为什么回调函数在类中声明时需要是静态的 [英] Why callback functions needs to be static when declared in class

问题描述

我试图在类中声明一个回调函数，然后在某个地方我读的函数需要是静态的，但它没有解释为什么？

I was trying to declare a callback function in class and then somewhere i read the function needs to be static but It didn't explain why?

#include <iostream>
using std::cout;
using std::endl;

class Test
{
public:
    Test() {}

    void my_func(void (*f)())
    {
        cout << "In My Function" << endl;
        f(); //Invoke callback function
    }

    static void callback_func()
    {cout << "In Callback function" << endl;}
};

int main()
{
    Test Obj;
    Obj.my_func(Obj.callback_func);
}

推荐答案

需要一个类实例被调用。
如果不提供要调用的实例，则无法调用成员函数。这使得有时很难使用。

A member function is a function that need a class instance to be called on. Members function cannot be called without providing the instance to call on to. That makes it harder to use sometimes.

静态函数几乎就像一个全局函数：它不需要一个类实例来调用。所以你只需要得到函数的指针就可以调用它。

A static function is almost like a global function : it don't need a class instance to be called on. So you only need to get the pointer to the function to be able to call it.

看看std :: function（或std :: tr1 :: function或boost :: function，如果你的编译器没有提供它），它在你的情况下是有用的，因为它允许你使用任何可调用（provide（）语法或操作符）作为回调，包括可调用对象和成员函数std :: bind或boost :: bind）。

Take a look to std::function (or std::tr1::function or boost::function if your compiler doesn't provide it yet), it's useful in your case as it allow you to use anything that is callable (providing () syntax or operator ) as callback, including callable objects and member functions (see std::bind or boost::bind for this case).

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