我用这个简单的方法调用转发类重新发明轮子？ [英] Am I reinventing the wheel with this trivial method call forwarding class?

问题描述

我刚刚发现自己创建了一个类

I just found myself creating a class

template <typename T> struct invoker {
  void operator()(T& it) const {it();}
};

，所以我可以传递 invoker< foo> 到不同的<$ c>重复调用 invoker< foo> :: operator（）（foo&） $ c> foo 的实例，让它转发这些调用到 foo 的 foo :: operator （）方法。

so I could pass an invoker<foo> to something (which isn't under my control) which wants to call invoker<foo>::operator()(foo&) on it repeatedly with different foo instances, to get it to forward those calls to the foo's foo::operator()() method.

我知道它只有几行，但这似乎是类似的东西，可能已经提供了STL功能，或 boost :: bind 不知何故。除非我看不到把戏，如果有一个。 （我相信我不是第一个使用这样的东西的人，它有一个名字吗？）

I know it's only a few lines, but this seems like the sort of thing which is probably already provided for by STL's functional, or boost::bind somehow. Except I can't see the trick, if there is one. (I'm sure I'm not the first person to use something very like this; does it have a name ?)

推荐答案

是的，你是重塑的轮子。 std :: mem_fun_ref做你想要的。

Yeah, you're reinventing the wheel. std::mem_fun_ref does what you want.

std::vector<foo> foos;

...

std::for_each(foos.begin(), foos.end(), std::mem_fun_ref(&foo::operator()));

或者：

std::vector<foo*> foos;

...

std::for_each(foos.begin(), foos.end(), std::mem_fun(&foo::operator()));

不用担心你的参数是ptr还是不是boost :: mem_fn 。

Not having to mess with whether your param is ptr or not is one great benefit of boost::mem_fn.

任何比这更复杂的事情，你开始遇到C ++ 03 binder的麻烦，需要一些更具表现力的东西，如boost.bind。

Anything much more complex than that though and you begin running into trouble with the C++03 binders and need something more expressive like boost.bind.

这篇关于我用这个简单的方法调用转发类重新发明轮子？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！