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tree-vrp.c (vrp_int_const_binop): Handle PLUS_EXPR and the *_DIV_EXPR codes correctly with overflow infinities.

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* tree-vrp.c (vrp_int_const_binop): Handle PLUS_EXPR and
	the *_DIV_EXPR codes correctly with overflow infinities.

From-SVN: r122820
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Ian Lance Taylor 2007-03-11 15:53:30 +00:00 committed by Ian Lance Taylor
parent bd91a8c4ed
commit d7419dec45
2 changed files with 27 additions and 7 deletions
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5
gcc/ChangeLog
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@ -1,3 +1,8 @@
2007-03-11 Ian Lance Taylor <iant@google.com>
* tree-vrp.c (vrp_int_const_binop): Handle PLUS_EXPR and
the *_DIV_EXPR codes correctly with overflow infinities.
2007-03-11 Ira Rosen <irar@il.ibm.com>
* tree-data-ref.c (analyze_offset): Add a return value (bool) to

29
gcc/tree-vrp.c
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@ -1519,15 +1519,26 @@ vrp_int_const_binop (enum tree_code code, tree val1, tree val2)
&& !supports_overflow_infinity (TREE_TYPE (res)))
return NULL_TREE;
/* We have to punt on subtracting infinities of the same sign,
since we can't tell what the sign of the result should
be. */
if (code == MINUS_EXPR
&& sgn1 == sgn2
/* We have to punt on adding infinities of different signs,
since we can't tell what the sign of the result should be.
Likewise for subtracting infinities of the same sign. */
if (((code == PLUS_EXPR && sgn1 != sgn2)
|| (code == MINUS_EXPR && sgn1 == sgn2))
&& is_overflow_infinity (val1)
&& is_overflow_infinity (val2))
return NULL_TREE;
/* Don't try to handle division or shifting of infinities. */
if ((code == TRUNC_DIV_EXPR
|| code == FLOOR_DIV_EXPR
|| code == CEIL_DIV_EXPR
|| code == EXACT_DIV_EXPR
|| code == ROUND_DIV_EXPR
|| code == RSHIFT_EXPR)
&& (is_overflow_infinity (val1)
|| is_overflow_infinity (val2)))
return NULL_TREE;
/* Notice that we only need to handle the restricted set of
operations handled by extract_range_from_binary_expr.
Among them, only multiplication, addition and subtraction
@ -1541,8 +1552,12 @@ vrp_int_const_binop (enum tree_code code, tree val1, tree val2)
if ((code == MULT_EXPR && sgn1 == sgn2)
/* For addition, the operands must be of the same sign
to yield an overflow. Its sign is therefore that
of one of the operands, for example the first. */
|| (code == PLUS_EXPR && sgn1 > 0)
of one of the operands, for example the first. For
infinite operands X + -INF is negative, not positive. */
|| (code == PLUS_EXPR
&& (sgn1 >= 0
? !is_negative_overflow_infinity (val2)
: is_positive_overflow_infinity (val2)))
/* For subtraction, non-infinite operands must be of
different signs to yield an overflow. Its sign is
therefore that of the first operand or the opposite of