必须调用对非静态成员函数的引用 [英] Reference to non-static member function must be called

问题描述

我使用C ++（不是C ++ 11）。我需要一个指向一个类中的函数的指针。我尝试做以下操作：

  void MyClass :: buttonClickedEvent（int buttonId）{
 //我需要对MyClass类的所有成员的访问
} 
 
 void MyClass :: setEvent（）{
 
 void（* func）（int）; 
 func = buttonClickedEvent; //<  - 对非静态成员函数的引用必须调用
 
} 
 
 setEvent（）; 
  

但是有一个错误：引用非静态成员函数必须调用。我应该如何做一个指向MyClass成员的指针？

解决方案

问题是 buttonClickedEvent 是一个成员函数，你需要一个指向成员的指针才能调用它。

尝试：

  void（MyClass :: * func）（int）; 
 func =& MyClass :: buttonClickedEvent; 
  

然后当你调用它，你需要一个 MyClass 这样做，例如 this ：

 （this-> * func）（< argument>）; 
  

http://www.codeguru.com/cpp/cpp/article.php/c17401/C-Tutorial-PointertoMember-Function.htm p>

I'm using C++ (not C++11). I need to make a pointer to a function inside a class. I try to do following:

void MyClass::buttonClickedEvent( int buttonId ) {
    // I need to have an access to all members of MyClass's class
}

void MyClass::setEvent() {

    void ( *func ) ( int ); 
    func = buttonClickedEvent; // <-- Reference to non static member function must be called

}

setEvent();

But there's an error: "Reference to non static member function must be called". What should I do to make a pointer to a member of MyClass?

解决方案

The problem is that buttonClickedEvent is a member function and you need a pointer to member in order to invoke it.

Try this:

void (MyClass::*func)(int);
func = &MyClass::buttonClickedEvent;

And then when you invoke it, you need an object of type MyClass to do so, for example this:

(this->*func)(<argument>);

http://www.codeguru.com/cpp/cpp/article.php/c17401/C-Tutorial-PointertoMember-Function.htm

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