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c++: Fix -fstrong-eval-order for operator &&, || and , [PR82959]

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P0145R3 added
"However, the operands are sequenced in the order prescribed for the built-in
operator" rule for overloaded operator calls when using the operator syntax.
op_is_ordered follows that, but added just the overloaded operators
added in that paper.  &&, || and comma operators had rules that
lhs is sequenced before rhs already in C++98.
The following patch adds those cases to op_is_ordered.

2021-03-03  Jakub Jelinek  <jakub@redhat.com>

	PR c++/82959
	* call.c (op_is_ordered): Handle TRUTH_ANDIF_EXPR, TRUTH_ORIF_EXPR
	and COMPOUND_EXPR.

	* g++.dg/cpp1z/eval-order10.C: New test.
This commit is contained in:
Jakub Jelinek 2021-03-03 16:12:23 +01:00
parent fa6092d2cd
commit 0b8fa12015
2 changed files with 36 additions and 0 deletions
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9
gcc/cp/call.c
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@ -6083,6 +6083,15 @@ op_is_ordered (tree_code code)
case LSHIFT_EXPR:
// 8. a >> b
case RSHIFT_EXPR:
// a && b
// Predates P0145R3.
case TRUTH_ANDIF_EXPR:
// a || b
// Predates P0145R3.
case TRUTH_ORIF_EXPR:
// a , b
// Predates P0145R3.
case COMPOUND_EXPR:
return (flag_strong_eval_order ? 1 : 0);
default:

27
gcc/testsuite/g++.dg/cpp1z/eval-order10.C Normal file
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@ -0,0 +1,27 @@
// PR c++/82959
// { dg-do run }
// { dg-additional-options -fstrong-eval-order }
struct A {};
void operator && (const A x, const A) {}
void operator || (const A x, const A) {}
void operator , (const A x, const A) {}
int i;
A f () { if (i != 0) __builtin_abort (); i = 1; return A (); }
A g () { if (i != 1) __builtin_abort (); i = 2; return A (); }
int
main ()
{
f () && g ();
if (i != 2) __builtin_abort ();
i = 0;
f () || g ();
if (i != 2) __builtin_abort ();
i = 0;
f (), g ();
if (i != 2) __builtin_abort ();
}