误解的函数指针-将其作为参数传递

问题描述

我想通过一个函数指针作为参数将A类的成员函数传递给B类.请告知这条路是否通往某处，并帮助我填补坑洼.

I want to pass a member function of class A to class B via a function pointer as argument. Please advise whether this road is leading somewhere and help me fill the pothole.

#include <iostream>


using namespace std;

class A{
public:

int dosomeA(int x){
    cout<< "doing some A to "<<x <<endl;
    return(0);
}
};

class B{
public:

B(int (*ptr)(int)){ptr(0);};
};

int main()
{
A a;
int (*APtr)(int)=&A::dosomeA;

B b(APtr);
return 0;
}

这段出色的代码让我陷入了编译器错误:

This brilliant piece of code leaves me with the compiler error:

首先，我要它进行编译.
其次，我不希望dosomeA处于静态状态.

Firstly I want it to compile.
Secondly I don't want dosomeA to be STATIC.

推荐答案

指向成员的指针与普通函数指针不同.由于编译器错误指示&A::dosomeA的类型实际上是int (A::*)(int)而不是int (*)(int).

Pointers to members are different from normal function pointers. As the compiler error indicates the type of &A::dosomeA is actually int (A::*)(int) and not int (*)(int).

在B的构造函数中，您需要一个A实例，以使用.*或->*运算符之一来调用该成员.

Inside B's constructor you need an instance of A to call the member on using one the .* or ->* operators.

E.g'

B(int(A::*ptr)(int))
{
    A atmp;
    (atmp.*ptr)(int);
}

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