What does overriding virtual function differs only by calling convention mean?(重写虚函数的区别仅在于调用约定意味着什么?)
问题描述
我正在尝试实现 未知.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:
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:usersseanmdesktop est estsource.cpp 6 1 test
Error 3 error C2695: 'testInterfaceImplementation::QueryInterface': overriding virtual function differs from 'IUnknown::QueryInterface' only by calling convention c:usersseanmdesktop est estsource.cpp 14 1 test
Error 4 error C2695: 'testInterfaceImplementation::Release': overriding virtual function differs from 'IUnknown::Release' only by calling convention c:usersseanmdesktop est estsource.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;
}
};
推荐答案
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__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,在这种情况下为 __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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本文标题为:重写虚函数的区别仅在于调用约定意味着什么?
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