C ++中的指针到成员 - > *和。*运算符是什么？ [英] What are the Pointer-to-Member ->* and .* Operators in C++?

问题描述

是，我看到此问题和 此常见问题（错误链接） 此常见问题，但我仍然不明白 - > * 和。* 意味着在C ++中。

这些页面提供有关

- > * / code>和。在C ++中，你需要什么时候使用它们 - >

我希望这个例子能够清楚的说明这个例子。 为你

  //我们有一个类
 struct X 
 {
 void f ）{} 
 void g（）{} 
}; 
 
 typedef void（X :: * pointer）（）; 
 // ok，让我们拿一个指针并赋值给它。 
 pointer somePointer =& X :: f; 
 //现在我想调用somePointer。但是，为此，我需要一个对象
 X x; 
 //现在我调用x上的成员函数，像这样
（x。* somePointer）（）; //将调用x.f（）
 //现在，假设x不是对象，而是对象的指针
 X * px = new X; 
 //我想调用px上的memfun指针。我使用 - > * 
（px  - > * somePointer）（）; //将调用px-> f（）; 
  

现在，您不能使用 x.somePointer c $ c>或 px-> somePointer（），因为在类X中没有这样的成员。为此，使用特殊成员函数指针调用语法...只是自己尝试几个例子，你会习惯它

Yes, I've seen this question and this FAQ (wrong link) this FAQ, but I still don't understand what ->* and .* mean in C++.
Those pages provide information about the operators (such as overloading), but don't seem to explain well what they are.

What are ->* and .* in C++, and when do you need to use them as compared to -> and .?

解决方案

I hope this example will clear things for you

//we have a class
struct X
{
   void f() {}
   void g() {}
};

typedef void (X::*pointer)();
//ok, let's take a pointer and assign f to it.
pointer somePointer = &X::f;
//now I want to call somePointer. But for that, I need an object
X x;
//now I call the member function on x like this
(x.*somePointer)(); //will call x.f()
//now, suppose x is not an object but a pointer to object
X* px = new X;
//I want to call the memfun pointer on px. I use ->*
(px ->* somePointer)(); //will call px->f();

Now, you can't use x.somePointer(), or px->somePointer() because there is no such member in class X. For that the special member function pointer call syntax is used... just try a few examples yourself ,you'll get used to it

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