覆盖虚拟函数的区别仅在于调用约定意味着什么? [英] What does overriding virtual function differs only by calling convention mean?

问题描述

我正在尝试实施 IUnknown .我按照发球台上的说明进行操作，但是没有用.当我尝试编译时，我得到:

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

来自此代码:

#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;
        }
};

推荐答案

为每种方法添加 STDMETHODCALLTYPE .

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

基类( IUnknown )方法被声明为 STDMETHODCALLTYPE (这是 __ stdcall 的宏).覆盖虚拟方法时，它必须具有与原始方法相同的调用约定，在本例中为 __ stdcall

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

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