c++ - 与类成员的函数调用？

Question

在我展示这篇文章底部的代码之前，我想谈谈我不想要的问题和修复。好吧，基本上我已经从头开始创建了一个 GUI，我想要的一个要求是允许组件有自己的点击执行，所以如果我点击一个按钮或选项卡等。它会调用 Component->Execute();那么通常你会做一些像 ids 的 switch 语句，如果那个组件 ID 等于 n 号那么它会执行这个 Action 。嗯，这对我来说有点愚蠢，我认为必须有更好的方法。我最终尝试在 JAVA 中加入一个功能，你可以像 Component.AddActionListener(new ActionListener( public void execute(ActionEvent ae) { }));或类似的东西，我认为此功能必须在 C++ 中实现。我最终遇到了将 void 函数存储到一个可以随时执行和随时修改的变量中。但是我没有注意到一个问题，那就是这只适用于静态函数。所以在下面你会看到我的问题。我已经通过使用指向 SomeClass 的指针修补了这个问题，但这意味着对每个类类型都有一个单独的函数调用，如果不执行以下策略，就没有办法将函数回调存储到非静态类成员吗？而不是像注释掉的代码那样做一个策略？

// main.cpp

#include <iostream> //system requires this.
#include "SomeClass.h"

void DoSomething1(void)
{
    std::cout << "We Called Static DoSomething1\n";
}

void DoSomething2(void)
{
    std::cout << "We Called Static DoSomething2\n";
}

int main()
{
    void (*function_call2)(SomeClass*);
    void (*function_call)() = DoSomething1; //This works No Problems!
    function_call(); //Will Call the DoSomething1(void);
    function_call = DoSomething2; //This works No Problems!
    function_call(); //Will Call the DoSomething2(void);

    SomeClass *some = new SomeClass(); //Create a SomeClass pointer;
    function_call = SomeClass::DoSomething3; //Static SomeClass::DoSomething3();
    function_call(); //Will Call the SomeClass::DoSomething3(void);
    //function_call = some->DoSomething4; //Non-Static SomeClass::DoSomething4 gives an error.
    //function_call(); //Not used because of error above.
    function_call2 = SomeClass::DoSomething5; //Store the SomeClass::DoSomething(SomeClass* some);
    function_call2(some); //Call out SomeClass::DoSomething5 which calls on SomeClass::DoSomething4's non static member.
    system("pause");
    return 0;
}

//SomeClass.hpp

#pragma once

#include <iostream>

class SomeClass 
{
    public:
        SomeClass();
        ~SomeClass();

    public:
        static void DoSomething3(void);
        void DoSomething4(void);
        static void DoSomething5(SomeClass* some);
};

//SomeClass.cpp

#include "SomeClass.h"

SomeClass::SomeClass(void)
{

}

SomeClass::~SomeClass(void)
{

}

void SomeClass::DoSomething3(void)
{
    std::cout << "We Called Static DoSomething3\n";
}

void SomeClass::DoSomething4(void)
{
    std::cout << "We Called Non-Static DoSomething4\n";
}

void SomeClass::DoSomething5(SomeClass *some)
{
    some->DoSomething4();
}

Secondary Fix for what I'll do not an exact answer I wanted but it meets my needs for now along with allowing additional features that would be complex over complicated without this not existed.

//组件.hpp

#pragma once

#include <iostream>
#include <windows.h>
#include <d3dx9.h>
#include <d3d9.h>

#include "Constants.hpp"
#include "ScreenState.hpp"
#include "ComponentType.hpp"

using namespace std;

class Component 
{

    static void EMPTY(void) { }
    static void EMPTY(int i) { }

    public:
    Component(void)
    {
        callback = EMPTY;
        callback2 = EMPTY;
        callback_id = -1;
    }

    Component* SetFunction(void (*callback)())
    {
        this->callback = callback;
        return this;
    }

    Component* SetFunction(void (*callback2)(int), int id)
    {
        this->callback_id = id;
        this->callback2 = callback2;
        return this;
    }

    void execute(void)
    {
        callback();
        callback2(callback_id);
    }

}

Answer 1

成员函数指针的语法如下:

struct Foo
{
    void bar(int, int);
    void zip(int, int);
};

Foo x;

void (Foo::*p)(int, int) = &Foo::bar;   // pointer

(x.*p)(1, 2);                           // invocation

p = &Foo::zip;

(x.*p)(3, 4);                           // invocation

注意函数调用中的额外括号，这是获得正确运算符优先级所必需的。成员取消引用运算符是 .*(还有来自实例指针的 ->*)。