什么是C ++中的> *运算符？ [英] What is ->* operator in C++?

问题描述

C ++继续让我吃惊。
今天我发现了关于 - > *运算符。它是可重载的，但我不知道如何调用它。我设法在我的类中重载它，但我没有如何调用它的线索。

  struct B {int a; }; 
 
 struct A 
 {
 typedef int（A :: * a_func）（void）; 
 B * p; 
 int a，b，c; 
 A（）{a = 0; } 
 A（int bb）{b = b; c = b; } 
 int operator +（int a）{return 2; } 
 int operator  - > *（a_func a）{return 99; } 
 int operator  - > *（int a）{return 94; } 
 int operator *（int a）{return 2; } 
 B * operator  - > （）{return p; } 
 
 
 int ff（）{return 4; } 
}; 
 
 
 void main（）
 {
 A a; 
 A * p =& a; 
 a + 2; 
} 
  

编辑：

感谢答案。要调用重载函数我写

  void main（）
 {
 A a; 
 A * p =& a; 
 a + 2; 
 a-> a; 
 A :: a_func f =& A :: ff; 
（& a-> * f）（）; 
（a-> * f）; // this 
} 
  

解决方案

code> - > * 运算符是二进制运算符（。不可重载）。它被解释为一个普通的二进制运算符，所以在你原来的情况下，为了调用该运算符，你必须做像

  A a; 
 B * p = a-> * 2; // call A :: operator-> *（int）
  

在Piotr的答案适用于内置运算符，而不是您的超载运算符。在您添加的示例中调用的还是内置运算符，而不是您重载的。为了调用重载的操作符，你必须做我在上面的例子中做的。

C++ continues to surprise me. Today i found out about the ->* operator. It is overloadable but i have no idea how to invoke it. I manage to overload it in my class but i have no clue how to call it.

struct B { int a; };

struct A
{
    typedef int (A::*a_func)(void);
    B *p;
    int a,b,c;
    A() { a=0; }
    A(int bb) { b=b; c=b; }
    int operator + (int a) { return 2; }
    int operator ->* (a_func a) { return 99; }
    int operator ->* (int a) { return 94; }
    int operator * (int a) { return 2; }
    B* operator -> () { return p; }


    int ff() { return 4; }
};


void main()
{
    A a;
    A*p = &a;
    a + 2;
}

edit:

Thanks to the answer. To call the overloaded function i write

void main()
{
    A a;
    A*p = &a;
    a + 2;
    a->a;
    A::a_func f = &A::ff;
    (&a->*f)();
    (a->*f); //this
}

解决方案

The overloaded ->* operator is a binary operator (while .* is not overloadable). It is interpreted as an ordinary binary operator, so in you original case in order to call that operator you have to do something like

A a;
B* p = a->*2; // calls A::operator->*(int)

What you read in the Piotr's answer applies to the built-in operators, not to your overloaded one. What you call in your added example is also the built-in operator, not your overloaded one. In order to call the overloaded operator you have to do what I do in my example above.

这篇关于什么是C ++中的> *运算符？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！