI says we can drop signals, so we can do this anyway. */
5517 /* If ARG1 is a constant, we can convert this to a multiply by the
5518 reciprocal. This does not have the same rounding properties,
5519 so only do this if -funsafe-math-optimizations. We can actually
5520 always safely do it if ARG1 is a power of two, but it's hard to
5521 tell if it is or not in a portable manner. */
5523 {
5528 /* Find the reciprocal if optimizing and the result is exact. */
5530 {
5531 REAL_VALUE_TYPE r;
5534 {
5537 }
5538 }
5539 }
5540 /* Convert A/B/C to A/(B*C). */
5543 {
5546 arg1)));
5547 }
5548 /* Convert A/(B/C) to (A/B)*C. */
5551 {
5556 }
5569 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5570 operation, EXACT_DIV_EXPR.
5572 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5573 At one time others generated faster code, it's not clear if they do
5574 after the last round to changes to the DIV code in expmed.c. */
5608 /* Since negative shift count is not well-defined,
5609 don't try to compute it in the compiler. */
5612 /* Rewrite an LROTATE_EXPR by a constant into an
5613 RROTATE_EXPR by a new constant. */
5615 {
5619 = const_binop
5626 }
5628 /* If we have a rotate of a bit operation with the rotate count and
5629 the second operand of the bit operation both constant,
5630 permute the two operations. */
5643 /* Two consecutive rotates adding up to the width of the mode can
5644 be ignored. */
5675 /* Note that the operand of this must be an int
5676 and its values must be 0 or 1.
5677 ("true" is a fixed value perhaps depending on the language,
5678 but we don't handle values other than 1 correctly yet.) */
5680 /* Avoid infinite recursion. */
5686 /* Note that the operands of this must be ints
5687 and their values must be 0 or 1.
5688 ("true" is a fixed value perhaps depending on the language.) */
5689 /* If first arg is constant zero, return it. */
5693 /* If either arg is constant true, drop it. */
5697 /* Preserve sequence points. */
5700 /* If second arg is constant zero, result is zero, but first arg
5701 must be evaluated. */
5704 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
5705 case will be handled here. */
5709 truth_andor:
5710 /* We only do these simplifications if we are optimizing. */
5714 /* Check for things like (A || B) && (A || C). We can convert this
5715 to A || (B && C). Note that either operator can be any of the four
5716 truth and/or operations and the transformation will still be
5717 valid. Also note that we only care about order for the
5718 ANDIF and ORIF operators. If B contains side effects, this
5719 might change the truth-value of A. */
5726 {
5746 /* This case if tricky because we must either have commutative
5747 operators or else A10 must not have side-effects. */
5753 a01));
5754 }
5756 /* See if we can build a range comparison. */
5760 /* Check for the possibility of merging component references. If our
5761 lhs is another similar operation, try to merge its rhs with our
5762 rhs. Then try to merge our lhs and rhs. */
5774 /* Note that the operands of this must be ints
5775 and their values must be 0 or true.
5776 ("true" is a fixed value perhaps depending on the language.) */
5777 /* If first arg is constant true, return it. */
5781 /* If either arg is constant zero, drop it. */
5785 /* Preserve sequence points. */
5788 /* If second arg is constant true, result is true, but we must
5789 evaluate first arg. */
5792 /* Likewise for first arg, but note this only occurs here for
5793 TRUTH_OR_EXPR. */
5799 /* If either arg is constant zero, drop it. */
5804 /* If either arg is constant true, this is a logical inversion. */
5818 {
5819 /* (-a) CMP (-b) -> b CMP a */
5824 /* (-a) CMP CST -> a swap(CMP) (-CST) */
5826 return
5832 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
5833 /* a CMP (-0) -> a CMP 0 */
5838 }
5840 /* If one arg is a constant integer, put it last. */
5843 {
5850 }
5852 /* Convert foo++ == CONST into ++foo == CONST + INCR.
5853 First, see if one arg is constant; find the constant arg
5854 and the other one. */
5855 {
5865 {
5866 /* This optimization is invalid for ordered comparisons
5867 if CONST+INCR overflows or if foo+incr might overflow.
5868 This optimization is invalid for floating point due to rounding.
5869 For pointer types we assume overflow doesn't happen. */
5873 {
5874 tree newconst
5878 /* Do not overwrite the current varop to be a preincrement,
5879 create a new node so that we won't confuse our caller who
5880 might create trees and throw them away, reusing the
5881 arguments that they passed to build. This shows up in
5882 the THEN or ELSE parts of ?: being postincrements. */
5887 /* If VAROP is a reference to a bitfield, we must mask
5888 the constant by the width of the field. */
5892 {
5893 int size
5895 (TREE_OPERAND
5899 tree folded_compare;
5901 /* First check whether the comparison would come out
5902 always the same. If we don't do that we would
5903 change the meaning with the masking. */
5907 else
5910 constop));
5925 mask)));
5926 }
5932 }
5933 }
5935 {
5939 {
5940 tree newconst
5944 /* Do not overwrite the current varop to be a predecrement,
5945 create a new node so that we won't confuse our caller who
5946 might create trees and throw them away, reusing the
5947 arguments that they passed to build. This shows up in
5948 the THEN or ELSE parts of ?: being postdecrements. */
5956 {
5957 int size
5959 (TREE_OPERAND
5963 tree folded_compare;
5968 else
5971 constop));
5986 mask)));
5987 }
5993 }
5994 }
5995 }
5997 /* Comparisons with the highest or lowest possible integer of
5998 the specified size will have known values and an unsigned
5999 <= 0x7fffffff can be simplified. */
6000 {
6008 {
6014 {
6018 arg0);
6026 arg0);
6033 }
6040 {
6044 arg0);
6052 arg0);
6059 }
6065 /* signed_type does not work on pointer types. */
6067 {
6069 {
6073 return fold
6077 }
6078 }
6084 {
6088 arg0);
6096 arg0);
6103 }
6104 }
6105 }
6107 /* Change X >= C to X > C-1 and X < C to X <= C-1 if C is positive. */
6111 {
6113 {
6128 }
6129 }
6131 /* An unsigned comparison against 0 can be simplified. */
6136 {
6138 {
6150 arg0);
6154 arg0);
6157 }
6158 }
6160 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
6161 a MINUS_EXPR of a constant, we can convert it into a comparison with
6162 a revised constant as long as no overflow occurs. */
6174 /* Similarly for a NEGATE_EXPR. */
6183 /* If we have X - Y == 0, we can convert that to X == Y and similarly
6184 for !=. Don't do this for ordered comparisons due to overflow. */
6190 /* If we are widening one operand of an integer comparison,
6191 see if the other operand is similarly being widened. Perhaps we
6192 can do the comparison in the narrower type. */
6202 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
6203 constant, we can simplify it. */
6210 /* If we are comparing an ABS_EXPR with a constant, we can
6211 convert all the cases into explicit comparisons, but they may
6212 well not be faster than doing the ABS and one comparison.
6213 But ABS (X) <= C is a range comparison, which becomes a subtraction
6214 and a comparison, and is probably faster. */
6226 /* If this is an EQ or NE comparison with zero and ARG0 is
6227 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
6228 two operations, but the latter can be done in one less insn
6229 on machines that have only two-operand insns or on which a
6230 constant cannot be the first operand. */
6233 {
6236 return
6244 integer_one_node)),
6245 arg1));
6248 return
6256 integer_one_node)),
6257 arg1));
6258 }
6260 /* If this is an NE or EQ comparison of zero against the result of a
6261 signed MOD operation whose second operand is a power of 2, make
6262 the MOD operation unsigned since it is simpler and equivalent. */
6271 {
6278 }
6280 /* If this is an NE comparison of zero with an AND of one, remove the
6281 comparison since the AND will give the correct value. */
6287 /* If we have (A & C) == C where C is a power of 2, convert this into
6288 (A & C) != 0. Similarly for NE_EXPR. */
6296 /* If we have (A & C) != 0 where C is the sign bit of A, convert
6297 this into A < 0. Similarly for (A & C) == 0 into A >= 0. */
6301 {
6305 {
6310 }
6311 }
6313 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6314 and similarly for >= into !=. */
6330 return
6337 /* Simplify comparison of something with itself. (For IEEE
6338 floating-point, we can only do some of these simplifications.) */
6340 {
6342 {
6353 /* For NE, we can only do this simplification if integer. */
6356 /* ... fall through ... */
6362 }
6363 }
6365 /* If we are comparing an expression that just has comparisons
6366 of two integer values, arithmetic expressions of those comparisons,
6367 and constants, we can simplify it. There are only three cases
6368 to check: the two values can either be equal, the first can be
6369 greater, or the second can be greater. Fold the expression for
6370 those three values. Since each value must be 0 or 1, we have
6371 eight possibilities, each of which corresponds to the constant 0
6372 or 1 or one of the six possible comparisons.
6374 This handles common cases like (a > b) == 0 but also handles
6375 expressions like ((x > y) - (y > x)) > 0, which supposedly
6376 occur in macroized code. */
6379 {
6384 /* Don't handle degenerate cases here; they should already
6385 have been handled anyway. */
6394 {
6398 /* We can't just pass T to eval_subst in case cval1 or cval2
6399 was the same as ARG1. */
6401 tree high_result
6404 arg1));
6405 tree equal_result
6408 arg1));
6409 tree low_result
6412 arg1));
6414 /* All three of these results should be 0 or 1. Confirm they
6415 are. Then use those values to select the proper code
6416 to use. */
6424 {
6425 /* Make a 3-bit mask with the high-order bit being the
6426 value for `>', the next for '=', and the low for '<'. */
6430 {
6432 /* Always false. */
6453 /* Always true. */
6455 }
6460 else
6462 }
6463 }
6464 }
6466 /* If this is a comparison of a field, we may be able to simplify it. */
6470 /* Handle the constant case even without -O
6471 to make sure the warnings are given. */
6473 {
6476 }
6478 /* If this is a comparison of complex values and either or both sides
6479 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6480 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6481 This may prevent needless evaluations. */
6488 {
6501 type,
6504 }
6506 /* Optimize comparisons of strlen vs zero to a compare of the
6507 first character of the string vs zero. To wit,
6508 strlen(ptr) == 0 => *ptr == 0
6509 strlen(ptr) != 0 => *ptr != 0
6510 Other cases should reduce to one of these two (or a constant)
6511 due to the return value of strlen being unsigned. */
6516 {
6518 tree arglist;
6530 integer_zero_node));
6531 }
6533 /* From here on, the only cases we handle are when the result is
6534 known to be a constant.
6536 To compute GT, swap the arguments and do LT.
6537 To compute GE, do LT and invert the result.
6538 To compute LE, swap the arguments, do LT and invert the result.
6539 To compute NE, do EQ and invert the result.
6541 Therefore, the code below must handle only EQ and LT. */
6544 {
6547 }
6549 /* Note that it is safe to invert for real values here because we
6550 will check below in the one case that it matters. */
6555 {
6558 }
6560 /* Compute a result for LT or EQ if args permit;
6561 otherwise return T. */
6563 {
6566 else
6571 }
6574 /* Assume a nonexplicit constant cannot equal an explicit one,
6575 since such code would be undefined anyway.
6576 Exception: on sysvr4, using #pragma weak,
6577 a label can come out as 0. */
6584 #endif
6585 /* Two real constants can be compared explicitly. */
6587 {
6588 /* If either operand is a NaN, the result is false with two
6589 exceptions: First, an NE_EXPR is true on NaNs, but that case
6590 is already handled correctly since we will be inverting the
6591 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6592 or a GE_EXPR into a LT_EXPR, we must return true so that it
6593 will be inverted into false. */
6603 else
6607 }
6621 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6622 so all simple results must be passed through pedantic_non_lvalue. */
6624 return pedantic_non_lvalue
6629 /* If the second operand is zero, invert the comparison and swap
6630 the second and third operands. Likewise if the second operand
6631 is constant and the third is not or if the third operand is
6632 equivalent to the first operand of the comparison. */
6640 {
6641 /* See if this can be inverted. If it can't, possibly because
6642 it was a floating-point inequality comparison, don't do
6643 anything. */
6647 {
6651 /* arg1 should be the first argument of the new T. */
6654 }
6655 }
6657 /* If we have A op B ? A : C, we may be able to convert this to a
6658 simpler expression, depending on the operation and the values
6659 of B and C. Signed zeros prevent all of these transformations,
6660 for reasons given above each one. */
6666 {
6672 /* If we have A op 0 ? A : -A, consider applying the following
6673 transformations:
6675 A == 0? A : -A same as -A
6676 A != 0? A : -A same as A
6677 A >= 0? A : -A same as abs (A)
6678 A > 0? A : -A same as abs (A)
6679 A <= 0? A : -A same as -abs (A)
6680 A < 0? A : -A same as -abs (A)
6682 None of these transformations work for modes with signed
6683 zeros. If A is +/-0, the first two transformations will
6684 change the sign of the result (from +0 to -0, or vice
6685 versa). The last four will fix the sign of the result,
6686 even though the original expressions could be positive or
6687 negative, depending on the sign of A.
6689 Note that all these transformations are correct if A is
6690 NaN, since the two alternatives (A and -A) are also NaNs. */
6697 {
6699 return
6700 pedantic_non_lvalue
6702 negate_expr
6704 arg1))));
6712 return pedantic_non_lvalue
6720 return pedantic_non_lvalue
6724 arg1)))));
6727 }
6729 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
6730 A == 0 ? A : 0 is always 0 unless A is -0. Note that
6731 both transformations are correct when A is NaN: A != 0
6732 is then true, and A == 0 is false. */
6735 {
6740 }
6742 /* Try some transformations of A op B ? A : B.
6744 A == B? A : B same as B
6745 A != B? A : B same as A
6746 A >= B? A : B same as max (A, B)
6747 A > B? A : B same as max (B, A)
6748 A <= B? A : B same as min (A, B)
6749 A < B? A : B same as min (B, A)
6751 As above, these transformations don't work in the presence
6752 of signed zeros. For example, if A and B are zeros of
6753 opposite sign, the first two transformations will change
6754 the sign of the result. In the last four, the original
6755 expressions give different results for (A=+0, B=-0) and
6756 (A=-0, B=+0), but the transformed expressions do not.
6758 The first two transformations are correct if either A or B
6759 is a NaN. In the first transformation, the condition will
6760 be false, and B will indeed be chosen. In the case of the
6761 second transformation, the condition A != B will be true,
6762 and A will be chosen.
6764 The conversions to max() and min() are not correct if B is
6765 a number and A is not. The conditions in the original
6766 expressions will be false, so all four give B. The min()
6767 and max() versions would give a NaN instead. */
6770 {
6775 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
6780 {
6787 /* In C++ a ?: expression can be an lvalue, so put the
6788 operand which will be used if they are equal first
6789 so that we can convert this back to the
6790 corresponding COND_EXPR. */
6792 return pedantic_non_lvalue
6802 return pedantic_non_lvalue
6811 }
6812 }
6814 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
6815 we might still be able to simplify this. For example,
6816 if C1 is one less or one more than C2, this might have started
6817 out as a MIN or MAX and been transformed by this function.
6818 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
6824 {
6826 /* We can replace A with C1 in this case. */
6833 /* If C1 is C2 + 1, this is min(A, C2). */
6838 return pedantic_non_lvalue
6843 /* If C1 is C2 - 1, this is min(A, C2). */
6848 return pedantic_non_lvalue
6853 /* If C1 is C2 - 1, this is max(A, C2). */
6858 return pedantic_non_lvalue
6863 /* If C1 is C2 + 1, this is max(A, C2). */
6868 return pedantic_non_lvalue
6875 }
6876 }
6878 /* If the second operand is simpler than the third, swap them
6879 since that produces better jump optimization results. */
6885 {
6886 /* See if this can be inverted. If it can't, possibly because
6887 it was a floating-point inequality comparison, don't do
6888 anything. */
6892 {
6896 /* arg1 should be the first argument of the new T. */
6899 }
6900 }
6902 /* Convert A ? 1 : 0 to simply A. */
6905 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
6906 call to fold will try to move the conversion inside
6907 a COND, which will recurse. In that case, the COND_EXPR
6908 is probably the best choice, so leave it alone. */
6912 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
6913 operation is simply A & 2. */
6927 /* When pedantic, a compound expression can be neither an lvalue
6928 nor an integer constant expression. */
6931 /* Don't let (0, 0) be null pointer constant. */
6973 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
6974 appropriate. */
6979 {
6983 tree arg01;
6994 {
7008 arg01));
7009 }
7012 }
7015 /* Check for a built-in function. */
7020 {
7024 }
7030 }
7032 /* Determine if first argument is a multiple of second argument. Return 0 if
7033 it is not, or we cannot easily determined it to be.
7035 An example of the sort of thing we care about (at this point; this routine
7036 could surely be made more general, and expanded to do what the *_DIV_EXPR's
7037 fold cases do now) is discovering that
7039 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7041 is a multiple of
7043 SAVE_EXPR (J * 8)
7045 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
7047 This code also handles discovering that
7049 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7051 is a multiple of 8 so we don't have to worry about dealing with a
7052 possible remainder.
7054 Note that we *look* inside a SAVE_EXPR only to determine how it was
7055 calculated; it is not safe for fold to do much of anything else with the
7056 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
7057 at run time. For example, the latter example above *cannot* be implemented
7058 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
7059 evaluation time of the original SAVE_EXPR is not necessarily the same at
7060 the time the new expression is evaluated. The only optimization of this
7061 sort that would be valid is changing
7063 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
7065 divided by 8 to
7067 SAVE_EXPR (I) * SAVE_EXPR (J)
7069 (where the same SAVE_EXPR (J) is used in the original and the
7070 transformed version). */
7072 static int
7074 tree type;
7075 tree top;
7076 tree bottom;
7077 {
7085 {
7097 {
7101 /* const_binop may not detect overflow correctly,
7102 so check for it explicitly here. */
7111 }
7115 /* Can't handle conversions from non-integral or wider integral type. */
7121 /* .. fall through ... */
7137 }
7138 }
7140 /* Return true if `t' is known to be non-negative. */
7142 int
7144 tree t;
7145 {
7147 {
7188 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7190 else
7191 /* We don't know sign of `t', so be conservative and return false. */
7193 }
7194 }
7196 /* Return true if `r' is known to be non-negative.
7197 Only handles constants at the moment. */
7199 int
7201 rtx r;
7202 {
7204 {
7214 {
7216 rtx elt;
7221 {
7225 }
7228 }
7232 /* These are always nonnegative. */
7237 }
7238 }