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