Question

In xp 32bit this line compiles with not problem however in vista 64bit this line:

m_FuncAddr = ::GetProcAddress (somthing); 

gives the following error

error C2440: ‘=’ : cannot convert from ‘FARPROC’ to ‘int (__cdecl *)(void)’

GetProcAddress is defined as

WINBASEAPI FARPROC WINAPI GetProcAddress (somthing) 

And m_FuncAddr as

int (WINAPI *m_FuncAddr)(); 

From what I understand both are stdcall’s.

To avoid the error I had to put

m_FuncAddr = (int (__cdecl *)(void))::GetProcAddress(somthing); 

My Question:

If both m_FuncAddr and GetProcAddress have the stdcall calling convention why do I have to ‘recall’ it with cdecl ?

Is it possible that the VS project setting ‘default calling convention (which is set to cdecl) over-rides the assignemet statment above ?

Thanks in advance!

[Edit]

To clerfiy the question:

On one side of the equation (say side 1) i have

int __stdcall * m_FuncAddr 

On other side (side 2)

INT_PTR far __stdcall GetProcAddress 

So how is it that i have to cast side 2 with cdecl if both are stdcalls ? Or am I not getting something ?

Answer 1

The return type should be INT_PTR (a 64-bit value in 64-bit builds). You shouldn’t cast around this error — the compiler is trying to tell you that something is wrong.

From WinDef.h:

#ifdef _WIN64 typedef INT_PTR (FAR WINAPI *FARPROC)(); 

So the declaration of m_FuncAddr should be:

INT_PTR (WINAPI *m_FuncAddr)();