C ++回调函数 [英] C++ Callback function

问题描述

我使用afxwin SetTimer函数设置计时器。这个函数接收三个参数，其中一个是回调函数。

I'm setting a timer using afxwin SetTimer function. This function receives three params one of which is a callback function.

我在一个类中设置了一个回调函数（我们称之为NS），并试图将它作为参数。然而，编译器抱怨这：

I set a callback function in a class (let's call it NS) and tried to pass it as an argument. However the compiler complains that this:

void (__stdcall NS::*) (HWND, UINT, UINT_PTR, DWORD);

与以下不同：

void (__stdcall *) (HWND, UINT, UINT_PTR, DWORD);

我该怎么办？

推荐答案

使函数 static *。否则它甚至不是一个函数，而是一个成员函数，这是一个非常不同的动物。此外，将作为回调在类中传递函数没有其他意义，如下所示：非静态成员函数绑定到对象实例的状态该类。

Make the function static*. Otherwise it's not even a function, but rather a member function, which is a very different animal. Moreover, it doesn't otherwise make sense to "pass the function in a class as a callback", as you put it: A non-static member function is tied to the state of an object instance of that class.

（此外，搜索此网站，此问题已被问过百万次，有很多好的替代品遍布整个SO。）

(Also, search this site, this question has been asked a million times over and there are many good alternatives spread out throughout SO.)

_{*）正如@James指出的，如果回调函数有一个C接口，这是不够的，在这种情况下，你还需要一个免费的 extern C包装函数。}

_{*) As @James points out, this is not sufficient if the callback function has a C interface, in which case you also need a free extern "C" wrapper function.}

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