如何调用指针到成员函数？ [英] How do I call a pointer-to-member-function?

问题描述

我收到一个编译错误（MS VS 2008），我只是不明白。在弄乱它了很多个小时后，这一切都是模糊的，我觉得有一些非常明显（和非常愚蠢），我错过了。这里是基本的代码：

  typedef int（C :: * PFN） 
 
 struct MAP_ENTRY 
 {
 int id; 
 PFN pfn; 
}; 
 
类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; 
} 
  

编译器声称在箭头处，term不评估为函数取1参数。为什么不？ PFN作为接受一个参数的函数原型化，MAP_ENTRY.pfn是PFN。

指针到成员函数类型的指针。为了通过这样的指针调用一个函数，你需要使用操作符 - > * 或者操作符。并提供 C 类型的对象作为左操作数。你没有。

我不知道应该在这里使用 C 只有你知道 - 但在你的例子中它可以 * this 。在这种情况下，调用可能如下

 （this-> * p-> pfn）（val）
  

为了使它看起来稍微有点复杂，可以引入一个中间变量

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

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;
    }

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?

解决方案

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.

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)

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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