C++中的函数指针

Question

我使用CreateThread函数在C++中编写类似C#BackgroundWorker的类.我的代码:

BackgroundWorker.h:

class BackgroundWorker
{
    private :
        HANDLE _threadHandle;
        unsigned int _threadCallcounter;
        DWORD _threadID;
    public:
        BackgroundWorker();
        ~BackgroundWorker();
        virtual DWORD WINAPI Function(LPVOID vpPram);
}

BackgroundWorker.cpp:

#include "BackgroundWorker.h"
void BackgroundWorker::DoWork()
{
    this->_threadHandle = CreateThread(NULL,
        0,this->Function,&this->_threadCallcounter,
        0, &this->_threadID); // !!!***This part throws an error***!!!
}

然后我创建了另一个派生自BackgroundWorker的类:

ListenThread.cpp:

class ListenThread :public BackgroundWorker
{
    DWORD WINAPI Function(LPVOID vpPram)
    {
        //DO somthing...
        return 0;
    }
};

但该行给了我以下错误:

非标准语法; 使用'&'创建指向成员的指针

Answer 1

您的错误消息意味着您需要将指针传递给函数 as &Function，而不是Functionin DoWork。

不幸的是修复这个问题没有帮助。CreateThread不适用于（非静态）成员函数。解决方案是创建一个静态方法来用作实际的线程启动函数。

检查这个例子：

#include <Windows.h>
#include <iostream>

class BackgroundWorker
{
private :
    HANDLE _threadHandle;
    DWORD _threadID;

public:
    static DWORD WINAPI StaticThreadStart(void * Param) {
        BackgroundWorker * This = static_cast<BackgroundWorker *>(Param);
        return This->Function();
    }

    virtual DWORD Function() {
        return 0;
    }

    void DoWork() {
        _threadHandle = CreateThread(NULL, 0, StaticThreadStart, this, 0, &_threadID);
    }
};

class ListenThread : public BackgroundWorker {
    DWORD Function() override {
        std::cout << "Working...\n";
        return 0;
    }
};

int main()
{
    ListenThread cl;
    cl.DoWork();

    // Call pause to wait for new thread in this example
    system("pause");
    return 0;
}