Template partial specialization for __stdcall function pointer

Question

typedef bool (*my_function_f)(int, double);
typedef bool (__stdcall *my_function_f2)(int, double);
//            ^^^^^^^^^

template class TFunction;

tem         


        

Answer 1

This is because (*) means default calling convention, which is __cdecl.

template<class R, class T0, class T1>
class TFunction<R(*)(T0,T1)>
{
  typedef R (*func_type)(T0,T1);
};

is actually equal to

template<class R, class T0, class T1>
class TFunction<R(__cdecl *)(T0,T1)>
{
  typedef R (__cdecl *func_type)(T0,T1);
};

which, of course, will not match an R(__stdcall *)(T0, T1) on Win32 where __stdcall is not ignored. If you want to partially specialize for function pointers then you will need a partial spec for every calling convention you want to accept.

Answer 2

No, this is by design. The calling convention is very much part of the function declaration, your template function uses the default calling convention. Which is not __stdcall unless you compile with /Gz. The default is /Gd, __cdecl.

The code compiles when you target x64 because it blissfully has only one calling convention.

Fix:

template<class R, class T0, class T1>
class TFunction<R (__stdcall *)(T0,T1)>
{
    // etc..
};

Answer 3

You have not specialized your template for the stdcall case, i.e. you need

template<class R, class T0, class T1>
class TFunction<R(__stdcall *)(T0,T1)>
{
  typedef R (*func_type)(T0,T1);
};

Not sure about the syntax, untested, but that should be the issue.