问题描述
考虑C ++ 11中的以下功能:
Consider the following function in C++11:
template<class Function, class... Args, typename ReturnType = /*SOMETHING*/>
inline ReturnType apply(Function&& f, const Args&... args);
我希望 ReturnType
等于结果类型 f(args ...)
我必须写些什么,而不是 / * SOMETHING * /
?
I want ReturnType
to be equal to the result type of f(args...)
What do I have to write instead of /*SOMETHING*/
?
推荐答案
我认为您应该使用 trailing-return-type 重写函数模板
I think you should rewrite your function template using trailing-return-type as:
template<class Function, class... Args>
inline auto apply(Function&& f, const Args&... args) -> decltype(f(args...))
{
typedef decltype(f(args...)) ReturnType;
//your code; you can use the above typedef.
}
请注意,如果您通过 args
为 Args&& ...
而不是 const Args&....
,那么它就是最好将 f
中的 std :: forward
用作:
Note that if you pass args
as Args&&...
instead of const Args&....
, then it is better to use std::forward
in f
as:
decltype(f(std::forward<Args>(args)...))
当您使用 const Args& ...
时,然后使用 std :: forward
When you use const Args&...
, then std::forward
doesn't make much sense (at least to me).
最好将 args
传递为 Args& ..
称为 universal-reference ,并使用 std :: forward
。
It is better to pass args
as Args&&..
called universal-reference and use std::forward
with it.
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