| blob | 033dbfcfc62da4dedbab772bfd895878c9db949b |
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 *));
88 HOST_WIDE_INT *,
109 static tree fold_binary_op_with_conditional_arg
113 /* The following constants represent a bit based encoding of GCC's
114 comparison operators. This encoding simplifies transformations
115 on relational comparison operators, such as AND and OR. */
116 #define COMPCODE_FALSE 0
117 #define COMPCODE_LT 1
118 #define COMPCODE_EQ 2
119 #define COMPCODE_LE 3
120 #define COMPCODE_GT 4
121 #define COMPCODE_NE 5
122 #define COMPCODE_GE 6
123 #define COMPCODE_TRUE 7
125 /* We know that A1 + B1 = SUM1, using 2's complement arithmetic and ignoring
126 overflow. Suppose A, B and SUM have the same respective signs as A1, B1,
127 and SUM1. Then this yields nonzero if overflow occurred during the
128 addition.
130 Overflow occurs if A and B have the same sign, but A and SUM differ in
131 sign. Use `^' to test whether signs differ, and `< 0' to isolate the
132 sign. */
133 #define OVERFLOW_SUM_SIGN(a, b, sum) ((~((a) ^ (b)) & ((a) ^ (sum))) < 0)
134 \f
135 /* To do constant folding on INTEGER_CST nodes requires two-word arithmetic.
136 We do that by representing the two-word integer in 4 words, with only
137 HOST_BITS_PER_WIDE_INT / 2 bits stored in each word, as a positive
138 number. The value of the word is LOWPART + HIGHPART * BASE. */
140 #define LOWPART(x) \
141 ((x) & (((unsigned HOST_WIDE_INT) 1 << (HOST_BITS_PER_WIDE_INT / 2)) - 1))
142 #define HIGHPART(x) \
143 ((unsigned HOST_WIDE_INT) (x) >> HOST_BITS_PER_WIDE_INT / 2)
144 #define BASE ((unsigned HOST_WIDE_INT) 1 << HOST_BITS_PER_WIDE_INT / 2)
146 /* Unpack a two-word integer into 4 words.
147 LOW and HI are the integer, as two `HOST_WIDE_INT' pieces.
148 WORDS points to the array of HOST_WIDE_INTs. */
150 static void
154 HOST_WIDE_INT hi;
155 {
160 }
162 /* Pack an array of 4 words into a two-word integer.
163 WORDS points to the array of words.
164 The integer is stored into *LOW and *HI as two `HOST_WIDE_INT' pieces. */
166 static void
171 {
174 }
175 \f
176 /* Make the integer constant T valid for its type by setting to 0 or 1 all
177 the bits in the constant that don't belong in the type.
179 Return 1 if a signed overflow occurs, 0 otherwise. If OVERFLOW is
180 nonzero, a signed overflow has already occurred in calculating T, so
181 propagate it.
183 Make the real constant T valid for its type by calling CHECK_FLOAT_VALUE,
184 if it exists. */
186 int
188 tree t;
190 {
192 HOST_WIDE_INT high;
196 {
197 #ifdef CHECK_FLOAT_VALUE
199 overflow);
200 #endif
202 }
212 else
215 /* First clear all bits that are beyond the type's precision. */
218 ;
222 else
223 {
227 }
229 /* Unsigned types do not suffer sign extension or overflow unless they
230 are a sizetype. */
236 /* If the value's sign bit is set, extend the sign. */
244 {
245 /* Value is negative:
246 set to 1 all the bits that are outside this type's precision. */
250 else
251 {
255 }
256 }
258 /* Return nonzero if signed overflow occurred. */
259 return
262 }
263 \f
264 /* Add two doubleword integers with doubleword result.
265 Each argument is given as two `HOST_WIDE_INT' pieces.
266 One argument is L1 and H1; the other, L2 and H2.
267 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
269 int
275 {
277 HOST_WIDE_INT h;
285 }
287 /* Negate a doubleword integer with doubleword result.
288 Return nonzero if the operation overflows, assuming it's signed.
289 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
290 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
292 int
295 HOST_WIDE_INT h1;
298 {
300 {
304 }
305 else
306 {
310 }
311 }
312 \f
313 /* Multiply two doubleword integers with doubleword result.
314 Return nonzero if the operation overflows, assuming it's signed.
315 Each argument is given as two `HOST_WIDE_INT' pieces.
316 One argument is L1 and H1; the other, L2 and H2.
317 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
319 int
325 {
340 {
343 {
345 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
347 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
351 }
353 }
357 /* Check for overflow by calculating the top half of the answer in full;
358 it should agree with the low half's sign bit. */
361 {
364 }
366 {
369 }
371 }
372 \f
373 /* Shift the doubleword integer in L1, H1 left by COUNT places
374 keeping only PREC bits of result.
375 Shift right if COUNT is negative.
376 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
377 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
379 void
387 {
391 {
394 }
396 #ifdef SHIFT_COUNT_TRUNCATED
399 #endif
402 {
403 /* Shifting by the host word size is undefined according to the
404 ANSI standard, so we must handle this as a special case. */
407 }
409 {
412 }
413 else
414 {
418 }
420 /* Sign extend all bits that are beyond the precision. */
428 ;
430 {
433 }
434 else
435 {
439 }
440 }
442 /* Shift the doubleword integer in L1, H1 right by COUNT places
443 keeping only PREC bits of result. COUNT must be positive.
444 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
445 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
447 void
455 {
458 signmask = (arith
462 #ifdef SHIFT_COUNT_TRUNCATED
465 #endif
468 {
469 /* Shifting by the host word size is undefined according to the
470 ANSI standard, so we must handle this as a special case. */
473 }
475 {
478 }
479 else
480 {
484 }
486 /* Zero / sign extend all bits that are beyond the precision. */
489 {
492 }
494 ;
496 {
499 }
500 else
501 {
505 }
506 }
507 \f
508 /* Rotate the doubleword integer in L1, H1 left by COUNT places
509 keeping only PREC bits of result.
510 Rotate right if COUNT is negative.
511 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
513 void
520 {
532 }
534 /* Rotate the doubleword integer in L1, H1 left by COUNT places
535 keeping only PREC bits of result. COUNT must be positive.
536 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
538 void
545 {
557 }
558 \f
559 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
560 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
561 CODE is a tree code for a kind of division, one of
562 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
563 or EXACT_DIV_EXPR
564 It controls how the quotient is rounded to an integer.
565 Return nonzero if the operation overflows.
566 UNS nonzero says do unsigned division. */
568 int
575 HOST_WIDE_INT hnum_orig;
577 HOST_WIDE_INT hden_orig;
580 {
596 /* calculate quotient sign and convert operands to unsigned. */
598 {
600 {
602 /* (minimum integer) / (-1) is the only overflow case. */
606 }
608 {
611 }
612 }
617 /* This unsigned division rounds toward zero. */
620 }
624 /* hden != 0 already checked. */
629 }
639 /* Special code for when the divisor < BASE. */
641 {
642 /* hnum != 0 already checked. */
644 {
648 }
649 }
650 else
651 {
652 /* Full double precision division,
653 with thanks to Don Knuth's "Seminumerical Algorithms". */
657 /* Find the highest non-zero divisor digit. */
660 {
663 }
665 /* Insure that the first digit of the divisor is at least BASE/2.
666 This is required by the quotient digit estimation algorithm. */
673 {
677 }
682 {
687 }
688 }
692 /* Main loop */
694 {
695 /* Guess the next quotient digit, quo_est, by dividing the first
696 two remaining dividend digits by the high order quotient digit.
697 quo_est is never low and is at most 2 high. */
704 else
707 /* Refine quo_est so it's usually correct, and at most one high. */
712 quo_est--;
714 /* Try QUO_EST as the quotient digit, by multiplying the
715 divisor by QUO_EST and subtracting from the remaining dividend.
716 Keep in mind that QUO_EST is the I - 1st digit. */
720 {
726 }
728 /* If quo_est was high by one, then num[i] went negative and
729 we need to correct things. */
731 {
732 quo_est--;
735 {
739 }
742 }
744 /* Store the quotient digit. */
746 }
747 }
751 finish_up:
752 /* if result is negative, make it so. */
756 /* compute trial remainder: rem = num - (quo * den) */
762 {
771 {
772 /* quo = quo - 1; */
775 }
776 else
783 {
786 }
787 else
793 {
799 /* Get absolute values */
805 /* If (2 * abs (lrem) >= abs (lden)) */
814 {
816 /* quo = quo - 1; */
819 else
820 /* quo = quo + 1; */
823 }
824 else
826 }
831 }
833 /* compute true remainder: rem = num - (quo * den) */
838 }
839 \f
840 /* Given T, an expression, return the negation of T. Allow for T to be
841 null, in which case return null. */
843 static tree
845 tree t;
846 {
847 tree type;
848 tree tem;
857 {
870 /* - (A - B) -> B - A */
880 }
883 }
884 \f
885 /* Split a tree IN into a constant, literal and variable parts that could be
886 combined with CODE to make IN. "constant" means an expression with
887 TREE_CONSTANT but that isn't an actual constant. CODE must be a
888 commutative arithmetic operation. Store the constant part into *CONP,
889 the literal in *LITP and return the variable part. If a part isn't
890 present, set it to null. If the tree does not decompose in this way,
891 return the entire tree as the variable part and the other parts as null.
893 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
894 case, we negate an operand that was subtracted. Except if it is a
895 literal for which we use *MINUS_LITP instead.
897 If NEGATE_P is true, we are negating all of IN, again except a literal
898 for which we use *MINUS_LITP instead.
900 If IN is itself a literal or constant, return it as appropriate.
902 Note that we do not guarantee that any of the three values will be the
903 same type as IN, but they will have the same signedness and mode. */
905 static tree
907 tree in;
911 {
918 /* Strip any conversions that don't change the machine mode or signedness. */
925 /* We can associate addition and subtraction together (even
926 though the C standard doesn't say so) for integers because
927 the value is not affected. For reals, the value might be
928 affected, so we can't. */
931 {
937 /* First see if either of the operands is a literal, then a constant. */
948 /* If we haven't dealt with either operand, this is not a case we can
949 decompose. Otherwise, VAR is either of the ones remaining, if any. */
954 else
957 /* Now do any needed negations. */
964 }
967 else
971 {
978 }
981 }
983 /* Re-associate trees split by the above function. T1 and T2 are either
984 expressions to associate or null. Return the new expression, if any. If
985 we build an operation, do it in TYPE and with CODE. */
987 static tree
991 tree type;
992 {
998 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
999 try to fold this since we will have infinite recursion. But do
1000 deal with any NEGATE_EXPRs. */
1003 {
1005 {
1012 }
1014 }
1017 }
1018 \f
1019 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1020 to produce a new constant.
1022 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1024 static tree
1029 {
1033 HOST_WIDE_INT hi;
1035 HOST_WIDE_INT garbageh;
1036 tree t;
1039 int is_sizetype
1050 {
1070 /* It's unclear from the C standard whether shifts can overflow.
1071 The following code ignores overflow; perhaps a C standard
1072 interpretation ruling is needed. */
1102 /* This is a shortcut for a common special case. */
1107 {
1113 }
1115 /* ... fall through ... */
1119 {
1122 }
1125 {
1128 }
1135 /* This is a shortcut for a common special case. */
1140 {
1145 }
1147 /* ... fall through ... */
1163 else
1169 else
1175 }
1177 /* If this is for a sizetype, can be represented as one (signed)
1178 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1179 constants. */
1185 else
1186 {
1189 }
1192 = ((notrunc
1199 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1200 So check if force_fit_type truncated the value. */
1211 }
1213 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1214 constant. We assume ARG1 and ARG2 have the same data type, or at least
1215 are the same kind of constant and the same machine mode.
1217 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1219 static tree
1224 {
1232 {
1233 REAL_VALUE_TYPE d1;
1234 REAL_VALUE_TYPE d2;
1235 REAL_VALUE_TYPE value;
1236 tree t;
1241 /* If either operand is a NaN, just return it. Otherwise, set up
1242 for floating-point trap; we return an overflow. */
1252 value));
1262 }
1264 {
1270 tree t;
1273 {
1293 notrunc),
1299 notrunc));
1303 {
1304 tree magsquared
1308 notrunc);
1311 const_binop
1316 notrunc),
1318 notrunc),
1319 notrunc),
1321 const_binop
1326 notrunc),
1328 notrunc),
1329 notrunc),
1331 }
1336 }
1338 }
1340 }
1342 /* These are the hash table functions for the hash table of INTEGER_CST
1343 nodes of a sizetype. */
1345 /* Return the hash code code X, an INTEGER_CST. */
1347 static hashval_t
1350 {
1356 }
1358 /* Return non-zero if the value represented by *X (an INTEGER_CST tree node)
1359 is the same as that given by *Y, which is the same. */
1361 static int
1365 {
1373 }
1374 \f
1375 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1376 bits are given by NUMBER and of the sizetype represented by KIND. */
1378 tree
1380 HOST_WIDE_INT number;
1382 {
1384 }
1386 /* Likewise, but the desired type is specified explicitly. */
1390 htab_t size_htab;
1392 tree
1394 HOST_WIDE_INT number;
1395 tree type;
1396 {
1400 {
1403 }
1405 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1406 hash table, we return the value from the hash table. Otherwise, we
1407 place that in the hash table and make a new node for the next time. */
1416 {
1422 }
1423 else
1425 }
1427 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1428 is a tree code. The type of the result is taken from the operands.
1429 Both must be the same type integer type and it must be a size type.
1430 If the operands are constant, so is the result. */
1432 tree
1436 {
1443 /* Handle the special case of two integer constants faster. */
1445 {
1446 /* And some specific cases even faster than that. */
1455 /* Handle general case of two integer constants. */
1457 }
1463 }
1465 /* Given two values, either both of sizetype or both of bitsizetype,
1466 compute the difference between the two values. Return the value
1467 in signed type corresponding to the type of the operands. */
1469 tree
1472 {
1474 tree ctype;
1480 /* If the type is already signed, just do the simple thing. */
1487 /* If either operand is not a constant, do the conversions to the signed
1488 type and subtract. The hardware will do the right thing with any
1489 overflow in the subtraction. */
1494 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1495 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1496 overflow) and negate (which can't either). Special-case a result
1497 of zero while we're here. */
1502 else
1505 }
1506 \f
1508 /* Given T, a tree representing type conversion of ARG1, a constant,
1509 return a constant tree representing the result of conversion. */
1511 static tree
1513 tree t;
1514 tree arg1;
1515 {
1520 {
1522 {
1523 /* If we would build a constant wider than GCC supports,
1524 leave the conversion unfolded. */
1528 /* If we are trying to make a sizetype for a small integer, use
1529 size_int to pick up cached types to reduce duplicate nodes. */
1535 /* Given an integer constant, make new constant with new type,
1536 appropriately sign-extended or truncated. */
1540 /* Indicate an overflow if (1) ARG1 already overflowed,
1541 or (2) force_fit_type indicates an overflow.
1542 Tell force_fit_type that an overflow has already occurred
1543 if ARG1 is a too-large unsigned value and T is signed.
1544 But don't indicate an overflow if converting a pointer. */
1554 }
1556 {
1557 /* Don't initialize these, use assignments.
1558 Initialized local aggregates don't work on old compilers. */
1559 REAL_VALUE_TYPE x;
1560 REAL_VALUE_TYPE l;
1561 REAL_VALUE_TYPE u;
1572 /* See if X will be in range after truncation towards 0.
1573 To compensate for truncation, move the bounds away from 0,
1574 but reject if X exactly equals the adjusted bounds. */
1578 /* If X is a NaN, use zero instead and show we have an overflow.
1579 Otherwise, range check. */
1583 && !no_upper_bound
1587 {
1591 }
1597 }
1599 }
1601 {
1605 {
1607 {
1608 /* We make a copy of ARG1 so that we don't modify an
1609 existing constant tree. */
1613 }
1624 }
1625 }
1628 }
1629 \f
1630 /* Return an expr equal to X but certainly not valid as an lvalue. */
1632 tree
1634 tree x;
1635 {
1636 tree result;
1638 /* These things are certainly not lvalues. */
1649 }
1651 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1652 Zero means allow extended lvalues. */
1656 /* When pedantic, return an expr equal to X but certainly not valid as a
1657 pedantic lvalue. Otherwise, return X. */
1659 tree
1661 tree x;
1662 {
1665 else
1667 }
1668 \f
1669 /* Given a tree comparison code, return the code that is the logical inverse
1670 of the given code. It is not safe to do this for floating-point
1671 comparisons, except for NE_EXPR and EQ_EXPR. */
1673 static enum tree_code
1676 {
1678 {
1693 }
1694 }
1696 /* Similar, but return the comparison that results if the operands are
1697 swapped. This is safe for floating-point. */
1699 static enum tree_code
1702 {
1704 {
1718 }
1719 }
1722 /* Convert a comparison tree code from an enum tree_code representation
1723 into a compcode bit-based encoding. This function is the inverse of
1724 compcode_to_comparison. */
1726 static int
1729 {
1731 {
1746 }
1747 }
1749 /* Convert a compcode bit-based encoding of a comparison operator back
1750 to GCC's enum tree_code representation. This function is the
1751 inverse of comparison_to_compcode. */
1753 static enum tree_code
1756 {
1758 {
1773 }
1774 }
1776 /* Return nonzero if CODE is a tree code that represents a truth value. */
1778 static int
1781 {
1786 }
1787 \f
1788 /* Return nonzero if two operands are necessarily equal.
1789 If ONLY_CONST is non-zero, only return non-zero for constants.
1790 This function tests whether the operands are indistinguishable;
1791 it does not test whether they are equal using C's == operation.
1792 The distinction is important for IEEE floating point, because
1793 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1794 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1796 int
1800 {
1801 /* If both types don't have the same signedness, then we can't consider
1802 them equal. We must check this before the STRIP_NOPS calls
1803 because they may change the signedness of the arguments. */
1811 /* This is needed for conversions and for COMPONENT_REF.
1812 Might as well play it safe and always test this. */
1818 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1819 We don't care about side effects in that case because the SAVE_EXPR
1820 takes care of that for us. In all other cases, two expressions are
1821 equal if they have no side effects. If we have two identical
1822 expressions with side effects that should be treated the same due
1823 to the only side effects being identical SAVE_EXPR's, that will
1824 be detected in the recursive calls below. */
1830 /* Next handle constant cases, those for which we can return 1 even
1831 if ONLY_CONST is set. */
1834 {
1847 {
1857 {
1862 }
1865 }
1869 only_const)
1871 only_const));
1884 }
1890 {
1892 /* Two conversions are equal only if signedness and modes match. */
1908 /* For commutative ops, allow the other order. */
1921 /* If either of the pointer (or reference) expressions we are dereferencing
1922 contain a side effect, these cannot be equal. */
1928 {
1950 }
1959 }
1960 }
1961 \f
1962 /* Similar to operand_equal_p, but see if ARG0 might have been made by
1963 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
1965 When in doubt, return 0. */
1967 static int
1970 tree other;
1971 {
1983 /* Discard any conversions that don't change the modes of ARG0 and ARG1
1984 and see if the inner values are the same. This removes any
1985 signedness comparison, which doesn't matter here. */
1992 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
1993 actual comparison operand, ARG0.
1995 First throw away any conversions to wider types
1996 already present in the operands. */
2005 {
2008 /* Make sure shorter operand is extended the right way
2009 to match the longer operand. */
2015 }
2018 }
2019 \f
2020 /* See if ARG is an expression that is either a comparison or is performing
2021 arithmetic on comparisons. The comparisons must only be comparing
2022 two different values, which will be stored in *CVAL1 and *CVAL2; if
2023 they are non-zero it means that some operands have already been found.
2024 No variables may be used anywhere else in the expression except in the
2025 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
2026 the expression and save_expr needs to be called with CVAL1 and CVAL2.
2028 If this is true, return 1. Otherwise, return zero. */
2030 static int
2032 tree arg;
2035 {
2039 /* We can handle some of the 'e' cases here. */
2049 {
2050 /* If we've already found a CVAL1 or CVAL2, this expression is
2051 two complex to handle. */
2057 }
2060 {
2083 /* First see if we can handle the first operand, then the second. For
2084 the second operand, we know *CVAL1 can't be zero. It must be that
2085 one side of the comparison is each of the values; test for the
2086 case where this isn't true by failing if the two operands
2087 are the same. */
2096 ;
2100 ;
2101 else
2105 ;
2109 ;
2110 else
2117 }
2118 }
2119 \f
2120 /* ARG is a tree that is known to contain just arithmetic operations and
2121 comparisons. Evaluate the operations in the tree substituting NEW0 for
2122 any occurrence of OLD0 as an operand of a comparison and likewise for
2123 NEW1 and OLD1. */
2125 static tree
2127 tree arg;
2129 {
2134 /* We can handle some of the 'e' cases here. */
2142 {
2157 {
2174 }
2175 /* fall through - ??? */
2178 {
2182 /* We need to check both for exact equality and tree equality. The
2183 former will be true if the operand has a side-effect. In that
2184 case, we know the operand occurred exactly once. */
2197 }
2201 }
2202 }
2203 \f
2204 /* Return a tree for the case when the result of an expression is RESULT
2205 converted to TYPE and OMITTED was previously an operand of the expression
2206 but is now not needed (e.g., we folded OMITTED * 0).
2208 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2209 the conversion of RESULT to TYPE. */
2211 static tree
2214 {
2221 }
2223 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2225 static tree
2228 {
2235 }
2236 \f
2237 /* Return a simplified tree node for the truth-negation of ARG. This
2238 never alters ARG itself. We assume that ARG is an operation that
2239 returns a truth value (0 or 1). */
2241 tree
2243 tree arg;
2244 {
2251 /* If this is a comparison, we can simply invert it, except for
2252 floating-point non-equality comparisons, in which case we just
2253 enclose a TRUTH_NOT_EXPR around what we have. */
2256 {
2258 && !flag_unsafe_math_optimizations
2262 else
2265 }
2268 {
2283 /* Here we can invert either operand. We invert the first operand
2284 unless the second operand is a TRUTH_NOT_EXPR in which case our
2285 result is the XOR of the first operand with the inside of the
2286 negation of the second operand. */
2291 else
2346 }
2350 }
2352 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2353 operands are another bit-wise operation with a common input. If so,
2354 distribute the bit operations to save an operation and possibly two if
2355 constants are involved. For example, convert
2356 (A | B) & (A | C) into A | (B & C)
2357 Further simplification will occur if B and C are constants.
2359 If this optimization cannot be done, 0 will be returned. */
2361 static tree
2364 tree type;
2366 {
2367 tree common;
2377 {
2381 }
2383 {
2387 }
2389 {
2393 }
2395 {
2399 }
2400 else
2405 }
2406 \f
2407 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2408 starting at BITPOS. The field is unsigned if UNSIGNEDP is non-zero. */
2410 static tree
2412 tree inner;
2413 tree type;
2416 {
2423 }
2425 /* Optimize a bit-field compare.
2427 There are two cases: First is a compare against a constant and the
2428 second is a comparison of two items where the fields are at the same
2429 bit position relative to the start of a chunk (byte, halfword, word)
2430 large enough to contain it. In these cases we can avoid the shift
2431 implicit in bitfield extractions.
2433 For constants, we emit a compare of the shifted constant with the
2434 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2435 compared. For two fields at the same position, we do the ANDs with the
2436 similar mask and compare the result of the ANDs.
2438 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2439 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2440 are the left and right operands of the comparison, respectively.
2442 If the optimization described above can be done, we return the resulting
2443 tree. Otherwise we return zero. */
2445 static tree
2448 tree compare_type;
2450 {
2459 tree mask;
2460 tree offset;
2462 /* Get all the information about the extractions being done. If the bit size
2463 if the same as the size of the underlying object, we aren't doing an
2464 extraction at all and so can do nothing. We also don't want to
2465 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2466 then will no longer be able to replace it. */
2474 {
2475 /* If this is not a constant, we can only do something if bit positions,
2476 sizes, and signedness are the same. */
2484 }
2486 /* See if we can find a mode to refer to this field. We should be able to,
2487 but fail if we can't. */
2496 /* Set signed and unsigned types of the precision of this mode for the
2497 shifts below. */
2501 /* Compute the bit position and size for the new reference and our offset
2502 within it. If the new reference is the same size as the original, we
2503 won't optimize anything, so return zero. */
2513 /* Make the mask to be used against the extracted field. */
2523 /* If not comparing with constant, just rework the comparison
2524 and return. */
2529 mask),
2533 mask));
2535 /* Otherwise, we are handling the constant case. See if the constant is too
2536 big for the field. Warn and return a tree of for 0 (false) if so. We do
2537 this not only for its own sake, but to avoid having to test for this
2538 error case below. If we didn't, we might generate wrong code.
2540 For unsigned fields, the constant shifted right by the field length should
2541 be all zero. For signed fields, the high-order bits should agree with
2542 the sign bit. */
2545 {
2549 {
2555 }
2556 }
2557 else
2558 {
2562 {
2568 }
2569 }
2571 /* Single-bit compares should always be against zero. */
2573 {
2576 }
2578 /* Make a new bitfield reference, shift the constant over the
2579 appropriate number of bits and mask it with the computed mask
2580 (in case this was a signed field). If we changed it, make a new one. */
2583 {
2586 }
2596 rhs);
2597 }
2598 \f
2599 /* Subroutine for fold_truthop: decode a field reference.
2601 If EXP is a comparison reference, we return the innermost reference.
2603 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2604 set to the starting bit number.
2606 If the innermost field can be completely contained in a mode-sized
2607 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2609 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2610 otherwise it is not changed.
2612 *PUNSIGNEDP is set to the signedness of the field.
2614 *PMASK is set to the mask used. This is either contained in a
2615 BIT_AND_EXPR or derived from the width of the field.
2617 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2619 Return 0 if this is not a component reference or is one that we can't
2620 do anything with. */
2622 static tree
2625 tree exp;
2631 {
2634 tree unsigned_type;
2637 /* All the optimizations using this function assume integer fields.
2638 There are problems with FP fields since the type_for_size call
2639 below can fail for, e.g., XFmode. */
2646 {
2652 }
2661 /* Compute the mask to access the bitfield. */
2671 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2679 }
2681 /* Return non-zero if MASK represents a mask of SIZE ones in the low-order
2682 bit positions. */
2684 static int
2686 tree mask;
2688 {
2691 tree tmask;
2696 return
2703 }
2705 /* Subroutine for fold: determine if VAL is the INTEGER_CONST that
2706 represents the sign bit of EXP's type. If EXP represents a sign
2707 or zero extension, also test VAL against the unextended type.
2708 The return value is the (sub)expression whose sign bit is VAL,
2709 or NULL_TREE otherwise. */
2711 static tree
2713 tree exp;
2714 tree val;
2715 {
2717 HOST_WIDE_INT hi;
2719 tree t;
2721 /* Tree EXP must have an integral type. */
2726 /* Tree VAL must be an integer constant. */
2733 {
2736 }
2737 else
2738 {
2741 }
2746 /* Handle extension from a narrower type. */
2752 }
2754 /* Subroutine for fold_truthop: determine if an operand is simple enough
2755 to be evaluated unconditionally. */
2757 static int
2759 tree exp;
2760 {
2761 /* Strip any conversions that don't change the machine mode. */
2773 /* Don't regard global variables as simple. They may be
2774 allocated in ways unknown to the compiler (shared memory,
2775 #pragma weak, etc). */
2778 /* Loading a static variable is unduly expensive, but global
2779 registers aren't expensive. */
2781 }
2782 \f
2783 /* The following functions are subroutines to fold_range_test and allow it to
2784 try to change a logical combination of comparisons into a range test.
2786 For example, both
2787 X == 2 || X == 3 || X == 4 || X == 5
2788 and
2789 X >= 2 && X <= 5
2790 are converted to
2791 (unsigned) (X - 2) <= 3
2793 We describe each set of comparisons as being either inside or outside
2794 a range, using a variable named like IN_P, and then describe the
2795 range with a lower and upper bound. If one of the bounds is omitted,
2796 it represents either the highest or lowest value of the type.
2798 In the comments below, we represent a range by two numbers in brackets
2799 preceded by a "+" to designate being inside that range, or a "-" to
2800 designate being outside that range, so the condition can be inverted by
2801 flipping the prefix. An omitted bound is represented by a "-". For
2802 example, "- [-, 10]" means being outside the range starting at the lowest
2803 possible value and ending at 10, in other words, being greater than 10.
2804 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2805 always false.
2807 We set up things so that the missing bounds are handled in a consistent
2808 manner so neither a missing bound nor "true" and "false" need to be
2809 handled using a special case. */
2811 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2812 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2813 and UPPER1_P are nonzero if the respective argument is an upper bound
2814 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2815 must be specified for a comparison. ARG1 will be converted to ARG0's
2816 type if both are specified. */
2818 static tree
2821 tree type;
2824 {
2825 tree tem;
2829 /* If neither arg represents infinity, do the normal operation.
2830 Else, if not a comparison, return infinity. Else handle the special
2831 comparison rules. Note that most of the cases below won't occur, but
2832 are handled for consistency. */
2835 {
2840 }
2845 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2846 for neither. In real maths, we cannot assume open ended ranges are
2847 the same. But, this is computer arithmetic, where numbers are finite.
2848 We can therefore make the transformation of any unbounded range with
2849 the value Z, Z being greater than any representable number. This permits
2850 us to treat unbounded ranges as equal. */
2854 {
2875 }
2878 }
2879 \f
2880 /* Given EXP, a logical expression, set the range it is testing into
2881 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2882 actually being tested. *PLOW and *PHIGH will be made of the same type
2883 as the returned expression. If EXP is not a comparison, we will most
2884 likely not be returning a useful value and range. */
2886 static tree
2888 tree exp;
2891 {
2898 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2899 and see if we can refine the range. Some of the cases below may not
2900 happen, but it doesn't seem worth worrying about this. We "continue"
2901 the outer loop when we've changed something; otherwise we "break"
2902 the switch, which will "break" the while. */
2907 {
2911 {
2922 }
2924 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2925 lose a cast by accident. */
2930 {
2937 /* We can only do something if the range is testing for zero
2938 and if the second operand is an integer constant. Note that
2939 saying something is "in" the range we make is done by
2940 complementing IN_P since it will set in the initial case of
2941 being not equal to zero; "out" is leaving it alone. */
2948 {
2969 }
2973 /* If this is an unsigned comparison, we also know that EXP is
2974 greater than or equal to zero. We base the range tests we make
2975 on that fact, so we record it here so we can parse existing
2976 range tests. */
2978 {
2981 NULL_TREE))
2986 /* If the high bound is missing, but we
2987 have a low bound, reverse the range so
2988 it goes from zero to the low bound minus 1. */
2990 {
2995 }
2996 }
3000 /* (-x) IN [a,b] -> x in [-b, -a] */
3010 /* ~ X -> -X - 1 */
3019 /* If EXP is signed, any overflow in the computation is undefined,
3020 so we don't worry about it so long as our computations on
3021 the bounds don't overflow. For unsigned, overflow is defined
3022 and this is exactly the right thing. */
3031 /* Check for an unsigned range which has wrapped around the maximum
3032 value thus making n_high < n_low, and normalize it. */
3034 {
3040 /* If the range is of the form +/- [ x+1, x ], we won't
3041 be able to normalize it. But then, it represents the
3042 whole range or the empty set, so make it
3043 +/- [ -, - ]. */
3047 else
3049 }
3050 else
3073 /* If we're converting from an unsigned to a signed type,
3074 we will be doing the comparison as unsigned. The tests above
3075 have already verified that LOW and HIGH are both positive.
3077 So we have to make sure that the original unsigned value will
3078 be interpreted as positive. */
3080 {
3083 tree high_positive;
3085 /* A range without an upper bound is, naturally, unbounded.
3086 Since convert would have cropped a very large value, use
3087 the max value for the destination type. */
3088 high_positive
3096 /* If the low bound is specified, "and" the range with the
3097 range for which the original unsigned value will be
3098 positive. */
3100 {
3104 high_positive))
3108 }
3109 else
3110 {
3111 /* Otherwise, "or" the range with the range of the input
3112 that will be interpreted as negative. */
3116 high_positive))
3120 }
3121 }
3129 }
3132 }
3134 /* If EXP is a constant, we can evaluate whether this is true or false. */
3136 {
3143 }
3147 }
3148 \f
3149 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3150 type, TYPE, return an expression to test if EXP is in (or out of, depending
3151 on IN_P) the range. */
3153 static tree
3155 tree type;
3156 tree exp;
3159 {
3161 tree value;
3180 {
3182 {
3186 }
3188 }
3190 /* Optimize (c>=1) && (c<=127) into (signed char)c > 0. */
3192 {
3194 HOST_WIDE_INT hi;
3199 {
3202 }
3203 else
3204 {
3207 }
3210 {
3212 {
3215 }
3218 }
3219 }
3228 }
3229 \f
3230 /* Given two ranges, see if we can merge them into one. Return 1 if we
3231 can, 0 if we can't. Set the output range into the specified parameters. */
3233 static int
3239 {
3243 tree tem;
3253 /* Make range 0 be the range that starts first, or ends last if they
3254 start at the same value. Swap them if it isn't. */
3257 || (lowequal
3260 {
3264 }
3266 /* Now flag two cases, whether the ranges are disjoint or whether the
3267 second range is totally subsumed in the first. Note that the tests
3268 below are simplified by the ones above. */
3274 /* We now have four cases, depending on whether we are including or
3275 excluding the two ranges. */
3277 {
3278 /* If they don't overlap, the result is false. If the second range
3279 is a subset it is the result. Otherwise, the range is from the start
3280 of the second to the end of the first. */
3285 else
3287 }
3290 {
3291 /* If they don't overlap, the result is the first range. If they are
3292 equal, the result is false. If the second range is a subset of the
3293 first, and the ranges begin at the same place, we go from just after
3294 the end of the first range to the end of the second. If the second
3295 range is not a subset of the first, or if it is a subset and both
3296 ranges end at the same place, the range starts at the start of the
3297 first range and ends just before the second range.
3298 Otherwise, we can't describe this as a single range. */
3304 {
3308 }
3310 {
3314 }
3315 else
3317 }
3320 {
3321 /* If they don't overlap, the result is the second range. If the second
3322 is a subset of the first, the result is false. Otherwise,
3323 the range starts just after the first range and ends at the
3324 end of the second. */
3329 else
3330 {
3334 }
3335 }
3337 else
3338 {
3339 /* The case where we are excluding both ranges. Here the complex case
3340 is if they don't overlap. In that case, the only time we have a
3341 range is if they are adjacent. If the second is a subset of the
3342 first, the result is the first. Otherwise, the range to exclude
3343 starts at the beginning of the first range and ends at the end of the
3344 second. */
3346 {
3353 else
3355 }
3358 else
3360 }
3364 }
3365 \f
3366 /* EXP is some logical combination of boolean tests. See if we can
3367 merge it into some range test. Return the new tree if so. */
3369 static tree
3371 tree exp;
3372 {
3379 tree tem;
3381 /* If this is an OR operation, invert both sides; we will invert
3382 again at the end. */
3386 /* If both expressions are the same, if we can merge the ranges, and we
3387 can build the range test, return it or it inverted. If one of the
3388 ranges is always true or always false, consider it to be the same
3389 expression as the other. */
3399 /* On machines where the branch cost is expensive, if this is a
3400 short-circuited branch and the underlying object on both sides
3401 is the same, make a non-short-circuit operation. */
3407 {
3408 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3409 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3410 which cases we can't do this. */
3419 {
3431 }
3432 }
3435 }
3436 \f
3437 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3438 bit value. Arrange things so the extra bits will be set to zero if and
3439 only if C is signed-extended to its full width. If MASK is nonzero,
3440 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3442 static tree
3444 tree c;
3447 tree mask;
3448 {
3451 tree temp;
3456 /* We work by getting just the sign bit into the low-order bit, then
3457 into the high-order bit, then sign-extend. We then XOR that value
3458 with C. */
3462 /* We must use a signed type in order to get an arithmetic right shift.
3463 However, we must also avoid introducing accidental overflows, so that
3464 a subsequent call to integer_zerop will work. Hence we must
3465 do the type conversion here. At this point, the constant is either
3466 zero or one, and the conversion to a signed type can never overflow.
3467 We could get an overflow if this conversion is done anywhere else. */
3475 /* If necessary, convert the type back to match the type of C. */
3480 }
3481 \f
3482 /* Find ways of folding logical expressions of LHS and RHS:
3483 Try to merge two comparisons to the same innermost item.
3484 Look for range tests like "ch >= '0' && ch <= '9'".
3485 Look for combinations of simple terms on machines with expensive branches
3486 and evaluate the RHS unconditionally.
3488 For example, if we have p->a == 2 && p->b == 4 and we can make an
3489 object large enough to span both A and B, we can do this with a comparison
3490 against the object ANDed with the a mask.
3492 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3493 operations to do this with one comparison.
3495 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3496 function and the one above.
3498 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3499 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3501 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3502 two operands.
3504 We return the simplified tree or 0 if no optimization is possible. */
3506 static tree
3510 {
3511 /* If this is the "or" of two comparisons, we can do something if
3512 the comparisons are NE_EXPR. If this is the "and", we can do something
3513 if the comparisons are EQ_EXPR. I.e.,
3514 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3516 WANTED_CODE is this operation code. For single bit fields, we can
3517 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3518 comparison for one-bit fields. */
3538 /* Start by getting the comparison codes. Fail if anything is volatile.
3539 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3540 it were surrounded with a NE_EXPR. */
3565 /* Simplify (x<y) && (x==y) into (x<=y) and related optimizations. */
3569 {
3574 {
3582 }
3585 {
3594 }
3595 else
3605 }
3607 /* If the RHS can be evaluated unconditionally and its operands are
3608 simple, it wins to evaluate the RHS unconditionally on machines
3609 with expensive branches. In this case, this isn't a comparison
3610 that can be merged. Avoid doing this if the RHS is a floating-point
3611 comparison since those can trap. */
3617 {
3618 /* Convert (a != 0) || (b != 0) into (a | b) != 0. */
3626 integer_zero_node);
3628 /* Convert (a == 0) && (b == 0) into (a | b) == 0. */
3636 integer_zero_node);
3639 }
3641 /* See if the comparisons can be merged. Then get all the parameters for
3642 each side. */
3666 /* It must be true that the inner operation on the lhs of each
3667 comparison must be the same if we are to be able to do anything.
3668 Then see if we have constants. If not, the same must be true for
3669 the rhs's. */
3680 else
3683 /* If either comparison code is not correct for our logical operation,
3684 fail. However, we can convert a one-bit comparison against zero into
3685 the opposite comparison against that bit being set in the field. */
3689 {
3691 {
3692 /* Make the left operand unsigned, since we are only interested
3693 in the value of one bit. Otherwise we are doing the wrong
3694 thing below. */
3697 }
3698 else
3700 }
3702 /* This is analogous to the code for l_const above. */
3704 {
3706 {
3709 }
3710 else
3712 }
3714 /* See if we can find a mode that contains both fields being compared on
3715 the left. If we can't, fail. Otherwise, update all constants and masks
3716 to be relative to a field of that size. */
3721 volatilep);
3731 {
3734 }
3742 {
3750 {
3754 wanted_code == NE_EXPR
3756 }
3757 }
3759 {
3767 {
3771 wanted_code == NE_EXPR
3773 }
3774 }
3776 /* If the right sides are not constant, do the same for it. Also,
3777 disallow this optimization if a size or signedness mismatch occurs
3778 between the left and right sides. */
3780 {
3783 /* Make sure the two fields on the right
3784 correspond to the left without being swapped. */
3792 volatilep);
3802 {
3805 }
3812 /* Make a mask that corresponds to both fields being compared.
3813 Do this for both items being compared. If the operands are the
3814 same size and the bits being compared are in the same position
3815 then we can do this by masking both and comparing the masked
3816 results. */
3820 {
3832 }
3834 /* There is still another way we can do something: If both pairs of
3835 fields being compared are adjacent, we may be able to make a wider
3836 field containing them both.
3838 Note that we still must mask the lhs/rhs expressions. Furthermore,
3839 the mask must be shifted to account for the shift done by
3840 make_bit_field_ref. */
3845 {
3846 tree type;
3858 /* Convert to the smaller type before masking out unwanted bits. */
3861 {
3863 {
3867 }
3869 {
3873 }
3874 }
3883 }
3886 }
3888 /* Handle the case of comparisons with constants. If there is something in
3889 common between the masks, those bits of the constants must be the same.
3890 If not, the condition is always false. Test for this to avoid generating
3891 incorrect code below. */
3896 {
3898 {
3901 }
3902 else
3903 {
3906 }
3907 }
3909 /* Construct the expression we will return. First get the component
3910 reference we will make. Unless the mask is all ones the width of
3911 that field, perform the mask operation. Then compare with the
3912 merged constant. */
3922 }
3923 \f
3924 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3925 constant. */
3927 static tree
3929 tree t;
3930 {
3935 tree minmax_const;
3937 tree inner;
3947 /* If something does not permit us to optimize, return the original tree. */
3955 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3956 and GT_EXPR, doing the rest with recursive calls using logical
3957 simplifications. */
3959 {
3961 return
3965 return
3967 optimize_minmax_comparison
3969 optimize_minmax_comparison
3974 /* MAX (X, 0) == 0 -> X <= 0 */
3978 /* MAX (X, 0) == 5 -> X == 5 */
3982 /* MAX (X, 0) == -1 -> false */
3986 /* MIN (X, 0) == 0 -> X >= 0 */
3990 /* MIN (X, 0) == 5 -> false */
3993 else
3994 /* MIN (X, 0) == -1 -> X == -1 */
3999 /* MAX (X, 0) > 0 -> X > 0
4000 MAX (X, 0) > 5 -> X > 5 */
4004 /* MAX (X, 0) > -1 -> true */
4008 /* MIN (X, 0) > 0 -> false
4009 MIN (X, 0) > 5 -> false */
4012 else
4013 /* MIN (X, 0) > -1 -> X > -1 */
4018 }
4019 }
4020 \f
4021 /* T is an integer expression that is being multiplied, divided, or taken a
4022 modulus (CODE says which and what kind of divide or modulus) by a
4023 constant C. See if we can eliminate that operation by folding it with
4024 other operations already in T. WIDE_TYPE, if non-null, is a type that
4025 should be used for the computation if wider than our type.
4027 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
4028 (X * 2) + (Y * 4). We must, however, be assured that either the original
4029 expression would not overflow or that overflow is undefined for the type
4030 in the language in question.
4032 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
4033 the machine has a multiply-accumulate insn or that this is part of an
4034 addressing calculation.
4036 If we return a non-null expression, it is an equivalent form of the
4037 original computation, but need not be in the original type. */
4039 static tree
4041 tree t;
4042 tree c;
4044 tree wide_type;
4045 {
4055 /* Don't deal with constants of zero here; they confuse the code below. */
4065 /* Note that we need not handle conditional operations here since fold
4066 already handles those cases. So just do arithmetic here. */
4068 {
4070 /* For a constant, we can always simplify if we are a multiply
4071 or (for divide and modulus) if it is a multiple of our constant. */
4078 /* If op0 is an expression ... */
4083 /* ... and is unsigned, and its type is smaller than ctype,
4084 then we cannot pass through as widening. */
4090 /* ... or its type is larger than ctype,
4091 then we cannot pass through this truncation. */
4096 /* Pass the constant down and see if we can make a simplification. If
4097 we can, replace this expression with the inner simplification for
4098 possible later conversion to our or some other type. */
4110 /* If widening the type changes the signedness, then we can't perform
4111 this optimization as that changes the result. */
4115 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
4118 {
4124 }
4134 /* If this has not been evaluated and the operand has no side effects,
4135 we can see if we can do something inside it and make a new one.
4136 Note that this test is overly conservative since we can do this
4137 if the only reason it had side effects is that it was another
4138 similar SAVE_EXPR, but that isn't worth bothering with. */
4141 wide_type)))
4142 {
4149 }
4153 /* If the second operand is constant, this is a multiplication
4154 or floor division, by a power of two, so we can treat it that
4155 way unless the multiplier or divisor overflows. */
4157 /* const_binop may not detect overflow correctly,
4158 so check for it explicitly here. */
4172 /* See if we can eliminate the operation on both sides. If we can, we
4173 can return a new PLUS or MINUS. If we can't, the only remaining
4174 cases where we can do anything are if the second operand is a
4175 constant. */
4180 /* If not multiplication, we can only do this if either operand
4181 is divisible by c. */
4187 /* If this was a subtraction, negate OP1 and set it to be an addition.
4188 This simplifies the logic below. */
4195 /* If either OP1 or C are negative, this optimization is not safe for
4196 some of the division and remainder types while for others we need
4197 to change the code. */
4199 {
4207 }
4209 /* If it's a multiply or a division/modulus operation of a multiple
4210 of our constant, do the operation and verify it doesn't overflow. */
4213 {
4217 }
4218 else
4221 /* If we have an unsigned type is not a sizetype, we cannot widen
4222 the operation since it will change the result if the original
4223 computation overflowed. */
4229 /* If we were able to eliminate our operation from the first side,
4230 apply our operation to the second side and reform the PLUS. */
4234 /* The last case is if we are a multiply. In that case, we can
4235 apply the distributive law to commute the multiply and addition
4236 if the multiplication of the constants doesn't overflow. */
4241 op1));
4246 /* We have a special case here if we are doing something like
4247 (C * 8) % 4 since we know that's zero. */
4254 /* ... fall through ... */
4258 /* If we can extract our operation from the LHS, do so and return a
4259 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4260 do something only if the second operand is a constant. */
4272 /* If these are the same operation types, we can associate them
4273 assuming no overflow. */
4280 /* If these operations "cancel" each other, we have the main
4281 optimizations of this pass, which occur when either constant is a
4282 multiple of the other, in which case we replace this with either an
4283 operation or CODE or TCODE.
4285 If we have an unsigned type that is not a sizetype, we cannot do
4286 this since it will change the result if the original computation
4287 overflowed. */
4294 {
4305 }
4310 }
4313 }
4314 \f
4315 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4316 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4317 that we may sometimes modify the tree. */
4319 static tree
4321 tree t;
4322 tree s;
4323 {
4326 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4331 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4332 don't bother handling any other types. */
4334 {
4338 }
4343 {
4346 }
4349 }
4350 \f
4351 /* Return a node which has the indicated constant VALUE (either 0 or
4352 1), and is of the indicated TYPE. */
4354 static tree
4357 tree type;
4358 {
4363 integer_zero_node);
4364 else
4365 {
4370 }
4371 }
4373 /* Utility function for the following routine, to see how complex a nesting of
4374 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4375 we don't care (to avoid spending too much time on complex expressions.). */
4377 static int
4379 tree expr;
4381 {
4392 }
4394 /* Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
4395 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4396 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4397 expression, and ARG to `a'. If COND_FIRST_P is non-zero, then the
4398 COND is the first argument to CODE; otherwise (as in the example
4399 given here), it is the second argument. TYPE is the type of the
4400 original expression. */
4402 static tree
4405 tree type;
4406 tree cond;
4407 tree arg;
4409 {
4413 /* In the end, we'll produce a COND_EXPR. Both arms of the
4414 conditional expression will be binary operations. The left-hand
4415 side of the expression to be executed if the condition is true
4416 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4417 of the expression to be executed if the condition is true will be
4418 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4419 but apply to the expression to be executed if the conditional is
4420 false. */
4425 /* These are the codes to use for the left-hand side and right-hand
4426 side of the COND_EXPR. Normally, they are the same as CODE. */
4429 /* And these are the types of the expressions. */
4434 {
4438 }
4439 else
4440 {
4444 }
4447 {
4451 /* If this operand throws an expression, then it does not make
4452 sense to try to perform a logical or arithmetic operation
4453 involving it. Instead of building `a + throw 3' for example,
4454 we simply build `a, throw 3'. */
4456 {
4460 }
4462 {
4466 }
4467 }
4468 else
4469 {
4474 }
4476 /* If ARG is complex we want to make sure we only evaluate
4477 it once. Though this is only required if it is volatile, it
4478 might be more efficient even if it is not. However, if we
4479 succeed in folding one part to a constant, we do not need
4480 to make this SAVE_EXPR. Since we do this optimization
4481 primarily to see if we do end up with constant and this
4482 SAVE_EXPR interferes with later optimizations, suppressing
4483 it when we can is important.
4485 If we are not in a function, we can't make a SAVE_EXPR, so don't
4486 try to do so. Don't try to see if the result is a constant
4487 if an arm is a COND_EXPR since we get exponential behavior
4488 in that case. */
4495 {
4505 }
4518 else
4520 }
4522 \f
4523 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4525 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4526 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4527 ADDEND is the same as X.
4529 X + 0 and X - 0 both give X when X is NaN, infinite, or non-zero
4530 and finite. The problematic cases are when X is zero, and its mode
4531 has signed zeros. In the case of rounding towards -infinity,
4532 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4533 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4535 static bool
4539 {
4543 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4547 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4552 /* The mode has signed zeros, and we have to honor their sign.
4553 In this situation, there is only one case we can return true for.
4554 X - 0 is the same as X unless rounding towards -infinity is
4555 supported. */
4557 }
4560 /* Perform constant folding and related simplification of EXPR.
4561 The related simplifications include x*1 => x, x*0 => 0, etc.,
4562 and application of the associative law.
4563 NOP_EXPR conversions may be removed freely (as long as we
4564 are careful not to change the C type of the overall expression)
4565 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4566 but we can constant-fold them if they have constant operands. */
4568 tree
4570 tree expr;
4571 {
4574 tree tem;
4580 /* WINS will be nonzero when the switch is done
4581 if all operands are constant. */
4584 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4585 Likewise for a SAVE_EXPR that's already been evaluated. */
4589 /* Return right away if a constant. */
4593 #ifdef MAX_INTEGER_COMPUTATION_MODE
4595 #endif
4598 {
4599 tree subop;
4601 /* Special case for conversion ops that can have fixed point args. */
4604 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4610 else
4615 )
4616 /* Note that TREE_CONSTANT isn't enough:
4617 static var addresses are constant but we can't
4618 do arithmetic on them. */
4620 }
4622 {
4626 {
4628 tree subop;
4634 {
4635 /* Signedness matters here. Perhaps we can refine this
4636 later. */
4638 }
4639 else
4640 /* Strip any conversions that don't change the mode. */
4645 else
4650 /* Note that TREE_CONSTANT isn't enough:
4651 static var addresses are constant but we can't
4652 do arithmetic on them. */
4659 }
4660 }
4662 /* If this is a commutative operation, and ARG0 is a constant, move it
4663 to ARG1 to reduce the number of tests below. */
4668 {
4673 }
4675 /* Now WINS is set as described above,
4676 ARG0 is the first operand of EXPR,
4677 and ARG1 is the second operand (if it has more than one operand).
4679 First check for cases where an arithmetic operation is applied to a
4680 compound, conditional, or comparison operation. Push the arithmetic
4681 operation inside the compound or conditional to see if any folding
4682 can then be done. Convert comparison to conditional for this purpose.
4683 The also optimizes non-constant cases that used to be done in
4684 expand_expr.
4686 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4687 one of the operands is a comparison and the other is a comparison, a
4688 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4689 code below would make the expression more complex. Change it to a
4690 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4691 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4703 {
4713 }
4716 {
4721 {
4726 /* If this was a conversion, and all we did was to move into
4727 inside the COND_EXPR, bring it back out. But leave it if
4728 it is a conversion from integer to integer and the
4729 result precision is no wider than a word since such a
4730 conversion is cheap and may be optimized away by combine,
4731 while it couldn't if it were outside the COND_EXPR. Then return
4732 so we don't get into an infinite recursion loop taking the
4733 conversion out and then back in. */
4743 && (INTEGRAL_TYPE_P
4754 }
4759 }
4762 {
4775 return
4789 return
4792 }
4803 {
4819 /* Other kinds of FIX are not handled properly by fold_convert. */
4824 /* Handle cases of two conversions in a row. */
4827 {
4847 /* In addition to the cases of two conversions in a row
4848 handled below, if we are converting something to its own
4849 type via an object of identical or wider precision, neither
4850 conversion is needed. */
4856 /* Likewise, if the intermediate and final types are either both
4857 float or both integer, we don't need the middle conversion if
4858 it is wider than the final type and doesn't change the signedness
4859 (for integers). Avoid this if the final type is a pointer
4860 since then we sometimes need the inner conversion. Likewise if
4861 the outer has a precision not equal to the size of its mode. */
4871 /* If we have a sign-extension of a zero-extended value, we can
4872 replace that by a single zero-extension. */
4878 /* Two conversions in a row are not needed unless:
4879 - some conversion is floating-point (overstrict for now), or
4880 - the intermediate type is narrower than both initial and
4881 final, or
4882 - the intermediate type and innermost type differ in signedness,
4883 and the outermost type is wider than the intermediate, or
4884 - the initial type is a pointer type and the precisions of the
4885 intermediate and final types differ, or
4886 - the final type is a pointer type and the precisions of the
4887 initial and intermediate types differ. */
4901 }
4905 /* Detect assigning a bitfield. */
4908 {
4909 /* Don't leave an assignment inside a conversion
4910 unless assigning a bitfield. */
4913 /* First do the assignment, then return converted constant. */
4917 }
4919 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
4920 constants (if x has signed type, the sign bit cannot be set
4921 in c). This folds extension into the BIT_AND_EXPR. */
4926 {
4936 <= HOST_BITS_PER_WIDE_INT
4938 {
4945 #ifdef LOAD_EXTEND_OP
4949 {
4953 }
4954 #endif
4955 }
4960 }
4963 {
4966 }
4977 {
4981 }
4990 {
4992 {
4994 HOST_WIDE_INT high;
5005 }
5008 }
5012 /* Convert - (a - b) to (b - a) for non-floating-point. */
5022 {
5024 {
5025 /* If the value is unsigned, then the absolute value is
5026 the same as the ordinary value. */
5029 /* Similarly, if the value is non-negative. */
5032 /* If the value is negative, then the absolute value is
5033 its negation. */
5034 else
5035 {
5037 HOST_WIDE_INT high;
5048 }
5049 }
5051 {
5055 }
5056 }
5083 {
5090 }
5096 /* A + (-B) -> A - B */
5099 /* (-A) + B -> B - A */
5103 {
5107 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
5108 with a constant, and the two constants have no bits in common,
5109 we should treat this as a BIT_IOR_EXPR since this may produce more
5110 simplifications. */
5118 {
5121 }
5123 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
5124 (plus (plus (mult) (mult)) (foo)) so that we can
5125 take advantage of the factoring cases below. */
5130 {
5135 else
5146 parg1));
5151 parg0));
5152 }
5155 {
5159 /* (A * C) + (B * C) -> (A+B) * C.
5160 We are most concerned about the case where C is a constant,
5161 but other combinations show up during loop reduction. Since
5162 it is not difficult, try all four possibilities. */
5179 /* No identical multiplicands; see if we can find a common
5180 power-of-two factor in non-power-of-two multiplies. This
5181 can help in multi-dimensional array access. */
5186 {
5191 /* Move min of absolute values to int11. */
5194 {
5198 }
5201 {
5206 }
5207 }
5212 same));
5213 }
5214 }
5216 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5220 /* Likewise if the operands are reversed. */
5224 bit_rotate:
5225 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5226 is a rotate of A by C1 bits. */
5227 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5228 is a rotate of A by B bits. */
5229 {
5238 {
5257 {
5265 TYPE_PRECISION
5270 ? LROTATE_EXPR
5273 }
5275 {
5283 TYPE_PRECISION
5288 ? LROTATE_EXPR
5291 }
5292 }
5293 }
5295 associate:
5296 /* In most languages, can't associate operations on floats through
5297 parentheses. Rather than remember where the parentheses were, we
5298 don't associate floats at all. It shouldn't matter much. However,
5299 associating multiplications is only very slightly inaccurate, so do
5300 that if -funsafe-math-optimizations is specified. */
5305 {
5309 /* Split both trees into variables, constants, and literals. Then
5310 associate each group together, the constants with literals,
5311 then the result with variables. This increases the chances of
5312 literals being recombined later and of generating relocatable
5313 expressions for the sum of a constant and literal. */
5318 /* Only do something if we found more than two objects. Otherwise,
5319 nothing has changed and we risk infinite recursion. */
5324 {
5325 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5334 /* Preserve the MINUS_EXPR if the negative part of the literal is
5335 greater than the positive part. Otherwise, the multiplicative
5336 folding code (i.e extract_muldiv) may be fooled in case
5337 unsigned constants are substracted, like in the following
5338 example: ((X*2 + 4) - 8U)/2. */
5340 {
5342 {
5346 }
5347 else
5348 {
5352 }
5353 }
5355 {
5359 else
5360 {
5365 }
5366 }
5370 }
5371 }
5373 binary:
5377 {
5378 /* The return value should always have
5379 the same type as the original expression. */
5384 }
5388 /* A - (-B) -> A + B */
5391 /* (-A) - CST -> (-CST) - A for floating point (what about ints ?) */
5393 return
5400 {
5406 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5407 about the case where C is a constant, just try one of the
5408 four possibilities. */
5418 }
5420 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5424 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5425 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5426 (-ARG1 + ARG0) reduces to -ARG1. */
5430 /* Fold &x - &x. This can happen from &x.foo - &x.
5431 This is unsafe for certain floats even in non-IEEE formats.
5432 In IEEE, it is unsafe because it does wrong for NaNs.
5433 Also note that operand_equal_p is always false if an operand
5434 is volatile. */
5443 /* (-A) * (-B) -> A * B */
5449 {
5455 /* (a * (1 << b)) is (a << b) */
5470 }
5471 else
5472 {
5473 /* Maybe fold x * 0 to 0. The expressions aren't the same
5474 when x is NaN, since x * 0 is also NaN. Nor are they the
5475 same in modes with signed zeros, since multiplying a
5476 negative value by 0 gives -0, not +0. */
5481 /* In IEEE floating point, x*1 is not equivalent to x for snans. */
5486 /* Transform x * -1.0 into -x. */
5491 /* x*2 is x+x */
5495 {
5498 }
5499 }
5503 bit_ior:
5512 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5514 This results in more efficient code for machines without a NAND
5515 instruction. Combine will canonicalize to the first form
5516 which will allow use of NAND instructions provided by the
5517 backend if they exist. */
5520 {
5525 }
5527 /* See if this can be simplified into a rotate first. If that
5528 is unsuccessful continue in the association code. */
5537 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5538 with a constant, and the two constants have no bits in common,
5539 we should treat this as a BIT_IOR_EXPR since this may produce more
5540 simplifications. */
5548 {
5551 }
5553 /* See if this can be simplified into a rotate first. If that
5554 is unsuccessful continue in the association code. */
5558 bit_and:
5566 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5569 {
5570 unsigned int prec
5577 }
5579 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5581 This results in more efficient code for machines without a NOR
5582 instruction. Combine will canonicalize to the first form
5583 which will allow use of NOR instructions provided by the
5584 backend if they exist. */
5587 {
5592 }
5602 {
5606 }
5610 /* Don't touch a floating-point divide by zero unless the mode
5611 of the constant can represent infinity. */
5617 /* (-A) / (-B) -> A / B */
5622 /* In IEEE floating point, x/1 is not equivalent to x for snans. */
5627 /* If ARG1 is a constant, we can convert this to a multiply by the
5628 reciprocal. This does not have the same rounding properties,
5629 so only do this if -funsafe-math-optimizations. We can actually
5630 always safely do it if ARG1 is a power of two, but it's hard to
5631 tell if it is or not in a portable manner. */
5633 {
5638 /* Find the reciprocal if optimizing and the result is exact. */
5640 {
5641 REAL_VALUE_TYPE r;
5644 {
5647 }
5648 }
5649 }
5650 /* Convert A/B/C to A/(B*C). */
5653 {
5656 arg1)));
5657 }
5658 /* Convert A/(B/C) to (A/B)*C. */
5661 {
5666 }
5679 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5680 operation, EXACT_DIV_EXPR.
5682 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5683 At one time others generated faster code, it's not clear if they do
5684 after the last round to changes to the DIV code in expmed.c. */
5718 /* Since negative shift count is not well-defined,
5719 don't try to compute it in the compiler. */
5722 /* Rewrite an LROTATE_EXPR by a constant into an
5723 RROTATE_EXPR by a new constant. */
5725 {
5729 = const_binop
5736 }
5738 /* If we have a rotate of a bit operation with the rotate count and
5739 the second operand of the bit operation both constant,
5740 permute the two operations. */
5753 /* Two consecutive rotates adding up to the width of the mode can
5754 be ignored. */
5785 /* Note that the operand of this must be an int
5786 and its values must be 0 or 1.
5787 ("true" is a fixed value perhaps depending on the language,
5788 but we don't handle values other than 1 correctly yet.) */
5790 /* Avoid infinite recursion. */
5796 /* Note that the operands of this must be ints
5797 and their values must be 0 or 1.
5798 ("true" is a fixed value perhaps depending on the language.) */
5799 /* If first arg is constant zero, return it. */
5803 /* If either arg is constant true, drop it. */
5807 /* Preserve sequence points. */
5810 /* If second arg is constant zero, result is zero, but first arg
5811 must be evaluated. */
5814 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
5815 case will be handled here. */
5819 truth_andor:
5820 /* We only do these simplifications if we are optimizing. */
5824 /* Check for things like (A || B) && (A || C). We can convert this
5825 to A || (B && C). Note that either operator can be any of the four
5826 truth and/or operations and the transformation will still be
5827 valid. Also note that we only care about order for the
5828 ANDIF and ORIF operators. If B contains side effects, this
5829 might change the truth-value of A. */
5836 {
5856 /* This case if tricky because we must either have commutative
5857 operators or else A10 must not have side-effects. */
5863 a01));
5864 }
5866 /* See if we can build a range comparison. */
5870 /* Check for the possibility of merging component references. If our
5871 lhs is another similar operation, try to merge its rhs with our
5872 rhs. Then try to merge our lhs and rhs. */
5884 /* Note that the operands of this must be ints
5885 and their values must be 0 or true.
5886 ("true" is a fixed value perhaps depending on the language.) */
5887 /* If first arg is constant true, return it. */
5891 /* If either arg is constant zero, drop it. */
5895 /* Preserve sequence points. */
5898 /* If second arg is constant true, result is true, but we must
5899 evaluate first arg. */
5902 /* Likewise for first arg, but note this only occurs here for
5903 TRUTH_OR_EXPR. */
5909 /* If either arg is constant zero, drop it. */
5914 /* If either arg is constant true, this is a logical inversion. */
5927 /* If one arg is a real or integer constant, put it last. */
5932 {
5939 }
5942 {
5943 /* (-a) CMP (-b) -> b CMP a */
5948 /* (-a) CMP CST -> a swap(CMP) (-CST) */
5950 return
5956 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
5957 /* a CMP (-0) -> a CMP 0 */
5963 /* If this is a comparison of a real constant with a PLUS_EXPR
5964 or a MINUS_EXPR of a real constant, we can convert it into a
5965 comparison with a revised real constant as long as no overflow
5966 occurs when unsafe_math_optimizations are enabled. */
5977 }
5979 /* Convert foo++ == CONST into ++foo == CONST + INCR.
5980 First, see if one arg is constant; find the constant arg
5981 and the other one. */
5982 {
5992 {
5993 /* This optimization is invalid for ordered comparisons
5994 if CONST+INCR overflows or if foo+incr might overflow.
5995 This optimization is invalid for floating point due to rounding.
5996 For pointer types we assume overflow doesn't happen. */
6000 {
6001 tree newconst
6005 /* Do not overwrite the current varop to be a preincrement,
6006 create a new node so that we won't confuse our caller who
6007 might create trees and throw them away, reusing the
6008 arguments that they passed to build. This shows up in
6009 the THEN or ELSE parts of ?: being postincrements. */
6014 /* If VAROP is a reference to a bitfield, we must mask
6015 the constant by the width of the field. */
6019 {
6020 int size
6022 (TREE_OPERAND
6026 tree folded_compare;
6028 /* First check whether the comparison would come out
6029 always the same. If we don't do that we would
6030 change the meaning with the masking. */
6034 else
6037 constop));
6052 mask)));
6053 }
6059 }
6060 }
6062 {
6066 {
6067 tree newconst
6071 /* Do not overwrite the current varop to be a predecrement,
6072 create a new node so that we won't confuse our caller who
6073 might create trees and throw them away, reusing the
6074 arguments that they passed to build. This shows up in
6075 the THEN or ELSE parts of ?: being postdecrements. */
6083 {
6084 int size
6086 (TREE_OPERAND
6090 tree folded_compare;
6095 else
6098 constop));
6113 mask)));
6114 }
6120 }
6121 }
6122 }
6124 /* Change X >= C to X > (C - 1) and X < C to X <= (C - 1) if C > 0.
6125 This transformation affects the cases which are handled in later
6126 optimizations involving comparisons with non-negative constants. */
6130 {
6132 {
6147 }
6148 }
6150 /* Comparisons with the highest or lowest possible integer of
6151 the specified size will have known values. */
6152 {
6160 {
6167 {
6170 }
6171 else
6172 {
6175 }
6180 {
6184 arg0);
6192 arg0);
6198 /* The GE_EXPR and LT_EXPR cases above are not normally
6199 reached because of previous transformations. */
6203 }
6207 {
6220 }
6224 {
6228 arg0);
6237 arg0);
6245 }
6249 {
6262 }
6267 /* signed_type does not work on pointer types. */
6269 {
6270 /* The following case also applies to X < signed_max+1
6271 and X >= signed_max+1 because previous transformations. */
6273 {
6277 return fold
6281 }
6282 }
6283 }
6284 }
6286 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
6287 a MINUS_EXPR of a constant, we can convert it into a comparison with
6288 a revised constant as long as no overflow occurs. */
6300 /* Similarly for a NEGATE_EXPR. */
6309 /* If we have X - Y == 0, we can convert that to X == Y and similarly
6310 for !=. Don't do this for ordered comparisons due to overflow. */
6316 /* If we are widening one operand of an integer comparison,
6317 see if the other operand is similarly being widened. Perhaps we
6318 can do the comparison in the narrower type. */
6328 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
6329 constant, we can simplify it. */
6336 /* If we are comparing an ABS_EXPR with a constant, we can
6337 convert all the cases into explicit comparisons, but they may
6338 well not be faster than doing the ABS and one comparison.
6339 But ABS (X) <= C is a range comparison, which becomes a subtraction
6340 and a comparison, and is probably faster. */
6352 /* If this is an EQ or NE comparison with zero and ARG0 is
6353 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
6354 two operations, but the latter can be done in one less insn
6355 on machines that have only two-operand insns or on which a
6356 constant cannot be the first operand. */
6359 {
6362 return
6370 integer_one_node)),
6371 arg1));
6374 return
6382 integer_one_node)),
6383 arg1));
6384 }
6386 /* If this is an NE or EQ comparison of zero against the result of a
6387 signed MOD operation whose second operand is a power of 2, make
6388 the MOD operation unsigned since it is simpler and equivalent. */
6397 {
6404 }
6406 /* If this is an NE comparison of zero with an AND of one, remove the
6407 comparison since the AND will give the correct value. */
6413 /* If we have (A & C) == C where C is a power of 2, convert this into
6414 (A & C) != 0. Similarly for NE_EXPR. */
6422 /* If we have (A & C) != 0 where C is the sign bit of A, convert
6423 this into A < 0. Similarly for (A & C) == 0 into A >= 0. */
6427 {
6431 {
6436 }
6437 }
6439 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6440 and similarly for >= into !=. */
6456 return
6463 /* Simplify comparison of something with itself. (For IEEE
6464 floating-point, we can only do some of these simplifications.) */
6466 {
6468 {
6479 /* For NE, we can only do this simplification if integer. */
6482 /* ... fall through ... */
6488 }
6489 }
6491 /* If we are comparing an expression that just has comparisons
6492 of two integer values, arithmetic expressions of those comparisons,
6493 and constants, we can simplify it. There are only three cases
6494 to check: the two values can either be equal, the first can be
6495 greater, or the second can be greater. Fold the expression for
6496 those three values. Since each value must be 0 or 1, we have
6497 eight possibilities, each of which corresponds to the constant 0
6498 or 1 or one of the six possible comparisons.
6500 This handles common cases like (a > b) == 0 but also handles
6501 expressions like ((x > y) - (y > x)) > 0, which supposedly
6502 occur in macroized code. */
6505 {
6510 /* Don't handle degenerate cases here; they should already
6511 have been handled anyway. */
6520 {
6524 /* We can't just pass T to eval_subst in case cval1 or cval2
6525 was the same as ARG1. */
6527 tree high_result
6530 arg1));
6531 tree equal_result
6534 arg1));
6535 tree low_result
6538 arg1));
6540 /* All three of these results should be 0 or 1. Confirm they
6541 are. Then use those values to select the proper code
6542 to use. */
6550 {
6551 /* Make a 3-bit mask with the high-order bit being the
6552 value for `>', the next for '=', and the low for '<'. */
6556 {
6558 /* Always false. */
6579 /* Always true. */
6581 }
6586 else
6588 }
6589 }
6590 }
6592 /* If this is a comparison of a field, we may be able to simplify it. */
6596 /* Handle the constant case even without -O
6597 to make sure the warnings are given. */
6599 {
6602 }
6604 /* If this is a comparison of complex values and either or both sides
6605 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6606 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6607 This may prevent needless evaluations. */
6614 {
6627 type,
6630 }
6632 /* Optimize comparisons of strlen vs zero to a compare of the
6633 first character of the string vs zero. To wit,
6634 strlen(ptr) == 0 => *ptr == 0
6635 strlen(ptr) != 0 => *ptr != 0
6636 Other cases should reduce to one of these two (or a constant)
6637 due to the return value of strlen being unsigned. */
6642 {
6644 tree arglist;
6656 integer_zero_node));
6657 }
6659 /* From here on, the only cases we handle are when the result is
6660 known to be a constant.
6662 To compute GT, swap the arguments and do LT.
6663 To compute GE, do LT and invert the result.
6664 To compute LE, swap the arguments, do LT and invert the result.
6665 To compute NE, do EQ and invert the result.
6667 Therefore, the code below must handle only EQ and LT. */
6670 {
6673 }
6675 /* Note that it is safe to invert for real values here because we
6676 will check below in the one case that it matters. */
6681 {
6684 }
6686 /* Compute a result for LT or EQ if args permit;
6687 otherwise return T. */
6689 {
6692 else
6697 }
6700 /* Assume a nonexplicit constant cannot equal an explicit one,
6701 since such code would be undefined anyway.
6702 Exception: on sysvr4, using #pragma weak,
6703 a label can come out as 0. */
6710 #endif
6711 /* Two real constants can be compared explicitly. */
6713 {
6714 /* If either operand is a NaN, the result is false with two
6715 exceptions: First, an NE_EXPR is true on NaNs, but that case
6716 is already handled correctly since we will be inverting the
6717 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6718 or a GE_EXPR into a LT_EXPR, we must return true so that it
6719 will be inverted into false. */
6729 else
6733 }
6747 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6748 so all simple results must be passed through pedantic_non_lvalue. */
6750 return pedantic_non_lvalue
6755 /* If the second operand is zero, invert the comparison and swap
6756 the second and third operands. Likewise if the second operand
6757 is constant and the third is not or if the third operand is
6758 equivalent to the first operand of the comparison. */
6766 {
6767 /* See if this can be inverted. If it can't, possibly because
6768 it was a floating-point inequality comparison, don't do
6769 anything. */
6773 {
6777 /* arg1 should be the first argument of the new T. */
6780 }
6781 }
6783 /* If we have A op B ? A : C, we may be able to convert this to a
6784 simpler expression, depending on the operation and the values
6785 of B and C. Signed zeros prevent all of these transformations,
6786 for reasons given above each one. */
6792 {
6798 /* If we have A op 0 ? A : -A, consider applying the following
6799 transformations:
6801 A == 0? A : -A same as -A
6802 A != 0? A : -A same as A
6803 A >= 0? A : -A same as abs (A)
6804 A > 0? A : -A same as abs (A)
6805 A <= 0? A : -A same as -abs (A)
6806 A < 0? A : -A same as -abs (A)
6808 None of these transformations work for modes with signed
6809 zeros. If A is +/-0, the first two transformations will
6810 change the sign of the result (from +0 to -0, or vice
6811 versa). The last four will fix the sign of the result,
6812 even though the original expressions could be positive or
6813 negative, depending on the sign of A.
6815 Note that all these transformations are correct if A is
6816 NaN, since the two alternatives (A and -A) are also NaNs. */
6823 {
6825 return
6826 pedantic_non_lvalue
6828 negate_expr
6830 arg1))));
6838 return pedantic_non_lvalue
6846 return pedantic_non_lvalue
6850 arg1)))));
6853 }
6855 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
6856 A == 0 ? A : 0 is always 0 unless A is -0. Note that
6857 both transformations are correct when A is NaN: A != 0
6858 is then true, and A == 0 is false. */
6861 {
6866 }
6868 /* Try some transformations of A op B ? A : B.
6870 A == B? A : B same as B
6871 A != B? A : B same as A
6872 A >= B? A : B same as max (A, B)
6873 A > B? A : B same as max (B, A)
6874 A <= B? A : B same as min (A, B)
6875 A < B? A : B same as min (B, A)
6877 As above, these transformations don't work in the presence
6878 of signed zeros. For example, if A and B are zeros of
6879 opposite sign, the first two transformations will change
6880 the sign of the result. In the last four, the original
6881 expressions give different results for (A=+0, B=-0) and
6882 (A=-0, B=+0), but the transformed expressions do not.
6884 The first two transformations are correct if either A or B
6885 is a NaN. In the first transformation, the condition will
6886 be false, and B will indeed be chosen. In the case of the
6887 second transformation, the condition A != B will be true,
6888 and A will be chosen.
6890 The conversions to max() and min() are not correct if B is
6891 a number and A is not. The conditions in the original
6892 expressions will be false, so all four give B. The min()
6893 and max() versions would give a NaN instead. */
6896 {
6901 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
6906 {
6913 /* In C++ a ?: expression can be an lvalue, so put the
6914 operand which will be used if they are equal first
6915 so that we can convert this back to the
6916 corresponding COND_EXPR. */
6918 return pedantic_non_lvalue
6928 return pedantic_non_lvalue
6937 }
6938 }
6940 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
6941 we might still be able to simplify this. For example,
6942 if C1 is one less or one more than C2, this might have started
6943 out as a MIN or MAX and been transformed by this function.
6944 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
6950 {
6952 /* We can replace A with C1 in this case. */
6959 /* If C1 is C2 + 1, this is min(A, C2). */
6964 return pedantic_non_lvalue
6969 /* If C1 is C2 - 1, this is min(A, C2). */
6974 return pedantic_non_lvalue
6979 /* If C1 is C2 - 1, this is max(A, C2). */
6984 return pedantic_non_lvalue
6989 /* If C1 is C2 + 1, this is max(A, C2). */
6994 return pedantic_non_lvalue
7001 }
7002 }
7004 /* If the second operand is simpler than the third, swap them
7005 since that produces better jump optimization results. */
7011 {
7012 /* See if this can be inverted. If it can't, possibly because
7013 it was a floating-point inequality comparison, don't do
7014 anything. */
7018 {
7022 /* arg1 should be the first argument of the new T. */
7025 }
7026 }
7028 /* Convert A ? 1 : 0 to simply A. */
7031 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
7032 call to fold will try to move the conversion inside
7033 a COND, which will recurse. In that case, the COND_EXPR
7034 is probably the best choice, so leave it alone. */
7038 /* Convert A ? 0 : 1 to !A. This prefers the use of NOT_EXPR
7039 over COND_EXPR in cases such as floating point comparisons. */
7046 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
7047 operation is simply A & 2. */
7058 /* Convert A ? B : 0 into A && B if A and B are truth values. */
7065 /* Convert A ? B : 1 into !A || B if A and B are truth values. */
7069 {
7070 /* Only perform transformation if ARG0 is easily inverted. */
7075 }
7080 /* When pedantic, a compound expression can be neither an lvalue
7081 nor an integer constant expression. */
7084 /* Don't let (0, 0) be null pointer constant. */
7126 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
7127 appropriate. */
7132 {
7136 tree arg01;
7147 {
7161 arg01));
7162 }
7165 }
7168 /* Check for a built-in function. */
7173 {
7177 }
7183 }
7185 /* Determine if first argument is a multiple of second argument. Return 0 if
7186 it is not, or we cannot easily determined it to be.
7188 An example of the sort of thing we care about (at this point; this routine
7189 could surely be made more general, and expanded to do what the *_DIV_EXPR's
7190 fold cases do now) is discovering that
7192 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7194 is a multiple of
7196 SAVE_EXPR (J * 8)
7198 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
7200 This code also handles discovering that
7202 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7204 is a multiple of 8 so we don't have to worry about dealing with a
7205 possible remainder.
7207 Note that we *look* inside a SAVE_EXPR only to determine how it was
7208 calculated; it is not safe for fold to do much of anything else with the
7209 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
7210 at run time. For example, the latter example above *cannot* be implemented
7211 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
7212 evaluation time of the original SAVE_EXPR is not necessarily the same at
7213 the time the new expression is evaluated. The only optimization of this
7214 sort that would be valid is changing
7216 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
7218 divided by 8 to
7220 SAVE_EXPR (I) * SAVE_EXPR (J)
7222 (where the same SAVE_EXPR (J) is used in the original and the
7223 transformed version). */
7225 static int
7227 tree type;
7228 tree top;
7229 tree bottom;
7230 {
7238 {
7250 {
7254 /* const_binop may not detect overflow correctly,
7255 so check for it explicitly here. */
7264 }
7268 /* Can't handle conversions from non-integral or wider integral type. */
7274 /* .. fall through ... */
7290 }
7291 }
7293 /* Return true if `t' is known to be non-negative. */
7295 int
7297 tree t;
7298 {
7300 {
7341 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7343 else
7344 /* We don't know sign of `t', so be conservative and return false. */
7346 }
7347 }
7349 /* Return true if `r' is known to be non-negative.
7350 Only handles constants at the moment. */
7352 int
7354 rtx r;
7355 {
7357 {
7367 {
7369 rtx elt;
7374 {
7378 }
7381 }
7385 /* These are always nonnegative. */
7390 }
7391 }