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From: "jim at meyering dot net" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug c/49234] New: -Wstrict-overflow gives obviously unwarranted warning
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=49234

           Summary: -Wstrict-overflow gives obviously unwarranted warning
           Product: gcc
           Version: 4.7.0
            Status: UNCONFIRMED
          Severity: minor
          Priority: P3
         Component: c
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: jim@meyering.net


Given this code, in file k.c, -O2 -Wstrict-overflow evokes a warning.
However, since the only values assigned to "state" are 0, 1 and 2,
gcc should be able to determine that no overflow is possible,
and hence should issue no warning:

char *
trim2 (char *d)
{
  int state = 0;
  char *r;

  int i;
  for (i = 0; d[i]; i++)
    {
      if (state == 0 && d[i] == ' ')
    continue;

      if (state == 0)              /* line 13 */
    state = 1;

      if (state == 1)
    {
      state = 2;
      r = d + i;
    }
    }

  if (state == 2)
    *r = '\0';

  return d;
}

$ gcc -O2 -Wstrict-overflow -c k.c
k.c: In function 'trim2':
k.c:13:10: warning: assuming signed overflow does not occur when simplifying
conditional to constant [-Wstrict-overflow]

For the record, until recently I would not have bothered enabling
-Wstrict-overflow, due to the high proportion of false-positives,
but since I've learned about the risk of this bug,
  http://gcc.gnu.org/bugzilla/show_bug.cgi?id=33498
I am now more inclined to use -Wstrict-overflow in spite of that.