[Bug c++/32291] New: -Wformat either too picky or not picky enough

["scovich at gmail dot com" <gcc-bugzilla@gcc.gnu.org>]

Compiling the following snippet with -Wformat (or -Wall) causes the compiler to
complain: "wformat-bug.C:8: warning: format '%u' expects type 'unsigned int',
but argument 2 has type 'uint32_t'"

The problem seems to be that stdint.h defines uint32_t as "long" in cygwin. I
realize that int != long on some platforms, but i686 isn't one of them. Why
should the user be forced to cast their uint32_t (read: 32-bit unsigned int) to
"unsigned int" before passing it to printf() when they are logically identical? 

On the other hand, there's no complaint about passing a signed integer into an
unsigned format or vice-versa, even though the output value might actually
change because of the oversight in those cases.

wformat.C:
=======================================
#include <cstdio>
#include <stdint.h>

int main() {
  uint32_t a = ~0;
  unsigned int b = a;
  uint32_t c = b;
  printf("%u\n", c); // warning (?)

  int d = c;
  unsigned e = d;
  printf("%d\n", e); // no warning
  printf("%u\n", d); // no warning
}


-- 
           Summary: -Wformat either too picky or not picky enough
           Product: gcc
           Version: 4.2.0
            Status: UNCONFIRMED
          Severity: minor
          Priority: P3
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: scovich at gmail dot com
GCC target triplet: i686-pc-cygwin


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=32291