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[Bug c/60838] bugos warning: initializer element is not a constant expression: int test = (0 ? (1,0) : 0) + 1;
- From: "mpolacek at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: Tue, 15 Apr 2014 08:00:13 +0000
- Subject: [Bug c/60838] bugos warning: initializer element is not a constant expression: int test = (0 ? (1,0) : 0) + 1;
- Auto-submitted: auto-generated
- References: <bug-60838-4 at http dot gcc dot gnu dot org/bugzilla/>
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=60838
Marek Polacek <mpolacek at gcc dot gnu.org> changed:
What |Removed |Added
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CC| |mpolacek at gcc dot gnu.org
--- Comment #2 from Marek Polacek <mpolacek at gcc dot gnu.org> ---
I actually don't think this is a bug. While C99/C11 say what you wrote about
Constant expressions, C89 says something else:
"Constant expressions shall not contain assignment, increment, decrement,
function-call, or comma operators, except when they are contained within the
operand of a sizeof operator."
And really:
int u = sizeof ((0 ? (3, 0) : 0) + 1);
issues no warnings even with -std=c89. The default is still -std=gnu89, so you
may want to use e.g. -std=gnu99.