c++ - 覆盖虚拟函数的区别仅在于调用约定意味着什么？

我正在尝试实现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:

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

从此代码:

#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;
        }
};

最佳答案

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)。当您覆盖虚拟方法时，它必须具有与原始方法相同的调用约定，在本例中为__stdcall