如何调用指向成员函数的指针? [英] How do I call a pointer-to-member-function?

问题描述

我收到了一个我不明白的编译错误 (MS VS 2008).搞了好几个小时之后，一切都变得模糊了，我觉得我错过了一些非常明显(而且非常愚蠢)的东西.这是基本代码:

I'm getting a compile error (MS VS 2008) that I just don't understand. After messing with it for many hours, it's all blurry and I feel like there's something very obvious (and very stupid) that I'm missing. Here's the essential code:

typedef int (C::*PFN)(int);

struct MAP_ENTRY
    {
    int id;
    PFN pfn;
    };

class C
    {
    ...
    int Dispatch(int, int);
    MAP_ENTRY *pMap;
    ...
    };

int C::Dispatch(int id, int val)
    {
    for (MAP_ENTRY *p = pMap; p->id != 0; ++p)
        {
        if (p->id == id)
            return p->pfn(val);  // <--- error here
        }
    return 0;
    }

编译器在箭头处声称术语不会评估为采用 1 个参数的函数".为什么不?PFN 的原型是一个带一个参数的函数，而 MAP_ENTRY.pfn 是一个 PFN.我在这里错过了什么?

The compiler claims at the arrow that the "term does not evaluate to a function taking 1 argument". Why not? PFN is prototyped as a function taking one argument, and MAP_ENTRY.pfn is a PFN. What am I missing here?

推荐答案

p->pfn 是指向成员函数类型的指针.为了通过这样的指针调用函数，您需要使用运算符 ->* 或运算符 .* 并提供 C 作为左操作数.你没有.

p->pfn is a pointer of pointer-to-member-function type. In order to call a function through such a pointer you need to use either operator ->* or operator .* and supply an object of type C as the left operand. You didn't.

我不知道应该在这里使用哪种 C 类型的对象 - 只有你知道 - 但在你的例子中它可能是 *这个.在这种情况下，调用可能如下所示

I don't know which object of type C is supposed to be used here - only you know that - but in your example it could be *this. In that case the call might look as follows

(this->*p->pfn)(val)

为了让它看起来不那么复杂，可以引入一个中间变量

In order to make it look a bit less convoluted, you can introduce an intermediate variable

PFN pfn = p->pfn;
(this->*pfn)(val);

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