| blob | 2e869017dee72d33d9f0734125f1dbf6f6016255 |
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, 2002,
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 int
185 tree t;
187 {
189 HOST_WIDE_INT high;
193 {
194 /* ??? Used to check for overflow here via CHECK_FLOAT_TYPE.
195 Consider doing it via real_convert now. */
197 }
207 else
210 /* First clear all bits that are beyond the type's precision. */
213 ;
217 else
218 {
222 }
224 /* Unsigned types do not suffer sign extension or overflow unless they
225 are a sizetype. */
231 /* If the value's sign bit is set, extend the sign. */
239 {
240 /* Value is negative:
241 set to 1 all the bits that are outside this type's precision. */
245 else
246 {
250 }
251 }
253 /* Return nonzero if signed overflow occurred. */
254 return
257 }
258 \f
259 /* Add two doubleword integers with doubleword result.
260 Each argument is given as two `HOST_WIDE_INT' pieces.
261 One argument is L1 and H1; the other, L2 and H2.
262 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
264 int
270 {
272 HOST_WIDE_INT h;
280 }
282 /* Negate a doubleword integer with doubleword result.
283 Return nonzero if the operation overflows, assuming it's signed.
284 The argument is given as two `HOST_WIDE_INT' pieces in L1 and H1.
285 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
287 int
290 HOST_WIDE_INT h1;
293 {
295 {
299 }
300 else
301 {
305 }
306 }
307 \f
308 /* Multiply two doubleword integers with doubleword result.
309 Return nonzero if the operation overflows, assuming it's signed.
310 Each argument is given as two `HOST_WIDE_INT' pieces.
311 One argument is L1 and H1; the other, L2 and H2.
312 The value is stored as two `HOST_WIDE_INT' pieces in *LV and *HV. */
314 int
320 {
335 {
338 {
340 /* This product is <= 0xFFFE0001, the sum <= 0xFFFF0000. */
342 /* Since prod[p] < 0xFFFF, this sum <= 0xFFFFFFFF. */
346 }
348 }
352 /* Check for overflow by calculating the top half of the answer in full;
353 it should agree with the low half's sign bit. */
356 {
359 }
361 {
364 }
366 }
367 \f
368 /* Shift the doubleword integer in L1, H1 left by COUNT places
369 keeping only PREC bits of result.
370 Shift right if COUNT is negative.
371 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
372 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
374 void
382 {
386 {
389 }
391 #ifdef SHIFT_COUNT_TRUNCATED
394 #endif
397 {
398 /* Shifting by the host word size is undefined according to the
399 ANSI standard, so we must handle this as a special case. */
402 }
404 {
407 }
408 else
409 {
413 }
415 /* Sign extend all bits that are beyond the precision. */
423 ;
425 {
428 }
429 else
430 {
434 }
435 }
437 /* Shift the doubleword integer in L1, H1 right by COUNT places
438 keeping only PREC bits of result. COUNT must be positive.
439 ARITH nonzero specifies arithmetic shifting; otherwise use logical shift.
440 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
442 void
450 {
453 signmask = (arith
457 #ifdef SHIFT_COUNT_TRUNCATED
460 #endif
463 {
464 /* Shifting by the host word size is undefined according to the
465 ANSI standard, so we must handle this as a special case. */
468 }
470 {
473 }
474 else
475 {
479 }
481 /* Zero / sign extend all bits that are beyond the precision. */
484 {
487 }
489 ;
491 {
494 }
495 else
496 {
500 }
501 }
502 \f
503 /* Rotate the doubleword integer in L1, H1 left by COUNT places
504 keeping only PREC bits of result.
505 Rotate right if COUNT is negative.
506 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
508 void
515 {
527 }
529 /* Rotate the doubleword integer in L1, H1 left by COUNT places
530 keeping only PREC bits of result. COUNT must be positive.
531 Store the value as two `HOST_WIDE_INT' pieces in *LV and *HV. */
533 void
540 {
552 }
553 \f
554 /* Divide doubleword integer LNUM, HNUM by doubleword integer LDEN, HDEN
555 for a quotient (stored in *LQUO, *HQUO) and remainder (in *LREM, *HREM).
556 CODE is a tree code for a kind of division, one of
557 TRUNC_DIV_EXPR, FLOOR_DIV_EXPR, CEIL_DIV_EXPR, ROUND_DIV_EXPR
558 or EXACT_DIV_EXPR
559 It controls how the quotient is rounded to an integer.
560 Return nonzero if the operation overflows.
561 UNS nonzero says do unsigned division. */
563 int
570 HOST_WIDE_INT hnum_orig;
572 HOST_WIDE_INT hden_orig;
575 {
591 /* calculate quotient sign and convert operands to unsigned. */
593 {
595 {
597 /* (minimum integer) / (-1) is the only overflow case. */
601 }
603 {
606 }
607 }
612 /* This unsigned division rounds toward zero. */
615 }
619 /* hden != 0 already checked. */
624 }
634 /* Special code for when the divisor < BASE. */
636 {
637 /* hnum != 0 already checked. */
639 {
643 }
644 }
645 else
646 {
647 /* Full double precision division,
648 with thanks to Don Knuth's "Seminumerical Algorithms". */
652 /* Find the highest nonzero divisor digit. */
655 {
658 }
660 /* Insure that the first digit of the divisor is at least BASE/2.
661 This is required by the quotient digit estimation algorithm. */
668 {
672 }
677 {
682 }
683 }
687 /* Main loop */
689 {
690 /* Guess the next quotient digit, quo_est, by dividing the first
691 two remaining dividend digits by the high order quotient digit.
692 quo_est is never low and is at most 2 high. */
699 else
702 /* Refine quo_est so it's usually correct, and at most one high. */
707 quo_est--;
709 /* Try QUO_EST as the quotient digit, by multiplying the
710 divisor by QUO_EST and subtracting from the remaining dividend.
711 Keep in mind that QUO_EST is the I - 1st digit. */
715 {
721 }
723 /* If quo_est was high by one, then num[i] went negative and
724 we need to correct things. */
726 {
727 quo_est--;
730 {
734 }
737 }
739 /* Store the quotient digit. */
741 }
742 }
746 finish_up:
747 /* if result is negative, make it so. */
751 /* compute trial remainder: rem = num - (quo * den) */
757 {
766 {
767 /* quo = quo - 1; */
770 }
771 else
778 {
781 }
782 else
788 {
794 /* Get absolute values */
800 /* If (2 * abs (lrem) >= abs (lden)) */
809 {
811 /* quo = quo - 1; */
814 else
815 /* quo = quo + 1; */
818 }
819 else
821 }
826 }
828 /* compute true remainder: rem = num - (quo * den) */
833 }
834 \f
835 /* Given T, an expression, return the negation of T. Allow for T to be
836 null, in which case return null. */
838 static tree
840 tree t;
841 {
842 tree type;
843 tree tem;
852 {
865 /* - (A - B) -> B - A */
875 }
878 }
879 \f
880 /* Split a tree IN into a constant, literal and variable parts that could be
881 combined with CODE to make IN. "constant" means an expression with
882 TREE_CONSTANT but that isn't an actual constant. CODE must be a
883 commutative arithmetic operation. Store the constant part into *CONP,
884 the literal in *LITP and return the variable part. If a part isn't
885 present, set it to null. If the tree does not decompose in this way,
886 return the entire tree as the variable part and the other parts as null.
888 If CODE is PLUS_EXPR we also split trees that use MINUS_EXPR. In that
889 case, we negate an operand that was subtracted. Except if it is a
890 literal for which we use *MINUS_LITP instead.
892 If NEGATE_P is true, we are negating all of IN, again except a literal
893 for which we use *MINUS_LITP instead.
895 If IN is itself a literal or constant, return it as appropriate.
897 Note that we do not guarantee that any of the three values will be the
898 same type as IN, but they will have the same signedness and mode. */
900 static tree
902 tree in;
906 {
913 /* Strip any conversions that don't change the machine mode or signedness. */
920 /* We can associate addition and subtraction together (even
921 though the C standard doesn't say so) for integers because
922 the value is not affected. For reals, the value might be
923 affected, so we can't. */
926 {
932 /* First see if either of the operands is a literal, then a constant. */
943 /* If we haven't dealt with either operand, this is not a case we can
944 decompose. Otherwise, VAR is either of the ones remaining, if any. */
949 else
952 /* Now do any needed negations. */
959 }
962 else
966 {
973 }
976 }
978 /* Re-associate trees split by the above function. T1 and T2 are either
979 expressions to associate or null. Return the new expression, if any. If
980 we build an operation, do it in TYPE and with CODE. */
982 static tree
986 tree type;
987 {
993 /* If either input is CODE, a PLUS_EXPR, or a MINUS_EXPR, don't
994 try to fold this since we will have infinite recursion. But do
995 deal with any NEGATE_EXPRs. */
998 {
1000 {
1007 }
1009 }
1012 }
1013 \f
1014 /* Combine two integer constants ARG1 and ARG2 under operation CODE
1015 to produce a new constant.
1017 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1019 static tree
1024 {
1028 HOST_WIDE_INT hi;
1030 HOST_WIDE_INT garbageh;
1031 tree t;
1034 int is_sizetype
1045 {
1065 /* It's unclear from the C standard whether shifts can overflow.
1066 The following code ignores overflow; perhaps a C standard
1067 interpretation ruling is needed. */
1097 /* This is a shortcut for a common special case. */
1102 {
1108 }
1110 /* ... fall through ... */
1114 {
1117 }
1120 {
1123 }
1130 /* This is a shortcut for a common special case. */
1135 {
1140 }
1142 /* ... fall through ... */
1158 else
1164 else
1170 }
1172 /* If this is for a sizetype, can be represented as one (signed)
1173 HOST_WIDE_INT word, and doesn't overflow, use size_int since it caches
1174 constants. */
1180 else
1181 {
1184 }
1187 = ((notrunc
1194 /* If we're doing a size calculation, unsigned arithmetic does overflow.
1195 So check if force_fit_type truncated the value. */
1206 }
1208 /* Combine two constants ARG1 and ARG2 under operation CODE to produce a new
1209 constant. We assume ARG1 and ARG2 have the same data type, or at least
1210 are the same kind of constant and the same machine mode.
1212 If NOTRUNC is nonzero, do not truncate the result to fit the data type. */
1214 static tree
1219 {
1227 {
1228 REAL_VALUE_TYPE d1;
1229 REAL_VALUE_TYPE d2;
1230 REAL_VALUE_TYPE value;
1231 tree t;
1236 /* If either operand is a NaN, just return it. Otherwise, set up
1237 for floating-point trap; we return an overflow. */
1247 value));
1257 }
1259 {
1265 tree t;
1268 {
1288 notrunc),
1294 notrunc));
1298 {
1299 tree magsquared
1303 notrunc);
1306 const_binop
1311 notrunc),
1313 notrunc),
1314 notrunc),
1316 const_binop
1321 notrunc),
1323 notrunc),
1324 notrunc),
1326 }
1331 }
1333 }
1335 }
1337 /* These are the hash table functions for the hash table of INTEGER_CST
1338 nodes of a sizetype. */
1340 /* Return the hash code code X, an INTEGER_CST. */
1342 static hashval_t
1345 {
1351 }
1353 /* Return nonzero if the value represented by *X (an INTEGER_CST tree node)
1354 is the same as that given by *Y, which is the same. */
1356 static int
1360 {
1368 }
1369 \f
1370 /* Return an INTEGER_CST with value whose low-order HOST_BITS_PER_WIDE_INT
1371 bits are given by NUMBER and of the sizetype represented by KIND. */
1373 tree
1375 HOST_WIDE_INT number;
1377 {
1379 }
1381 /* Likewise, but the desired type is specified explicitly. */
1385 htab_t size_htab;
1387 tree
1389 HOST_WIDE_INT number;
1390 tree type;
1391 {
1395 {
1398 }
1400 /* Adjust NEW_CONST to be the constant we want. If it's already in the
1401 hash table, we return the value from the hash table. Otherwise, we
1402 place that in the hash table and make a new node for the next time. */
1411 {
1417 }
1418 else
1420 }
1422 /* Combine operands OP1 and OP2 with arithmetic operation CODE. CODE
1423 is a tree code. The type of the result is taken from the operands.
1424 Both must be the same type integer type and it must be a size type.
1425 If the operands are constant, so is the result. */
1427 tree
1431 {
1438 /* Handle the special case of two integer constants faster. */
1440 {
1441 /* And some specific cases even faster than that. */
1450 /* Handle general case of two integer constants. */
1452 }
1458 }
1460 /* Given two values, either both of sizetype or both of bitsizetype,
1461 compute the difference between the two values. Return the value
1462 in signed type corresponding to the type of the operands. */
1464 tree
1467 {
1469 tree ctype;
1475 /* If the type is already signed, just do the simple thing. */
1482 /* If either operand is not a constant, do the conversions to the signed
1483 type and subtract. The hardware will do the right thing with any
1484 overflow in the subtraction. */
1489 /* If ARG0 is larger than ARG1, subtract and return the result in CTYPE.
1490 Otherwise, subtract the other way, convert to CTYPE (we know that can't
1491 overflow) and negate (which can't either). Special-case a result
1492 of zero while we're here. */
1497 else
1500 }
1501 \f
1503 /* Given T, a tree representing type conversion of ARG1, a constant,
1504 return a constant tree representing the result of conversion. */
1506 static tree
1508 tree t;
1509 tree arg1;
1510 {
1515 {
1517 {
1518 /* If we would build a constant wider than GCC supports,
1519 leave the conversion unfolded. */
1523 /* If we are trying to make a sizetype for a small integer, use
1524 size_int to pick up cached types to reduce duplicate nodes. */
1530 /* Given an integer constant, make new constant with new type,
1531 appropriately sign-extended or truncated. */
1535 /* Indicate an overflow if (1) ARG1 already overflowed,
1536 or (2) force_fit_type indicates an overflow.
1537 Tell force_fit_type that an overflow has already occurred
1538 if ARG1 is a too-large unsigned value and T is signed.
1539 But don't indicate an overflow if converting a pointer. */
1549 }
1551 {
1552 /* Don't initialize these, use assignments.
1553 Initialized local aggregates don't work on old compilers. */
1554 REAL_VALUE_TYPE x;
1555 REAL_VALUE_TYPE l;
1556 REAL_VALUE_TYPE u;
1567 /* See if X will be in range after truncation towards 0.
1568 To compensate for truncation, move the bounds away from 0,
1569 but reject if X exactly equals the adjusted bounds. */
1573 /* If X is a NaN, use zero instead and show we have an overflow.
1574 Otherwise, range check. */
1578 && !no_upper_bound
1582 {
1586 }
1592 }
1594 }
1596 {
1600 {
1602 {
1603 /* We make a copy of ARG1 so that we don't modify an
1604 existing constant tree. */
1608 }
1619 }
1620 }
1623 }
1624 \f
1625 /* Return an expr equal to X but certainly not valid as an lvalue. */
1627 tree
1629 tree x;
1630 {
1631 tree result;
1633 /* These things are certainly not lvalues. */
1644 }
1646 /* Nonzero means lvalues are limited to those valid in pedantic ANSI C.
1647 Zero means allow extended lvalues. */
1651 /* When pedantic, return an expr equal to X but certainly not valid as a
1652 pedantic lvalue. Otherwise, return X. */
1654 tree
1656 tree x;
1657 {
1660 else
1662 }
1663 \f
1664 /* Given a tree comparison code, return the code that is the logical inverse
1665 of the given code. It is not safe to do this for floating-point
1666 comparisons, except for NE_EXPR and EQ_EXPR. */
1668 static enum tree_code
1671 {
1673 {
1688 }
1689 }
1691 /* Similar, but return the comparison that results if the operands are
1692 swapped. This is safe for floating-point. */
1694 static enum tree_code
1697 {
1699 {
1713 }
1714 }
1717 /* Convert a comparison tree code from an enum tree_code representation
1718 into a compcode bit-based encoding. This function is the inverse of
1719 compcode_to_comparison. */
1721 static int
1724 {
1726 {
1741 }
1742 }
1744 /* Convert a compcode bit-based encoding of a comparison operator back
1745 to GCC's enum tree_code representation. This function is the
1746 inverse of comparison_to_compcode. */
1748 static enum tree_code
1751 {
1753 {
1768 }
1769 }
1771 /* Return nonzero if CODE is a tree code that represents a truth value. */
1773 static int
1776 {
1781 }
1782 \f
1783 /* Return nonzero if two operands are necessarily equal.
1784 If ONLY_CONST is nonzero, only return nonzero for constants.
1785 This function tests whether the operands are indistinguishable;
1786 it does not test whether they are equal using C's == operation.
1787 The distinction is important for IEEE floating point, because
1788 (1) -0.0 and 0.0 are distinguishable, but -0.0==0.0, and
1789 (2) two NaNs may be indistinguishable, but NaN!=NaN. */
1791 int
1795 {
1796 /* If both types don't have the same signedness, then we can't consider
1797 them equal. We must check this before the STRIP_NOPS calls
1798 because they may change the signedness of the arguments. */
1806 /* This is needed for conversions and for COMPONENT_REF.
1807 Might as well play it safe and always test this. */
1813 /* If ARG0 and ARG1 are the same SAVE_EXPR, they are necessarily equal.
1814 We don't care about side effects in that case because the SAVE_EXPR
1815 takes care of that for us. In all other cases, two expressions are
1816 equal if they have no side effects. If we have two identical
1817 expressions with side effects that should be treated the same due
1818 to the only side effects being identical SAVE_EXPR's, that will
1819 be detected in the recursive calls below. */
1825 /* Next handle constant cases, those for which we can return 1 even
1826 if ONLY_CONST is set. */
1829 {
1842 {
1852 {
1857 }
1860 }
1864 only_const)
1866 only_const));
1879 }
1885 {
1887 /* Two conversions are equal only if signedness and modes match. */
1903 /* For commutative ops, allow the other order. */
1916 /* If either of the pointer (or reference) expressions we are dereferencing
1917 contain a side effect, these cannot be equal. */
1923 {
1945 }
1954 }
1955 }
1956 \f
1957 /* Similar to operand_equal_p, but see if ARG0 might have been made by
1958 shorten_compare from ARG1 when ARG1 was being compared with OTHER.
1960 When in doubt, return 0. */
1962 static int
1965 tree other;
1966 {
1978 /* Discard any conversions that don't change the modes of ARG0 and ARG1
1979 and see if the inner values are the same. This removes any
1980 signedness comparison, which doesn't matter here. */
1987 /* Duplicate what shorten_compare does to ARG1 and see if that gives the
1988 actual comparison operand, ARG0.
1990 First throw away any conversions to wider types
1991 already present in the operands. */
2000 {
2003 /* Make sure shorter operand is extended the right way
2004 to match the longer operand. */
2010 }
2013 }
2014 \f
2015 /* See if ARG is an expression that is either a comparison or is performing
2016 arithmetic on comparisons. The comparisons must only be comparing
2017 two different values, which will be stored in *CVAL1 and *CVAL2; if
2018 they are nonzero it means that some operands have already been found.
2019 No variables may be used anywhere else in the expression except in the
2020 comparisons. If SAVE_P is true it means we removed a SAVE_EXPR around
2021 the expression and save_expr needs to be called with CVAL1 and CVAL2.
2023 If this is true, return 1. Otherwise, return zero. */
2025 static int
2027 tree arg;
2030 {
2034 /* We can handle some of the 'e' cases here. */
2044 {
2045 /* If we've already found a CVAL1 or CVAL2, this expression is
2046 two complex to handle. */
2052 }
2055 {
2078 /* First see if we can handle the first operand, then the second. For
2079 the second operand, we know *CVAL1 can't be zero. It must be that
2080 one side of the comparison is each of the values; test for the
2081 case where this isn't true by failing if the two operands
2082 are the same. */
2091 ;
2095 ;
2096 else
2100 ;
2104 ;
2105 else
2112 }
2113 }
2114 \f
2115 /* ARG is a tree that is known to contain just arithmetic operations and
2116 comparisons. Evaluate the operations in the tree substituting NEW0 for
2117 any occurrence of OLD0 as an operand of a comparison and likewise for
2118 NEW1 and OLD1. */
2120 static tree
2122 tree arg;
2124 {
2129 /* We can handle some of the 'e' cases here. */
2137 {
2152 {
2169 }
2170 /* fall through - ??? */
2173 {
2177 /* We need to check both for exact equality and tree equality. The
2178 former will be true if the operand has a side-effect. In that
2179 case, we know the operand occurred exactly once. */
2192 }
2196 }
2197 }
2198 \f
2199 /* Return a tree for the case when the result of an expression is RESULT
2200 converted to TYPE and OMITTED was previously an operand of the expression
2201 but is now not needed (e.g., we folded OMITTED * 0).
2203 If OMITTED has side effects, we must evaluate it. Otherwise, just do
2204 the conversion of RESULT to TYPE. */
2206 static tree
2209 {
2216 }
2218 /* Similar, but call pedantic_non_lvalue instead of non_lvalue. */
2220 static tree
2223 {
2230 }
2231 \f
2232 /* Return a simplified tree node for the truth-negation of ARG. This
2233 never alters ARG itself. We assume that ARG is an operation that
2234 returns a truth value (0 or 1). */
2236 tree
2238 tree arg;
2239 {
2246 /* If this is a comparison, we can simply invert it, except for
2247 floating-point non-equality comparisons, in which case we just
2248 enclose a TRUTH_NOT_EXPR around what we have. */
2251 {
2253 && !flag_unsafe_math_optimizations
2257 else
2260 }
2263 {
2278 /* Here we can invert either operand. We invert the first operand
2279 unless the second operand is a TRUTH_NOT_EXPR in which case our
2280 result is the XOR of the first operand with the inside of the
2281 negation of the second operand. */
2286 else
2341 }
2345 }
2347 /* Given a bit-wise operation CODE applied to ARG0 and ARG1, see if both
2348 operands are another bit-wise operation with a common input. If so,
2349 distribute the bit operations to save an operation and possibly two if
2350 constants are involved. For example, convert
2351 (A | B) & (A | C) into A | (B & C)
2352 Further simplification will occur if B and C are constants.
2354 If this optimization cannot be done, 0 will be returned. */
2356 static tree
2359 tree type;
2361 {
2362 tree common;
2372 {
2376 }
2378 {
2382 }
2384 {
2388 }
2390 {
2394 }
2395 else
2400 }
2401 \f
2402 /* Return a BIT_FIELD_REF of type TYPE to refer to BITSIZE bits of INNER
2403 starting at BITPOS. The field is unsigned if UNSIGNEDP is nonzero. */
2405 static tree
2407 tree inner;
2408 tree type;
2411 {
2418 }
2420 /* Optimize a bit-field compare.
2422 There are two cases: First is a compare against a constant and the
2423 second is a comparison of two items where the fields are at the same
2424 bit position relative to the start of a chunk (byte, halfword, word)
2425 large enough to contain it. In these cases we can avoid the shift
2426 implicit in bitfield extractions.
2428 For constants, we emit a compare of the shifted constant with the
2429 BIT_AND_EXPR of a mask and a byte, halfword, or word of the operand being
2430 compared. For two fields at the same position, we do the ANDs with the
2431 similar mask and compare the result of the ANDs.
2433 CODE is the comparison code, known to be either NE_EXPR or EQ_EXPR.
2434 COMPARE_TYPE is the type of the comparison, and LHS and RHS
2435 are the left and right operands of the comparison, respectively.
2437 If the optimization described above can be done, we return the resulting
2438 tree. Otherwise we return zero. */
2440 static tree
2443 tree compare_type;
2445 {
2454 tree mask;
2455 tree offset;
2457 /* Get all the information about the extractions being done. If the bit size
2458 if the same as the size of the underlying object, we aren't doing an
2459 extraction at all and so can do nothing. We also don't want to
2460 do anything if the inner expression is a PLACEHOLDER_EXPR since we
2461 then will no longer be able to replace it. */
2469 {
2470 /* If this is not a constant, we can only do something if bit positions,
2471 sizes, and signedness are the same. */
2479 }
2481 /* See if we can find a mode to refer to this field. We should be able to,
2482 but fail if we can't. */
2491 /* Set signed and unsigned types of the precision of this mode for the
2492 shifts below. */
2496 /* Compute the bit position and size for the new reference and our offset
2497 within it. If the new reference is the same size as the original, we
2498 won't optimize anything, so return zero. */
2508 /* Make the mask to be used against the extracted field. */
2518 /* If not comparing with constant, just rework the comparison
2519 and return. */
2524 mask),
2528 mask));
2530 /* Otherwise, we are handling the constant case. See if the constant is too
2531 big for the field. Warn and return a tree of for 0 (false) if so. We do
2532 this not only for its own sake, but to avoid having to test for this
2533 error case below. If we didn't, we might generate wrong code.
2535 For unsigned fields, the constant shifted right by the field length should
2536 be all zero. For signed fields, the high-order bits should agree with
2537 the sign bit. */
2540 {
2544 {
2550 }
2551 }
2552 else
2553 {
2557 {
2563 }
2564 }
2566 /* Single-bit compares should always be against zero. */
2568 {
2571 }
2573 /* Make a new bitfield reference, shift the constant over the
2574 appropriate number of bits and mask it with the computed mask
2575 (in case this was a signed field). If we changed it, make a new one. */
2578 {
2581 }
2591 rhs);
2592 }
2593 \f
2594 /* Subroutine for fold_truthop: decode a field reference.
2596 If EXP is a comparison reference, we return the innermost reference.
2598 *PBITSIZE is set to the number of bits in the reference, *PBITPOS is
2599 set to the starting bit number.
2601 If the innermost field can be completely contained in a mode-sized
2602 unit, *PMODE is set to that mode. Otherwise, it is set to VOIDmode.
2604 *PVOLATILEP is set to 1 if the any expression encountered is volatile;
2605 otherwise it is not changed.
2607 *PUNSIGNEDP is set to the signedness of the field.
2609 *PMASK is set to the mask used. This is either contained in a
2610 BIT_AND_EXPR or derived from the width of the field.
2612 *PAND_MASK is set to the mask found in a BIT_AND_EXPR, if any.
2614 Return 0 if this is not a component reference or is one that we can't
2615 do anything with. */
2617 static tree
2620 tree exp;
2626 {
2629 tree unsigned_type;
2632 /* All the optimizations using this function assume integer fields.
2633 There are problems with FP fields since the type_for_size call
2634 below can fail for, e.g., XFmode. */
2641 {
2647 }
2656 /* Compute the mask to access the bitfield. */
2666 /* Merge it with the mask we found in the BIT_AND_EXPR, if any. */
2674 }
2676 /* Return nonzero if MASK represents a mask of SIZE ones in the low-order
2677 bit positions. */
2679 static int
2681 tree mask;
2683 {
2686 tree tmask;
2691 return
2698 }
2700 /* Subroutine for fold: determine if VAL is the INTEGER_CONST that
2701 represents the sign bit of EXP's type. If EXP represents a sign
2702 or zero extension, also test VAL against the unextended type.
2703 The return value is the (sub)expression whose sign bit is VAL,
2704 or NULL_TREE otherwise. */
2706 static tree
2708 tree exp;
2709 tree val;
2710 {
2712 HOST_WIDE_INT hi;
2714 tree t;
2716 /* Tree EXP must have an integral type. */
2721 /* Tree VAL must be an integer constant. */
2728 {
2731 }
2732 else
2733 {
2736 }
2741 /* Handle extension from a narrower type. */
2747 }
2749 /* Subroutine for fold_truthop: determine if an operand is simple enough
2750 to be evaluated unconditionally. */
2752 static int
2754 tree exp;
2755 {
2756 /* Strip any conversions that don't change the machine mode. */
2768 /* Don't regard global variables as simple. They may be
2769 allocated in ways unknown to the compiler (shared memory,
2770 #pragma weak, etc). */
2773 /* Loading a static variable is unduly expensive, but global
2774 registers aren't expensive. */
2776 }
2777 \f
2778 /* The following functions are subroutines to fold_range_test and allow it to
2779 try to change a logical combination of comparisons into a range test.
2781 For example, both
2782 X == 2 || X == 3 || X == 4 || X == 5
2783 and
2784 X >= 2 && X <= 5
2785 are converted to
2786 (unsigned) (X - 2) <= 3
2788 We describe each set of comparisons as being either inside or outside
2789 a range, using a variable named like IN_P, and then describe the
2790 range with a lower and upper bound. If one of the bounds is omitted,
2791 it represents either the highest or lowest value of the type.
2793 In the comments below, we represent a range by two numbers in brackets
2794 preceded by a "+" to designate being inside that range, or a "-" to
2795 designate being outside that range, so the condition can be inverted by
2796 flipping the prefix. An omitted bound is represented by a "-". For
2797 example, "- [-, 10]" means being outside the range starting at the lowest
2798 possible value and ending at 10, in other words, being greater than 10.
2799 The range "+ [-, -]" is always true and hence the range "- [-, -]" is
2800 always false.
2802 We set up things so that the missing bounds are handled in a consistent
2803 manner so neither a missing bound nor "true" and "false" need to be
2804 handled using a special case. */
2806 /* Return the result of applying CODE to ARG0 and ARG1, but handle the case
2807 of ARG0 and/or ARG1 being omitted, meaning an unlimited range. UPPER0_P
2808 and UPPER1_P are nonzero if the respective argument is an upper bound
2809 and zero for a lower. TYPE, if nonzero, is the type of the result; it
2810 must be specified for a comparison. ARG1 will be converted to ARG0's
2811 type if both are specified. */
2813 static tree
2816 tree type;
2819 {
2820 tree tem;
2824 /* If neither arg represents infinity, do the normal operation.
2825 Else, if not a comparison, return infinity. Else handle the special
2826 comparison rules. Note that most of the cases below won't occur, but
2827 are handled for consistency. */
2830 {
2835 }
2840 /* Set SGN[01] to -1 if ARG[01] is a lower bound, 1 for upper, and 0
2841 for neither. In real maths, we cannot assume open ended ranges are
2842 the same. But, this is computer arithmetic, where numbers are finite.
2843 We can therefore make the transformation of any unbounded range with
2844 the value Z, Z being greater than any representable number. This permits
2845 us to treat unbounded ranges as equal. */
2849 {
2870 }
2873 }
2874 \f
2875 /* Given EXP, a logical expression, set the range it is testing into
2876 variables denoted by PIN_P, PLOW, and PHIGH. Return the expression
2877 actually being tested. *PLOW and *PHIGH will be made of the same type
2878 as the returned expression. If EXP is not a comparison, we will most
2879 likely not be returning a useful value and range. */
2881 static tree
2883 tree exp;
2886 {
2893 /* Start with simply saying "EXP != 0" and then look at the code of EXP
2894 and see if we can refine the range. Some of the cases below may not
2895 happen, but it doesn't seem worth worrying about this. We "continue"
2896 the outer loop when we've changed something; otherwise we "break"
2897 the switch, which will "break" the while. */
2902 {
2906 {
2917 }
2919 /* Set ORIG_TYPE as soon as TYPE is non-null so that we do not
2920 lose a cast by accident. */
2925 {
2932 /* We can only do something if the range is testing for zero
2933 and if the second operand is an integer constant. Note that
2934 saying something is "in" the range we make is done by
2935 complementing IN_P since it will set in the initial case of
2936 being not equal to zero; "out" is leaving it alone. */
2943 {
2964 }
2968 /* If this is an unsigned comparison, we also know that EXP is
2969 greater than or equal to zero. We base the range tests we make
2970 on that fact, so we record it here so we can parse existing
2971 range tests. */
2973 {
2976 NULL_TREE))
2981 /* If the high bound is missing, but we
2982 have a low bound, reverse the range so
2983 it goes from zero to the low bound minus 1. */
2985 {
2990 }
2991 }
2995 /* (-x) IN [a,b] -> x in [-b, -a] */
3005 /* ~ X -> -X - 1 */
3014 /* If EXP is signed, any overflow in the computation is undefined,
3015 so we don't worry about it so long as our computations on
3016 the bounds don't overflow. For unsigned, overflow is defined
3017 and this is exactly the right thing. */
3026 /* Check for an unsigned range which has wrapped around the maximum
3027 value thus making n_high < n_low, and normalize it. */
3029 {
3035 /* If the range is of the form +/- [ x+1, x ], we won't
3036 be able to normalize it. But then, it represents the
3037 whole range or the empty set, so make it
3038 +/- [ -, - ]. */
3042 else
3044 }
3045 else
3068 /* If we're converting from an unsigned to a signed type,
3069 we will be doing the comparison as unsigned. The tests above
3070 have already verified that LOW and HIGH are both positive.
3072 So we have to make sure that the original unsigned value will
3073 be interpreted as positive. */
3075 {
3078 tree high_positive;
3080 /* A range without an upper bound is, naturally, unbounded.
3081 Since convert would have cropped a very large value, use
3082 the max value for the destination type. */
3083 high_positive
3092 /* If the low bound is specified, "and" the range with the
3093 range for which the original unsigned value will be
3094 positive. */
3096 {
3100 high_positive))
3104 }
3105 else
3106 {
3107 /* Otherwise, "or" the range with the range of the input
3108 that will be interpreted as negative. */
3112 high_positive))
3116 }
3117 }
3125 }
3128 }
3130 /* If EXP is a constant, we can evaluate whether this is true or false. */
3132 {
3139 }
3143 }
3144 \f
3145 /* Given a range, LOW, HIGH, and IN_P, an expression, EXP, and a result
3146 type, TYPE, return an expression to test if EXP is in (or out of, depending
3147 on IN_P) the range. */
3149 static tree
3151 tree type;
3152 tree exp;
3155 {
3157 tree value;
3176 {
3178 {
3182 }
3184 }
3186 /* Optimize (c>=1) && (c<=127) into (signed char)c > 0. */
3188 {
3190 HOST_WIDE_INT hi;
3195 {
3198 }
3199 else
3200 {
3203 }
3206 {
3208 {
3211 }
3214 }
3215 }
3224 }
3225 \f
3226 /* Given two ranges, see if we can merge them into one. Return 1 if we
3227 can, 0 if we can't. Set the output range into the specified parameters. */
3229 static int
3235 {
3239 tree tem;
3249 /* Make range 0 be the range that starts first, or ends last if they
3250 start at the same value. Swap them if it isn't. */
3253 || (lowequal
3256 {
3260 }
3262 /* Now flag two cases, whether the ranges are disjoint or whether the
3263 second range is totally subsumed in the first. Note that the tests
3264 below are simplified by the ones above. */
3270 /* We now have four cases, depending on whether we are including or
3271 excluding the two ranges. */
3273 {
3274 /* If they don't overlap, the result is false. If the second range
3275 is a subset it is the result. Otherwise, the range is from the start
3276 of the second to the end of the first. */
3281 else
3283 }
3286 {
3287 /* If they don't overlap, the result is the first range. If they are
3288 equal, the result is false. If the second range is a subset of the
3289 first, and the ranges begin at the same place, we go from just after
3290 the end of the first range to the end of the second. If the second
3291 range is not a subset of the first, or if it is a subset and both
3292 ranges end at the same place, the range starts at the start of the
3293 first range and ends just before the second range.
3294 Otherwise, we can't describe this as a single range. */
3300 {
3304 }
3306 {
3310 }
3311 else
3313 }
3316 {
3317 /* If they don't overlap, the result is the second range. If the second
3318 is a subset of the first, the result is false. Otherwise,
3319 the range starts just after the first range and ends at the
3320 end of the second. */
3325 else
3326 {
3330 }
3331 }
3333 else
3334 {
3335 /* The case where we are excluding both ranges. Here the complex case
3336 is if they don't overlap. In that case, the only time we have a
3337 range is if they are adjacent. If the second is a subset of the
3338 first, the result is the first. Otherwise, the range to exclude
3339 starts at the beginning of the first range and ends at the end of the
3340 second. */
3342 {
3349 else
3351 }
3354 else
3356 }
3360 }
3361 \f
3362 /* EXP is some logical combination of boolean tests. See if we can
3363 merge it into some range test. Return the new tree if so. */
3365 static tree
3367 tree exp;
3368 {
3375 tree tem;
3377 /* If this is an OR operation, invert both sides; we will invert
3378 again at the end. */
3382 /* If both expressions are the same, if we can merge the ranges, and we
3383 can build the range test, return it or it inverted. If one of the
3384 ranges is always true or always false, consider it to be the same
3385 expression as the other. */
3395 /* On machines where the branch cost is expensive, if this is a
3396 short-circuited branch and the underlying object on both sides
3397 is the same, make a non-short-circuit operation. */
3403 {
3404 /* If simple enough, just rewrite. Otherwise, make a SAVE_EXPR
3405 unless we are at top level or LHS contains a PLACEHOLDER_EXPR, in
3406 which cases we can't do this. */
3415 {
3427 }
3428 }
3431 }
3432 \f
3433 /* Subroutine for fold_truthop: C is an INTEGER_CST interpreted as a P
3434 bit value. Arrange things so the extra bits will be set to zero if and
3435 only if C is signed-extended to its full width. If MASK is nonzero,
3436 it is an INTEGER_CST that should be AND'ed with the extra bits. */
3438 static tree
3440 tree c;
3443 tree mask;
3444 {
3447 tree temp;
3452 /* We work by getting just the sign bit into the low-order bit, then
3453 into the high-order bit, then sign-extend. We then XOR that value
3454 with C. */
3458 /* We must use a signed type in order to get an arithmetic right shift.
3459 However, we must also avoid introducing accidental overflows, so that
3460 a subsequent call to integer_zerop will work. Hence we must
3461 do the type conversion here. At this point, the constant is either
3462 zero or one, and the conversion to a signed type can never overflow.
3463 We could get an overflow if this conversion is done anywhere else. */
3471 /* If necessary, convert the type back to match the type of C. */
3476 }
3477 \f
3478 /* Find ways of folding logical expressions of LHS and RHS:
3479 Try to merge two comparisons to the same innermost item.
3480 Look for range tests like "ch >= '0' && ch <= '9'".
3481 Look for combinations of simple terms on machines with expensive branches
3482 and evaluate the RHS unconditionally.
3484 For example, if we have p->a == 2 && p->b == 4 and we can make an
3485 object large enough to span both A and B, we can do this with a comparison
3486 against the object ANDed with the a mask.
3488 If we have p->a == q->a && p->b == q->b, we may be able to use bit masking
3489 operations to do this with one comparison.
3491 We check for both normal comparisons and the BIT_AND_EXPRs made this by
3492 function and the one above.
3494 CODE is the logical operation being done. It can be TRUTH_ANDIF_EXPR,
3495 TRUTH_AND_EXPR, TRUTH_ORIF_EXPR, or TRUTH_OR_EXPR.
3497 TRUTH_TYPE is the type of the logical operand and LHS and RHS are its
3498 two operands.
3500 We return the simplified tree or 0 if no optimization is possible. */
3502 static tree
3506 {
3507 /* If this is the "or" of two comparisons, we can do something if
3508 the comparisons are NE_EXPR. If this is the "and", we can do something
3509 if the comparisons are EQ_EXPR. I.e.,
3510 (a->b == 2 && a->c == 4) can become (a->new == NEW).
3512 WANTED_CODE is this operation code. For single bit fields, we can
3513 convert EQ_EXPR to NE_EXPR so we need not reject the "wrong"
3514 comparison for one-bit fields. */
3534 /* Start by getting the comparison codes. Fail if anything is volatile.
3535 If one operand is a BIT_AND_EXPR with the constant one, treat it as if
3536 it were surrounded with a NE_EXPR. */
3561 /* Simplify (x<y) && (x==y) into (x<=y) and related optimizations. */
3565 {
3570 {
3578 }
3581 {
3590 }
3591 else
3601 }
3603 /* If the RHS can be evaluated unconditionally and its operands are
3604 simple, it wins to evaluate the RHS unconditionally on machines
3605 with expensive branches. In this case, this isn't a comparison
3606 that can be merged. Avoid doing this if the RHS is a floating-point
3607 comparison since those can trap. */
3613 {
3614 /* Convert (a != 0) || (b != 0) into (a | b) != 0. */
3622 integer_zero_node);
3624 /* Convert (a == 0) && (b == 0) into (a | b) == 0. */
3632 integer_zero_node);
3635 }
3637 /* See if the comparisons can be merged. Then get all the parameters for
3638 each side. */
3662 /* It must be true that the inner operation on the lhs of each
3663 comparison must be the same if we are to be able to do anything.
3664 Then see if we have constants. If not, the same must be true for
3665 the rhs's. */
3676 else
3679 /* If either comparison code is not correct for our logical operation,
3680 fail. However, we can convert a one-bit comparison against zero into
3681 the opposite comparison against that bit being set in the field. */
3685 {
3687 {
3688 /* Make the left operand unsigned, since we are only interested
3689 in the value of one bit. Otherwise we are doing the wrong
3690 thing below. */
3693 }
3694 else
3696 }
3698 /* This is analogous to the code for l_const above. */
3700 {
3702 {
3705 }
3706 else
3708 }
3710 /* After this point all optimizations will generate bit-field
3711 references, which we might not want. */
3715 /* See if we can find a mode that contains both fields being compared on
3716 the left. If we can't, fail. Otherwise, update all constants and masks
3717 to be relative to a field of that size. */
3722 volatilep);
3732 {
3735 }
3743 {
3751 {
3755 wanted_code == NE_EXPR
3757 }
3758 }
3760 {
3768 {
3772 wanted_code == NE_EXPR
3774 }
3775 }
3777 /* If the right sides are not constant, do the same for it. Also,
3778 disallow this optimization if a size or signedness mismatch occurs
3779 between the left and right sides. */
3781 {
3784 /* Make sure the two fields on the right
3785 correspond to the left without being swapped. */
3793 volatilep);
3803 {
3806 }
3813 /* Make a mask that corresponds to both fields being compared.
3814 Do this for both items being compared. If the operands are the
3815 same size and the bits being compared are in the same position
3816 then we can do this by masking both and comparing the masked
3817 results. */
3821 {
3833 }
3835 /* There is still another way we can do something: If both pairs of
3836 fields being compared are adjacent, we may be able to make a wider
3837 field containing them both.
3839 Note that we still must mask the lhs/rhs expressions. Furthermore,
3840 the mask must be shifted to account for the shift done by
3841 make_bit_field_ref. */
3846 {
3847 tree type;
3859 /* Convert to the smaller type before masking out unwanted bits. */
3862 {
3864 {
3868 }
3870 {
3874 }
3875 }
3884 }
3887 }
3889 /* Handle the case of comparisons with constants. If there is something in
3890 common between the masks, those bits of the constants must be the same.
3891 If not, the condition is always false. Test for this to avoid generating
3892 incorrect code below. */
3897 {
3899 {
3902 }
3903 else
3904 {
3907 }
3908 }
3910 /* Construct the expression we will return. First get the component
3911 reference we will make. Unless the mask is all ones the width of
3912 that field, perform the mask operation. Then compare with the
3913 merged constant. */
3923 }
3924 \f
3925 /* Optimize T, which is a comparison of a MIN_EXPR or MAX_EXPR with a
3926 constant. */
3928 static tree
3930 tree t;
3931 {
3936 tree minmax_const;
3938 tree inner;
3948 /* If something does not permit us to optimize, return the original tree. */
3956 /* Now handle all the various comparison codes. We only handle EQ_EXPR
3957 and GT_EXPR, doing the rest with recursive calls using logical
3958 simplifications. */
3960 {
3962 return
3966 return
3968 optimize_minmax_comparison
3970 optimize_minmax_comparison
3975 /* MAX (X, 0) == 0 -> X <= 0 */
3979 /* MAX (X, 0) == 5 -> X == 5 */
3983 /* MAX (X, 0) == -1 -> false */
3987 /* MIN (X, 0) == 0 -> X >= 0 */
3991 /* MIN (X, 0) == 5 -> false */
3994 else
3995 /* MIN (X, 0) == -1 -> X == -1 */
4000 /* MAX (X, 0) > 0 -> X > 0
4001 MAX (X, 0) > 5 -> X > 5 */
4005 /* MAX (X, 0) > -1 -> true */
4009 /* MIN (X, 0) > 0 -> false
4010 MIN (X, 0) > 5 -> false */
4013 else
4014 /* MIN (X, 0) > -1 -> X > -1 */
4019 }
4020 }
4021 \f
4022 /* T is an integer expression that is being multiplied, divided, or taken a
4023 modulus (CODE says which and what kind of divide or modulus) by a
4024 constant C. See if we can eliminate that operation by folding it with
4025 other operations already in T. WIDE_TYPE, if non-null, is a type that
4026 should be used for the computation if wider than our type.
4028 For example, if we are dividing (X * 8) + (Y * 16) by 4, we can return
4029 (X * 2) + (Y * 4). We must, however, be assured that either the original
4030 expression would not overflow or that overflow is undefined for the type
4031 in the language in question.
4033 We also canonicalize (X + 7) * 4 into X * 4 + 28 in the hope that either
4034 the machine has a multiply-accumulate insn or that this is part of an
4035 addressing calculation.
4037 If we return a non-null expression, it is an equivalent form of the
4038 original computation, but need not be in the original type. */
4040 static tree
4042 tree t;
4043 tree c;
4045 tree wide_type;
4046 {
4056 /* Don't deal with constants of zero here; they confuse the code below. */
4066 /* Note that we need not handle conditional operations here since fold
4067 already handles those cases. So just do arithmetic here. */
4069 {
4071 /* For a constant, we can always simplify if we are a multiply
4072 or (for divide and modulus) if it is a multiple of our constant. */
4079 /* If op0 is an expression ... */
4084 /* ... and is unsigned, and its type is smaller than ctype,
4085 then we cannot pass through as widening. */
4091 /* ... or its type is larger than ctype,
4092 then we cannot pass through this truncation. */
4097 /* Pass the constant down and see if we can make a simplification. If
4098 we can, replace this expression with the inner simplification for
4099 possible later conversion to our or some other type. */
4111 /* If widening the type changes the signedness, then we can't perform
4112 this optimization as that changes the result. */
4116 /* MIN (a, b) / 5 -> MIN (a / 5, b / 5) */
4119 {
4125 }
4135 /* If this has not been evaluated and the operand has no side effects,
4136 we can see if we can do something inside it and make a new one.
4137 Note that this test is overly conservative since we can do this
4138 if the only reason it had side effects is that it was another
4139 similar SAVE_EXPR, but that isn't worth bothering with. */
4142 wide_type)))
4143 {
4150 }
4154 /* If the second operand is constant, this is a multiplication
4155 or floor division, by a power of two, so we can treat it that
4156 way unless the multiplier or divisor overflows. */
4158 /* const_binop may not detect overflow correctly,
4159 so check for it explicitly here. */
4173 /* See if we can eliminate the operation on both sides. If we can, we
4174 can return a new PLUS or MINUS. If we can't, the only remaining
4175 cases where we can do anything are if the second operand is a
4176 constant. */
4181 /* If not multiplication, we can only do this if either operand
4182 is divisible by c. */
4188 /* If this was a subtraction, negate OP1 and set it to be an addition.
4189 This simplifies the logic below. */
4196 /* If either OP1 or C are negative, this optimization is not safe for
4197 some of the division and remainder types while for others we need
4198 to change the code. */
4200 {
4208 }
4210 /* If it's a multiply or a division/modulus operation of a multiple
4211 of our constant, do the operation and verify it doesn't overflow. */
4214 {
4218 }
4219 else
4222 /* If we have an unsigned type is not a sizetype, we cannot widen
4223 the operation since it will change the result if the original
4224 computation overflowed. */
4230 /* If we were able to eliminate our operation from the first side,
4231 apply our operation to the second side and reform the PLUS. */
4235 /* The last case is if we are a multiply. In that case, we can
4236 apply the distributive law to commute the multiply and addition
4237 if the multiplication of the constants doesn't overflow. */
4242 op1));
4247 /* We have a special case here if we are doing something like
4248 (C * 8) % 4 since we know that's zero. */
4255 /* ... fall through ... */
4259 /* If we can extract our operation from the LHS, do so and return a
4260 new operation. Likewise for the RHS from a MULT_EXPR. Otherwise,
4261 do something only if the second operand is a constant. */
4273 /* If these are the same operation types, we can associate them
4274 assuming no overflow. */
4281 /* If these operations "cancel" each other, we have the main
4282 optimizations of this pass, which occur when either constant is a
4283 multiple of the other, in which case we replace this with either an
4284 operation or CODE or TCODE.
4286 If we have an unsigned type that is not a sizetype, we cannot do
4287 this since it will change the result if the original computation
4288 overflowed. */
4295 {
4306 }
4311 }
4314 }
4315 \f
4316 /* If T contains a COMPOUND_EXPR which was inserted merely to evaluate
4317 S, a SAVE_EXPR, return the expression actually being evaluated. Note
4318 that we may sometimes modify the tree. */
4320 static tree
4322 tree t;
4323 tree s;
4324 {
4327 /* See if this is the COMPOUND_EXPR we want to eliminate. */
4332 /* See if this is a COND_EXPR or a simple arithmetic operator. We
4333 don't bother handling any other types. */
4335 {
4339 }
4344 {
4347 }
4350 }
4351 \f
4352 /* Return a node which has the indicated constant VALUE (either 0 or
4353 1), and is of the indicated TYPE. */
4355 static tree
4358 tree type;
4359 {
4364 integer_zero_node);
4365 else
4366 {
4371 }
4372 }
4374 /* Utility function for the following routine, to see how complex a nesting of
4375 COND_EXPRs can be. EXPR is the expression and LIMIT is a count beyond which
4376 we don't care (to avoid spending too much time on complex expressions.). */
4378 static int
4380 tree expr;
4382 {
4393 }
4395 /* Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
4396 Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'. Here
4397 CODE corresponds to the `+', COND to the `(b ? x : y)' or `(x < y)'
4398 expression, and ARG to `a'. If COND_FIRST_P is nonzero, then the
4399 COND is the first argument to CODE; otherwise (as in the example
4400 given here), it is the second argument. TYPE is the type of the
4401 original expression. */
4403 static tree
4406 tree type;
4407 tree cond;
4408 tree arg;
4410 {
4414 /* In the end, we'll produce a COND_EXPR. Both arms of the
4415 conditional expression will be binary operations. The left-hand
4416 side of the expression to be executed if the condition is true
4417 will be pointed to by TRUE_LHS. Similarly, the right-hand side
4418 of the expression to be executed if the condition is true will be
4419 pointed to by TRUE_RHS. FALSE_LHS and FALSE_RHS are analogous --
4420 but apply to the expression to be executed if the conditional is
4421 false. */
4426 /* These are the codes to use for the left-hand side and right-hand
4427 side of the COND_EXPR. Normally, they are the same as CODE. */
4430 /* And these are the types of the expressions. */
4435 {
4439 }
4440 else
4441 {
4445 }
4448 {
4452 /* If this operand throws an expression, then it does not make
4453 sense to try to perform a logical or arithmetic operation
4454 involving it. Instead of building `a + throw 3' for example,
4455 we simply build `a, throw 3'. */
4457 {
4461 }
4463 {
4467 }
4468 }
4469 else
4470 {
4475 }
4477 /* If ARG is complex we want to make sure we only evaluate
4478 it once. Though this is only required if it is volatile, it
4479 might be more efficient even if it is not. However, if we
4480 succeed in folding one part to a constant, we do not need
4481 to make this SAVE_EXPR. Since we do this optimization
4482 primarily to see if we do end up with constant and this
4483 SAVE_EXPR interferes with later optimizations, suppressing
4484 it when we can is important.
4486 If we are not in a function, we can't make a SAVE_EXPR, so don't
4487 try to do so. Don't try to see if the result is a constant
4488 if an arm is a COND_EXPR since we get exponential behavior
4489 in that case. */
4496 {
4506 }
4519 else
4521 }
4523 \f
4524 /* Subroutine of fold() that checks for the addition of +/- 0.0.
4526 If !NEGATE, return true if ADDEND is +/-0.0 and, for all X of type
4527 TYPE, X + ADDEND is the same as X. If NEGATE, return true if X -
4528 ADDEND is the same as X.
4530 X + 0 and X - 0 both give X when X is NaN, infinite, or nonzero
4531 and finite. The problematic cases are when X is zero, and its mode
4532 has signed zeros. In the case of rounding towards -infinity,
4533 X - 0 is not the same as X because 0 - 0 is -0. In other rounding
4534 modes, X + 0 is not the same as X because -0 + 0 is 0. */
4536 static bool
4540 {
4544 /* Allow the fold if zeros aren't signed, or their sign isn't important. */
4548 /* Treat x + -0 as x - 0 and x - -0 as x + 0. */
4553 /* The mode has signed zeros, and we have to honor their sign.
4554 In this situation, there is only one case we can return true for.
4555 X - 0 is the same as X unless rounding towards -infinity is
4556 supported. */
4558 }
4561 /* Perform constant folding and related simplification of EXPR.
4562 The related simplifications include x*1 => x, x*0 => 0, etc.,
4563 and application of the associative law.
4564 NOP_EXPR conversions may be removed freely (as long as we
4565 are careful not to change the C type of the overall expression)
4566 We cannot simplify through a CONVERT_EXPR, FIX_EXPR or FLOAT_EXPR,
4567 but we can constant-fold them if they have constant operands. */
4569 tree
4571 tree expr;
4572 {
4575 tree tem;
4581 /* WINS will be nonzero when the switch is done
4582 if all operands are constant. */
4585 /* Don't try to process an RTL_EXPR since its operands aren't trees.
4586 Likewise for a SAVE_EXPR that's already been evaluated. */
4590 /* Return right away if a constant. */
4594 #ifdef MAX_INTEGER_COMPUTATION_MODE
4596 #endif
4599 {
4600 tree subop;
4602 /* Special case for conversion ops that can have fixed point args. */
4605 /* Don't use STRIP_NOPS, because signedness of argument type matters. */
4611 else
4616 )
4617 /* Note that TREE_CONSTANT isn't enough:
4618 static var addresses are constant but we can't
4619 do arithmetic on them. */
4621 }
4623 {
4627 {
4629 tree subop;
4635 {
4636 /* Signedness matters here. Perhaps we can refine this
4637 later. */
4639 }
4640 else
4641 /* Strip any conversions that don't change the mode. */
4646 else
4651 /* Note that TREE_CONSTANT isn't enough:
4652 static var addresses are constant but we can't
4653 do arithmetic on them. */
4660 }
4661 }
4663 /* If this is a commutative operation, and ARG0 is a constant, move it
4664 to ARG1 to reduce the number of tests below. */
4669 {
4674 }
4676 /* Now WINS is set as described above,
4677 ARG0 is the first operand of EXPR,
4678 and ARG1 is the second operand (if it has more than one operand).
4680 First check for cases where an arithmetic operation is applied to a
4681 compound, conditional, or comparison operation. Push the arithmetic
4682 operation inside the compound or conditional to see if any folding
4683 can then be done. Convert comparison to conditional for this purpose.
4684 The also optimizes non-constant cases that used to be done in
4685 expand_expr.
4687 Before we do that, see if this is a BIT_AND_EXPR or a BIT_IOR_EXPR,
4688 one of the operands is a comparison and the other is a comparison, a
4689 BIT_AND_EXPR with the constant 1, or a truth value. In that case, the
4690 code below would make the expression more complex. Change it to a
4691 TRUTH_{AND,OR}_EXPR. Likewise, convert a similar NE_EXPR to
4692 TRUTH_XOR_EXPR and an EQ_EXPR to the inversion of a TRUTH_XOR_EXPR. */
4704 {
4714 }
4717 {
4722 {
4727 /* If this was a conversion, and all we did was to move into
4728 inside the COND_EXPR, bring it back out. But leave it if
4729 it is a conversion from integer to integer and the
4730 result precision is no wider than a word since such a
4731 conversion is cheap and may be optimized away by combine,
4732 while it couldn't if it were outside the COND_EXPR. Then return
4733 so we don't get into an infinite recursion loop taking the
4734 conversion out and then back in. */
4744 && (INTEGRAL_TYPE_P
4755 }
4760 }
4763 {
4776 return
4790 return
4793 }
4804 {
4820 /* Other kinds of FIX are not handled properly by fold_convert. */
4825 /* Handle cases of two conversions in a row. */
4828 {
4848 /* In addition to the cases of two conversions in a row
4849 handled below, if we are converting something to its own
4850 type via an object of identical or wider precision, neither
4851 conversion is needed. */
4857 /* Likewise, if the intermediate and final types are either both
4858 float or both integer, we don't need the middle conversion if
4859 it is wider than the final type and doesn't change the signedness
4860 (for integers). Avoid this if the final type is a pointer
4861 since then we sometimes need the inner conversion. Likewise if
4862 the outer has a precision not equal to the size of its mode. */
4872 /* If we have a sign-extension of a zero-extended value, we can
4873 replace that by a single zero-extension. */
4879 /* Two conversions in a row are not needed unless:
4880 - some conversion is floating-point (overstrict for now), or
4881 - the intermediate type is narrower than both initial and
4882 final, or
4883 - the intermediate type and innermost type differ in signedness,
4884 and the outermost type is wider than the intermediate, or
4885 - the initial type is a pointer type and the precisions of the
4886 intermediate and final types differ, or
4887 - the final type is a pointer type and the precisions of the
4888 initial and intermediate types differ. */
4902 }
4906 /* Detect assigning a bitfield. */
4909 {
4910 /* Don't leave an assignment inside a conversion
4911 unless assigning a bitfield. */
4914 /* First do the assignment, then return converted constant. */
4918 }
4920 /* Convert (T)(x & c) into (T)x & (T)c, if c is an integer
4921 constants (if x has signed type, the sign bit cannot be set
4922 in c). This folds extension into the BIT_AND_EXPR. */
4927 {
4937 <= HOST_BITS_PER_WIDE_INT
4939 {
4946 #ifdef LOAD_EXTEND_OP
4950 {
4954 }
4955 #endif
4956 }
4961 }
4964 {
4967 }
4978 {
4982 }
4991 {
4993 {
4995 HOST_WIDE_INT high;
5006 }
5009 }
5013 /* Convert - (a - b) to (b - a) for non-floating-point. */
5023 {
5025 {
5026 /* If the value is unsigned, then the absolute value is
5027 the same as the ordinary value. */
5030 /* Similarly, if the value is non-negative. */
5033 /* If the value is negative, then the absolute value is
5034 its negation. */
5035 else
5036 {
5038 HOST_WIDE_INT high;
5049 }
5050 }
5052 {
5056 }
5057 }
5084 {
5091 }
5097 /* A + (-B) -> A - B */
5100 /* (-A) + B -> B - A */
5104 {
5108 /* If we are adding two BIT_AND_EXPR's, both of which are and'ing
5109 with a constant, and the two constants have no bits in common,
5110 we should treat this as a BIT_IOR_EXPR since this may produce more
5111 simplifications. */
5119 {
5122 }
5124 /* Reassociate (plus (plus (mult) (foo)) (mult)) as
5125 (plus (plus (mult) (mult)) (foo)) so that we can
5126 take advantage of the factoring cases below. */
5131 {
5136 else
5147 parg1));
5152 parg0));
5153 }
5156 {
5160 /* (A * C) + (B * C) -> (A+B) * C.
5161 We are most concerned about the case where C is a constant,
5162 but other combinations show up during loop reduction. Since
5163 it is not difficult, try all four possibilities. */
5180 /* No identical multiplicands; see if we can find a common
5181 power-of-two factor in non-power-of-two multiplies. This
5182 can help in multi-dimensional array access. */
5187 {
5192 /* Move min of absolute values to int11. */
5195 {
5199 }
5202 {
5207 }
5208 }
5213 same));
5214 }
5215 }
5217 /* See if ARG1 is zero and X + ARG1 reduces to X. */
5221 /* Likewise if the operands are reversed. */
5225 bit_rotate:
5226 /* (A << C1) + (A >> C2) if A is unsigned and C1+C2 is the size of A
5227 is a rotate of A by C1 bits. */
5228 /* (A << B) + (A >> (Z - B)) if A is unsigned and Z is the size of A
5229 is a rotate of A by B bits. */
5230 {
5239 {
5258 {
5266 TYPE_PRECISION
5271 ? LROTATE_EXPR
5274 }
5276 {
5284 TYPE_PRECISION
5289 ? LROTATE_EXPR
5292 }
5293 }
5294 }
5296 associate:
5297 /* In most languages, can't associate operations on floats through
5298 parentheses. Rather than remember where the parentheses were, we
5299 don't associate floats at all. It shouldn't matter much. However,
5300 associating multiplications is only very slightly inaccurate, so do
5301 that if -funsafe-math-optimizations is specified. */
5306 {
5310 /* Split both trees into variables, constants, and literals. Then
5311 associate each group together, the constants with literals,
5312 then the result with variables. This increases the chances of
5313 literals being recombined later and of generating relocatable
5314 expressions for the sum of a constant and literal. */
5319 /* Only do something if we found more than two objects. Otherwise,
5320 nothing has changed and we risk infinite recursion. */
5325 {
5326 /* Recombine MINUS_EXPR operands by using PLUS_EXPR. */
5335 /* Preserve the MINUS_EXPR if the negative part of the literal is
5336 greater than the positive part. Otherwise, the multiplicative
5337 folding code (i.e extract_muldiv) may be fooled in case
5338 unsigned constants are substracted, like in the following
5339 example: ((X*2 + 4) - 8U)/2. */
5341 {
5343 {
5347 }
5348 else
5349 {
5353 }
5354 }
5356 {
5360 else
5361 {
5366 }
5367 }
5371 }
5372 }
5374 binary:
5378 {
5379 /* The return value should always have
5380 the same type as the original expression. */
5385 }
5389 /* A - (-B) -> A + B */
5392 /* (-A) - CST -> (-CST) - A for floating point (what about ints ?) */
5394 return
5401 {
5407 /* (A * C) - (B * C) -> (A-B) * C. Since we are most concerned
5408 about the case where C is a constant, just try one of the
5409 four possibilities. */
5419 }
5421 /* See if ARG1 is zero and X - ARG1 reduces to X. */
5425 /* (ARG0 - ARG1) is the same as (-ARG1 + ARG0). So check whether
5426 ARG0 is zero and X + ARG0 reduces to X, since that would mean
5427 (-ARG1 + ARG0) reduces to -ARG1. */
5431 /* Fold &x - &x. This can happen from &x.foo - &x.
5432 This is unsafe for certain floats even in non-IEEE formats.
5433 In IEEE, it is unsafe because it does wrong for NaNs.
5434 Also note that operand_equal_p is always false if an operand
5435 is volatile. */
5444 /* (-A) * (-B) -> A * B */
5450 {
5456 /* (a * (1 << b)) is (a << b) */
5471 }
5472 else
5473 {
5474 /* Maybe fold x * 0 to 0. The expressions aren't the same
5475 when x is NaN, since x * 0 is also NaN. Nor are they the
5476 same in modes with signed zeros, since multiplying a
5477 negative value by 0 gives -0, not +0. */
5482 /* In IEEE floating point, x*1 is not equivalent to x for snans. */
5487 /* Transform x * -1.0 into -x. */
5492 /* x*2 is x+x */
5496 {
5499 }
5500 }
5504 bit_ior:
5513 /* Convert (or (not arg0) (not arg1)) to (not (and (arg0) (arg1))).
5515 This results in more efficient code for machines without a NAND
5516 instruction. Combine will canonicalize to the first form
5517 which will allow use of NAND instructions provided by the
5518 backend if they exist. */
5521 {
5526 }
5528 /* See if this can be simplified into a rotate first. If that
5529 is unsuccessful continue in the association code. */
5538 /* If we are XORing two BIT_AND_EXPR's, both of which are and'ing
5539 with a constant, and the two constants have no bits in common,
5540 we should treat this as a BIT_IOR_EXPR since this may produce more
5541 simplifications. */
5549 {
5552 }
5554 /* See if this can be simplified into a rotate first. If that
5555 is unsuccessful continue in the association code. */
5559 bit_and:
5567 /* Simplify ((int)c & 0x377) into (int)c, if c is unsigned char. */
5570 {
5571 unsigned int prec
5578 }
5580 /* Convert (and (not arg0) (not arg1)) to (not (or (arg0) (arg1))).
5582 This results in more efficient code for machines without a NOR
5583 instruction. Combine will canonicalize to the first form
5584 which will allow use of NOR instructions provided by the
5585 backend if they exist. */
5588 {
5593 }
5603 {
5607 }
5611 /* Don't touch a floating-point divide by zero unless the mode
5612 of the constant can represent infinity. */
5618 /* (-A) / (-B) -> A / B */
5623 /* In IEEE floating point, x/1 is not equivalent to x for snans. */
5628 /* If ARG1 is a constant, we can convert this to a multiply by the
5629 reciprocal. This does not have the same rounding properties,
5630 so only do this if -funsafe-math-optimizations. We can actually
5631 always safely do it if ARG1 is a power of two, but it's hard to
5632 tell if it is or not in a portable manner. */
5634 {
5639 /* Find the reciprocal if optimizing and the result is exact. */
5641 {
5642 REAL_VALUE_TYPE r;
5645 {
5648 }
5649 }
5650 }
5651 /* Convert A/B/C to A/(B*C). */
5654 {
5657 arg1)));
5658 }
5659 /* Convert A/(B/C) to (A/B)*C. */
5662 {
5667 }
5680 /* If arg0 is a multiple of arg1, then rewrite to the fastest div
5681 operation, EXACT_DIV_EXPR.
5683 Note that only CEIL_DIV_EXPR and FLOOR_DIV_EXPR are rewritten now.
5684 At one time others generated faster code, it's not clear if they do
5685 after the last round to changes to the DIV code in expmed.c. */
5719 /* Since negative shift count is not well-defined,
5720 don't try to compute it in the compiler. */
5723 /* Rewrite an LROTATE_EXPR by a constant into an
5724 RROTATE_EXPR by a new constant. */
5726 {
5730 = const_binop
5737 }
5739 /* If we have a rotate of a bit operation with the rotate count and
5740 the second operand of the bit operation both constant,
5741 permute the two operations. */
5754 /* Two consecutive rotates adding up to the width of the mode can
5755 be ignored. */
5786 /* Note that the operand of this must be an int
5787 and its values must be 0 or 1.
5788 ("true" is a fixed value perhaps depending on the language,
5789 but we don't handle values other than 1 correctly yet.) */
5791 /* Avoid infinite recursion. */
5797 /* Note that the operands of this must be ints
5798 and their values must be 0 or 1.
5799 ("true" is a fixed value perhaps depending on the language.) */
5800 /* If first arg is constant zero, return it. */
5804 /* If either arg is constant true, drop it. */
5808 /* Preserve sequence points. */
5811 /* If second arg is constant zero, result is zero, but first arg
5812 must be evaluated. */
5815 /* Likewise for first arg, but note that only the TRUTH_AND_EXPR
5816 case will be handled here. */
5820 truth_andor:
5821 /* We only do these simplifications if we are optimizing. */
5825 /* Check for things like (A || B) && (A || C). We can convert this
5826 to A || (B && C). Note that either operator can be any of the four
5827 truth and/or operations and the transformation will still be
5828 valid. Also note that we only care about order for the
5829 ANDIF and ORIF operators. If B contains side effects, this
5830 might change the truth-value of A. */
5837 {
5857 /* This case if tricky because we must either have commutative
5858 operators or else A10 must not have side-effects. */
5864 a01));
5865 }
5867 /* See if we can build a range comparison. */
5871 /* Check for the possibility of merging component references. If our
5872 lhs is another similar operation, try to merge its rhs with our
5873 rhs. Then try to merge our lhs and rhs. */
5885 /* Note that the operands of this must be ints
5886 and their values must be 0 or true.
5887 ("true" is a fixed value perhaps depending on the language.) */
5888 /* If first arg is constant true, return it. */
5892 /* If either arg is constant zero, drop it. */
5896 /* Preserve sequence points. */
5899 /* If second arg is constant true, result is true, but we must
5900 evaluate first arg. */
5903 /* Likewise for first arg, but note this only occurs here for
5904 TRUTH_OR_EXPR. */
5910 /* If either arg is constant zero, drop it. */
5915 /* If either arg is constant true, this is a logical inversion. */
5928 /* If one arg is a real or integer constant, put it last. */
5933 {
5940 }
5943 {
5944 /* (-a) CMP (-b) -> b CMP a */
5949 /* (-a) CMP CST -> a swap(CMP) (-CST) */
5951 return
5957 /* IEEE doesn't distinguish +0 and -0 in comparisons. */
5958 /* a CMP (-0) -> a CMP 0 */
5964 /* If this is a comparison of a real constant with a PLUS_EXPR
5965 or a MINUS_EXPR of a real constant, we can convert it into a
5966 comparison with a revised real constant as long as no overflow
5967 occurs when unsafe_math_optimizations are enabled. */
5978 }
5980 /* Convert foo++ == CONST into ++foo == CONST + INCR.
5981 First, see if one arg is constant; find the constant arg
5982 and the other one. */
5983 {
5993 {
5994 /* This optimization is invalid for ordered comparisons
5995 if CONST+INCR overflows or if foo+incr might overflow.
5996 This optimization is invalid for floating point due to rounding.
5997 For pointer types we assume overflow doesn't happen. */
6001 {
6002 tree newconst
6006 /* Do not overwrite the current varop to be a preincrement,
6007 create a new node so that we won't confuse our caller who
6008 might create trees and throw them away, reusing the
6009 arguments that they passed to build. This shows up in
6010 the THEN or ELSE parts of ?: being postincrements. */
6015 /* If VAROP is a reference to a bitfield, we must mask
6016 the constant by the width of the field. */
6020 {
6021 int size
6023 (TREE_OPERAND
6027 tree folded_compare;
6029 /* First check whether the comparison would come out
6030 always the same. If we don't do that we would
6031 change the meaning with the masking. */
6035 else
6038 constop));
6053 mask)));
6054 }
6060 }
6061 }
6063 {
6067 {
6068 tree newconst
6072 /* Do not overwrite the current varop to be a predecrement,
6073 create a new node so that we won't confuse our caller who
6074 might create trees and throw them away, reusing the
6075 arguments that they passed to build. This shows up in
6076 the THEN or ELSE parts of ?: being postdecrements. */
6084 {
6085 int size
6087 (TREE_OPERAND
6091 tree folded_compare;
6096 else
6099 constop));
6114 mask)));
6115 }
6121 }
6122 }
6123 }
6125 /* Change X >= C to X > (C - 1) and X < C to X <= (C - 1) if C > 0.
6126 This transformation affects the cases which are handled in later
6127 optimizations involving comparisons with non-negative constants. */
6131 {
6133 {
6148 }
6149 }
6151 /* Comparisons with the highest or lowest possible integer of
6152 the specified size will have known values. */
6153 {
6161 {
6168 {
6171 }
6172 else
6173 {
6176 }
6181 {
6185 arg0);
6193 arg0);
6199 /* The GE_EXPR and LT_EXPR cases above are not normally
6200 reached because of previous transformations. */
6204 }
6208 {
6221 }
6225 {
6229 arg0);
6238 arg0);
6246 }
6250 {
6263 }
6268 /* signed_type does not work on pointer types. */
6270 {
6271 /* The following case also applies to X < signed_max+1
6272 and X >= signed_max+1 because previous transformations. */
6274 {
6278 return fold
6282 }
6283 }
6284 }
6285 }
6287 /* If this is an EQ or NE comparison of a constant with a PLUS_EXPR or
6288 a MINUS_EXPR of a constant, we can convert it into a comparison with
6289 a revised constant as long as no overflow occurs. */
6301 /* Similarly for a NEGATE_EXPR. */
6310 /* If we have X - Y == 0, we can convert that to X == Y and similarly
6311 for !=. Don't do this for ordered comparisons due to overflow. */
6317 /* If we are widening one operand of an integer comparison,
6318 see if the other operand is similarly being widened. Perhaps we
6319 can do the comparison in the narrower type. */
6329 /* If this is comparing a constant with a MIN_EXPR or a MAX_EXPR of a
6330 constant, we can simplify it. */
6337 /* If we are comparing an ABS_EXPR with a constant, we can
6338 convert all the cases into explicit comparisons, but they may
6339 well not be faster than doing the ABS and one comparison.
6340 But ABS (X) <= C is a range comparison, which becomes a subtraction
6341 and a comparison, and is probably faster. */
6353 /* If this is an EQ or NE comparison with zero and ARG0 is
6354 (1 << foo) & bar, convert it to (bar >> foo) & 1. Both require
6355 two operations, but the latter can be done in one less insn
6356 on machines that have only two-operand insns or on which a
6357 constant cannot be the first operand. */
6360 {
6363 return
6371 integer_one_node)),
6372 arg1));
6375 return
6383 integer_one_node)),
6384 arg1));
6385 }
6387 /* If this is an NE or EQ comparison of zero against the result of a
6388 signed MOD operation whose second operand is a power of 2, make
6389 the MOD operation unsigned since it is simpler and equivalent. */
6398 {
6405 }
6407 /* If this is an NE comparison of zero with an AND of one, remove the
6408 comparison since the AND will give the correct value. */
6414 /* If we have (A & C) == C where C is a power of 2, convert this into
6415 (A & C) != 0. Similarly for NE_EXPR. */
6423 /* If we have (A & C) != 0 where C is the sign bit of A, convert
6424 this into A < 0. Similarly for (A & C) == 0 into A >= 0. */
6428 {
6432 {
6437 }
6438 }
6440 /* If X is unsigned, convert X < (1 << Y) into X >> Y == 0
6441 and similarly for >= into !=. */
6457 return
6464 /* Simplify comparison of something with itself. (For IEEE
6465 floating-point, we can only do some of these simplifications.) */
6467 {
6469 {
6480 /* For NE, we can only do this simplification if integer. */
6483 /* ... fall through ... */
6489 }
6490 }
6492 /* If we are comparing an expression that just has comparisons
6493 of two integer values, arithmetic expressions of those comparisons,
6494 and constants, we can simplify it. There are only three cases
6495 to check: the two values can either be equal, the first can be
6496 greater, or the second can be greater. Fold the expression for
6497 those three values. Since each value must be 0 or 1, we have
6498 eight possibilities, each of which corresponds to the constant 0
6499 or 1 or one of the six possible comparisons.
6501 This handles common cases like (a > b) == 0 but also handles
6502 expressions like ((x > y) - (y > x)) > 0, which supposedly
6503 occur in macroized code. */
6506 {
6511 /* Don't handle degenerate cases here; they should already
6512 have been handled anyway. */
6521 {
6525 /* We can't just pass T to eval_subst in case cval1 or cval2
6526 was the same as ARG1. */
6528 tree high_result
6531 arg1));
6532 tree equal_result
6535 arg1));
6536 tree low_result
6539 arg1));
6541 /* All three of these results should be 0 or 1. Confirm they
6542 are. Then use those values to select the proper code
6543 to use. */
6551 {
6552 /* Make a 3-bit mask with the high-order bit being the
6553 value for `>', the next for '=', and the low for '<'. */
6557 {
6559 /* Always false. */
6580 /* Always true. */
6582 }
6587 else
6589 }
6590 }
6591 }
6593 /* If this is a comparison of a field, we may be able to simplify it. */
6598 /* Handle the constant case even without -O
6599 to make sure the warnings are given. */
6601 {
6604 }
6606 /* If this is a comparison of complex values and either or both sides
6607 are a COMPLEX_EXPR or COMPLEX_CST, it is best to split up the
6608 comparisons and join them with a TRUTH_ANDIF_EXPR or TRUTH_ORIF_EXPR.
6609 This may prevent needless evaluations. */
6616 {
6629 type,
6632 }
6634 /* Optimize comparisons of strlen vs zero to a compare of the
6635 first character of the string vs zero. To wit,
6636 strlen(ptr) == 0 => *ptr == 0
6637 strlen(ptr) != 0 => *ptr != 0
6638 Other cases should reduce to one of these two (or a constant)
6639 due to the return value of strlen being unsigned. */
6644 {
6646 tree arglist;
6658 integer_zero_node));
6659 }
6661 /* From here on, the only cases we handle are when the result is
6662 known to be a constant.
6664 To compute GT, swap the arguments and do LT.
6665 To compute GE, do LT and invert the result.
6666 To compute LE, swap the arguments, do LT and invert the result.
6667 To compute NE, do EQ and invert the result.
6669 Therefore, the code below must handle only EQ and LT. */
6672 {
6675 }
6677 /* Note that it is safe to invert for real values here because we
6678 will check below in the one case that it matters. */
6683 {
6686 }
6688 /* Compute a result for LT or EQ if args permit;
6689 otherwise return T. */
6691 {
6694 else
6699 }
6702 /* Assume a nonexplicit constant cannot equal an explicit one,
6703 since such code would be undefined anyway.
6704 Exception: on sysvr4, using #pragma weak,
6705 a label can come out as 0. */
6712 #endif
6713 /* Two real constants can be compared explicitly. */
6715 {
6716 /* If either operand is a NaN, the result is false with two
6717 exceptions: First, an NE_EXPR is true on NaNs, but that case
6718 is already handled correctly since we will be inverting the
6719 result for NE_EXPR. Second, if we had inverted a LE_EXPR
6720 or a GE_EXPR into a LT_EXPR, we must return true so that it
6721 will be inverted into false. */
6731 else
6735 }
6749 /* Pedantic ANSI C says that a conditional expression is never an lvalue,
6750 so all simple results must be passed through pedantic_non_lvalue. */
6752 return pedantic_non_lvalue
6757 /* If the second operand is zero, invert the comparison and swap
6758 the second and third operands. Likewise if the second operand
6759 is constant and the third is not or if the third operand is
6760 equivalent to the first operand of the comparison. */
6768 {
6769 /* See if this can be inverted. If it can't, possibly because
6770 it was a floating-point inequality comparison, don't do
6771 anything. */
6775 {
6779 /* arg1 should be the first argument of the new T. */
6782 }
6783 }
6785 /* If we have A op B ? A : C, we may be able to convert this to a
6786 simpler expression, depending on the operation and the values
6787 of B and C. Signed zeros prevent all of these transformations,
6788 for reasons given above each one. */
6794 {
6800 /* If we have A op 0 ? A : -A, consider applying the following
6801 transformations:
6803 A == 0? A : -A same as -A
6804 A != 0? A : -A same as A
6805 A >= 0? A : -A same as abs (A)
6806 A > 0? A : -A same as abs (A)
6807 A <= 0? A : -A same as -abs (A)
6808 A < 0? A : -A same as -abs (A)
6810 None of these transformations work for modes with signed
6811 zeros. If A is +/-0, the first two transformations will
6812 change the sign of the result (from +0 to -0, or vice
6813 versa). The last four will fix the sign of the result,
6814 even though the original expressions could be positive or
6815 negative, depending on the sign of A.
6817 Note that all these transformations are correct if A is
6818 NaN, since the two alternatives (A and -A) are also NaNs. */
6825 {
6827 return
6828 pedantic_non_lvalue
6830 negate_expr
6832 arg1))));
6840 return pedantic_non_lvalue
6848 return pedantic_non_lvalue
6852 arg1)))));
6855 }
6857 /* A != 0 ? A : 0 is simply A, unless A is -0. Likewise
6858 A == 0 ? A : 0 is always 0 unless A is -0. Note that
6859 both transformations are correct when A is NaN: A != 0
6860 is then true, and A == 0 is false. */
6863 {
6868 }
6870 /* Try some transformations of A op B ? A : B.
6872 A == B? A : B same as B
6873 A != B? A : B same as A
6874 A >= B? A : B same as max (A, B)
6875 A > B? A : B same as max (B, A)
6876 A <= B? A : B same as min (A, B)
6877 A < B? A : B same as min (B, A)
6879 As above, these transformations don't work in the presence
6880 of signed zeros. For example, if A and B are zeros of
6881 opposite sign, the first two transformations will change
6882 the sign of the result. In the last four, the original
6883 expressions give different results for (A=+0, B=-0) and
6884 (A=-0, B=+0), but the transformed expressions do not.
6886 The first two transformations are correct if either A or B
6887 is a NaN. In the first transformation, the condition will
6888 be false, and B will indeed be chosen. In the case of the
6889 second transformation, the condition A != B will be true,
6890 and A will be chosen.
6892 The conversions to max() and min() are not correct if B is
6893 a number and A is not. The conditions in the original
6894 expressions will be false, so all four give B. The min()
6895 and max() versions would give a NaN instead. */
6898 {
6903 /* Avoid adding NOP_EXPRs in case this is an lvalue. */
6908 {
6915 /* In C++ a ?: expression can be an lvalue, so put the
6916 operand which will be used if they are equal first
6917 so that we can convert this back to the
6918 corresponding COND_EXPR. */
6920 return pedantic_non_lvalue
6930 return pedantic_non_lvalue
6939 }
6940 }
6942 /* If this is A op C1 ? A : C2 with C1 and C2 constant integers,
6943 we might still be able to simplify this. For example,
6944 if C1 is one less or one more than C2, this might have started
6945 out as a MIN or MAX and been transformed by this function.
6946 Only good for INTEGER_TYPEs, because we need TYPE_MAX_VALUE. */
6952 {
6954 /* We can replace A with C1 in this case. */
6961 /* If C1 is C2 + 1, this is min(A, C2). */
6966 return pedantic_non_lvalue
6971 /* If C1 is C2 - 1, this is min(A, C2). */
6976 return pedantic_non_lvalue
6981 /* If C1 is C2 - 1, this is max(A, C2). */
6986 return pedantic_non_lvalue
6991 /* If C1 is C2 + 1, this is max(A, C2). */
6996 return pedantic_non_lvalue
7003 }
7004 }
7006 /* If the second operand is simpler than the third, swap them
7007 since that produces better jump optimization results. */
7013 {
7014 /* See if this can be inverted. If it can't, possibly because
7015 it was a floating-point inequality comparison, don't do
7016 anything. */
7020 {
7024 /* arg1 should be the first argument of the new T. */
7027 }
7028 }
7030 /* Convert A ? 1 : 0 to simply A. */
7033 /* If we try to convert TREE_OPERAND (t, 0) to our type, the
7034 call to fold will try to move the conversion inside
7035 a COND, which will recurse. In that case, the COND_EXPR
7036 is probably the best choice, so leave it alone. */
7040 /* Convert A ? 0 : 1 to !A. This prefers the use of NOT_EXPR
7041 over COND_EXPR in cases such as floating point comparisons. */
7048 /* Look for expressions of the form A & 2 ? 2 : 0. The result of this
7049 operation is simply A & 2. */
7060 /* Convert A ? B : 0 into A && B if A and B are truth values. */
7067 /* Convert A ? B : 1 into !A || B if A and B are truth values. */
7071 {
7072 /* Only perform transformation if ARG0 is easily inverted. */
7077 }
7082 /* When pedantic, a compound expression can be neither an lvalue
7083 nor an integer constant expression. */
7086 /* Don't let (0, 0) be null pointer constant. */
7128 /* Pull arithmetic ops out of the CLEANUP_POINT_EXPR where
7129 appropriate. */
7134 {
7138 tree arg01;
7149 {
7163 arg01));
7164 }
7167 }
7170 /* Check for a built-in function. */
7175 {
7179 }
7185 }
7187 /* Determine if first argument is a multiple of second argument. Return 0 if
7188 it is not, or we cannot easily determined it to be.
7190 An example of the sort of thing we care about (at this point; this routine
7191 could surely be made more general, and expanded to do what the *_DIV_EXPR's
7192 fold cases do now) is discovering that
7194 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7196 is a multiple of
7198 SAVE_EXPR (J * 8)
7200 when we know that the two SAVE_EXPR (J * 8) nodes are the same node.
7202 This code also handles discovering that
7204 SAVE_EXPR (I) * SAVE_EXPR (J * 8)
7206 is a multiple of 8 so we don't have to worry about dealing with a
7207 possible remainder.
7209 Note that we *look* inside a SAVE_EXPR only to determine how it was
7210 calculated; it is not safe for fold to do much of anything else with the
7211 internals of a SAVE_EXPR, since it cannot know when it will be evaluated
7212 at run time. For example, the latter example above *cannot* be implemented
7213 as SAVE_EXPR (I) * J or any variant thereof, since the value of J at
7214 evaluation time of the original SAVE_EXPR is not necessarily the same at
7215 the time the new expression is evaluated. The only optimization of this
7216 sort that would be valid is changing
7218 SAVE_EXPR (I) * SAVE_EXPR (SAVE_EXPR (J) * 8)
7220 divided by 8 to
7222 SAVE_EXPR (I) * SAVE_EXPR (J)
7224 (where the same SAVE_EXPR (J) is used in the original and the
7225 transformed version). */
7227 static int
7229 tree type;
7230 tree top;
7231 tree bottom;
7232 {
7240 {
7252 {
7256 /* const_binop may not detect overflow correctly,
7257 so check for it explicitly here. */
7266 }
7270 /* Can't handle conversions from non-integral or wider integral type. */
7276 /* .. fall through ... */
7292 }
7293 }
7295 /* Return true if `t' is known to be non-negative. */
7297 int
7299 tree t;
7300 {
7302 {
7343 /* Truth values evaluate to 0 or 1, which is nonnegative. */
7345 else
7346 /* We don't know sign of `t', so be conservative and return false. */
7348 }
7349 }
7351 /* Return true if `r' is known to be non-negative.
7352 Only handles constants at the moment. */
7354 int
7356 rtx r;
7357 {
7359 {
7369 {
7371 rtx elt;
7376 {
7380 }
7383 }
7387 /* These are always nonnegative. */
7392 }
7393 }