[Bug middle-end/55217] False -Wstrict-overflow warning

[msebor at gcc dot gnu.org]

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 ‘f’:
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]