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[Bug middle-end/55217] False -Wstrict-overflow warning
- From: "msebor at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: Wed, 11 Nov 2015 22:16:22 +0000
- Subject: [Bug middle-end/55217] False -Wstrict-overflow warning
- Auto-submitted: auto-generated
- References: <bug-55217-4 at http dot gcc dot gnu dot org/bugzilla/>
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=55217
Martin Sebor <msebor at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Last reconfirmed| |2015-11-11
CC| |msebor at gcc dot gnu.org
--- Comment #6 from Martin Sebor <msebor at gcc dot gnu.org> ---
6.0.0 20151111 issues the following slightly different warnings which disappear
when the increment of r in the first loop is made undonditional. Since the
condition should always be true (r can never be zero), the warning on that line
seems pointless (though not necessarily incorrect -- GCC is right to assume
there's no overflow).
I don't really know what to make of this so I'm leaving it UNCONFIRMED and for
someone else to look into in more depth.
$ cat u.c && /home/msebor/build/gcc-trunk/gcc/xgcc -B
/home/msebor/build/gcc-trunk/gcc -O2 -S -Wstrict-overflow=3 -o/dev/null
-std=c99 u.c
void h(int *s);
void f(int n, int s)
{
int r = 1;
for (int i = 1; i < n; i++)
if (r)
r++;
if (r * s >= s + 3) // warning here
for (int j = 0; j < r; j++)
h(&s);
}
u.c: In function âfâ:
u.c:7:20: warning: assuming signed overflow does not occur when simplifying
conditional to constant [-Wstrict-overflow]
if (r)
^
u.c:10:17: warning: assuming signed overflow does not occur when simplifying
conditional to constant [-Wstrict-overflow]
for (int j = 0; j < r; j++)
^
u.c:10:17: warning: assuming signed overflow does not occur when simplifying
conditional to constant [-Wstrict-overflow]