函数指向类成员的指针 [英] function pointer to a class member

问题描述

我想这样做：

class A {
    void *(*func)(void *);

    A(void *(*function)(void *)){
        func = function;
    }
}

class B {
    void *real_func(void *);
    A ptr;
    B()
    :ptr(&real_func)
    {
        ...
    }
}

但我收到此错误：

错误：ISO C ++禁止使用非限定或
括号非静态成员函数的地址形成指向成员
函数的指针。

error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function.

有人知道如何初始化函数指向同一类中的函数成员

Someone knows how to initialize the function pointer to a function member in the same class???

谢谢！

Carlos

推荐答案

由于 real_func 成员函数，其类型不能为 void *（*）（）。相反，它是 void *（B :: *）（）所以你需要相应地声明 func p>

Since real_func is not a static member function, its type cannot be void *(*)(). Instead, it is void *(B::*)() so you need to declare func accordingly:

void *(B::*func)();

// call it like this
pointer_to_b->*func();

如果你很小心，你也可以使用指针A作为基类，确保指向A的指针指向B的实例：

If you are careful, you can also use pointer to A as the base class, but you must make sure that the pointer to A points to an instance of B:

void *(A::*func)();

但是，现在大多数情况下只是复制虚拟成员函数的功能。所以我建议你改用：

At this point, however, you are mostly just replicating the functionality of virtual member functions. So I would recommend you use that instead:

class A {
    virtual void *func() = 0;
};

class B {
    void *func() {
        // ...
    }
};

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