如何像未定义的函数一样传递一个未定义的方法

问题描述

鉴于我传递了未定义的函数：

  void foo（char，short） ; 
  

我我收到错误：

我如何像函数一样传递这种方法？

解决方案

如果您只想使用 decltype ，您只需要额外的一个& ：

  decltype（m（& bar :: foo））
  pre> 
Given that I was passing the undefined function:
void foo(char, short);
I learned how to obtain the type tuple of the arguments by calling decltype(m(foo)) with this function:
template <typename Ret, typename... Args>
tuple<Args...> m(Ret(Args...));
I would now like to pass an undefined method:
struct bar { void foo(char, short); };
I had tried rewriting m like:
template <typename Ret, typename C, typename... Args>
tuple<Args...> m(Ret(C::*)(Args...));
But when I try to call this similarly with decltype(m(bar::foo)) I get the error:
How can I pass this method like I did for the function?
 解决方案 
If you only want to use decltype on it, you simply need an extra &:
decltype(m(&bar::foo))
                        
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