In C, using a previously undeclared function constitutes an implicit declaration of the function. In an implicit declaration, the return type is int if I recall correctly. Now, GCC has built-in definitions for some standard functions. If an implicit declaration does not match the built-in definition, you get this warning. To fix the problem, you … Read more
I usually write a macro like this: #define UNUSED(x) (void)(x) You can use this macro for all your unused parameters. (Note that this works on any compiler.) For example: void f(int x) { UNUSED(x); … }
You can’t. The manual for GCC 4.4.0 is only comprehensive for that version, but it does list all the possible warnings for 4.4.0. They’re not all on the page you link to though, for instance some language-specific options are on the pages for C++ options or Obj-C options. To find them all you’re better off … Read more
You may try to include library headers using -isystem instead of -I. This will make them “system headers” and GCC won’t report warnings for them.
Forget casts. Use memcpy. float xhalf = 0.5f*x; uint32_t i; assert(sizeof(x) == sizeof(i)); std::memcpy(&i, &x, sizeof(i)); i = 0x5f375a86 – (i>>1); std::memcpy(&x, &i, sizeof(i)); x = x*(1.5f – xhalf*x*x); return x; The original code tries to initialize the int32_t by first accessing the float object through an int32_t pointer, which is where the rules are … Read more
You can put it in “(void)var;” expression (does nothing) so that a compiler sees it is used. This is portable between compilers. E.g. void foo(int param1, int param2) { (void)param2; bar(param1); } Or, #define UNUSED(expr) do { (void)(expr); } while (0) … void foo(int param1, int param2) { UNUSED(param2); bar(param1); }