将合格的非静态成员函数作为函数指针传递 [英] Passing a qualified non-static member function as a function pointer

问题描述

我有一个外部库中的函数，我不能更改以下签名：

  void registerResizeCallback ）（int，int））
  

我想传入一个成员函数作为回调，回调需要修改实例变量。

显然，这是不可能的简单：

  registerResizeCallback（& Window :: Resize）; 
  

所以我不知道如何解决这个问题。

解决方案

检查[33.2]如何将指向成员函数的指针传递到信号处理程序，X事件回调，启动线程/任务的系统调用等？ C ++常见问题解答Lite：

不要。

如果没有对象可以直接调用它，则无法直接执行。

<对于现有软件的补丁，使用顶层（非成员）函数作为包装器，它接受通过某种其他技术获得的对象。

I have a function in an external library that I cannot change with the following signature:

void registerResizeCallback(void (*)(int, int))

I want to pass in a member function as the callback, as my callback needs to modify instance variables.

Obviously this isn't possible with a simple:

registerResizeCallback(&Window::Resize);

so I'm not really sure how to solve the problem.

解决方案

Check "[33.2] How do I pass a pointer-to-member-function to a signal handler, X event callback, system call that starts a thread/task, etc?" at the C++ FAQ Lite:

Don't.

Because a member function is meaningless without an object to invoke it on, you can't do this directly

...

As a patch for existing software, use a top-level (non-member) function as a wrapper which takes an object obtained through some other technique.

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