What does overriding virtual function differs only by calling convention mean?

Question

I am trying to implement IUnknown. I followed the instruction to the tee but it isn't working. When I try to compile I get:

Error   2   error C2695: 'testInterfaceImplementation::AddRef': overriding virtual function differs from 'IUnknown::AddRef' only by calling convention  c:\users\seanm\desktop\test\test\source.cpp 6   1   test
Error   3   error C2695: 'testInterfaceImplementation::QueryInterface': overriding virtual function differs from 'IUnknown::QueryInterface' only by calling convention  c:\users\seanm\desktop\test\test\source.cpp 14  1   test
Error   4   error C2695: 'testInterfaceImplementation::Release': overriding virtual function differs from 'IUnknown::Release' only by calling convention    c:\users\seanm\desktop\test\test\source.cpp 22  1   test

from this code:

#include <Windows.h>
#include <tchar.h>

class testInterfaceImplementation : public IUnknown {
    protected:
        ULONG AddRef() 
        {
            MessageBox(NULL,
                _T("TEST1"),
                _T("TEST1"),
                NULL);
            return 0;
        }
        HRESULT QueryInterface(IN REFIID riid, OUT void **ppvObject) 
        {
            MessageBox(NULL,
                _T("TEST2"),
                _T("TEST2"),
                NULL);
            return S_OK;
        }
        ULONG Release() {
            MessageBox(NULL,
                _T("TEST3"),
                _T("TEST3"),
                NULL);
            return 0;
        }
};

Answer 1

Add STDMETHODCALLTYPE for each of the methods.

ULONG STDMETHODCALLTYPE AddRef() 
HRESULT STDMETHODCALLTYPE QueryInterface(IN REFIID riid, OUT void **ppvObject) 
ULONG STDMETHODCALLTYPE Release() 

The base class(IUnknown) methods are declared as STDMETHODCALLTYPE (which is a macro for __stdcall). When you override a virtual method, it has to have the same calling convention as the original which in this case is __stdcall