Function pointer and calling convention

Question

float __stdcall (*pFunc)(float a, float b) = (float (__stdcall *)(float,float))0x411280;

How to declare a function pointer with calling convention? The

Answer 1

__fastcall is the optimized one (fastest calling convention) but not used for an unknown reason

Try:

int (__fastcall *myfunction)(int,float);

Answer 2

The trick is placing the __stdcall inside the parentheses like this:

float (__stdcall *pFunc)(float a, float b) = (float (__stdcall *)(float,float))0x411280;

Of course, you are recommended to use a typedef instead, but the same trick applies:

typedef float (__stdcall *FuncType)(float a, float b);

Answer 3

int (__cdecl **func)() is the most accepted form by compilers. Some compilers accept int __cdecl (**func)(int) and int (**__cdecl func)(int) or even int (*__cdecl *func)(int). Clang accepts all 3 forms and correctly interprets them as the type int ((**))(int) __attribute__((cdecl))

Introducing a const pointer on clang also does not change the picture and any arrangement is accepted so long as const is after the pointer

const int __cdecl (*const func)();
const int (__cdecl *const func)();
const int (*const __cdecl func)();