c++ - 重写虚函数的区别仅在于调用约定意味着什么？

Question

我正在尝试实现IUnknown。我按照指示到发球台，但它不起作用。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;
        }
};

Answer 1

为每个方法添加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（它是 的宏__stdcall）。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，在这种情况下是__stdcall