std::result_of doesn't compile

Question

I'm trying to use std::result_of to determine the return type of a callable object:

template <typename T>
std::result_of<T()>::type CallableWrapper(T callableObj) {
    return callableObj();
}

Somewhere else in the code:

auto i = CallableWrapper([](){return 1;});

This code doesn't compile for some reason. I will appreciate if someone would tell me why.

you are missing a `typename` keyword in front of `std::result_of` — Piotr Skotnicki, Aug 02 '15 at 16:32
@PiotrSkotnicki Thanks! This works perfectly. Why is the extra `typename` required? — Mr. Anderson, Aug 02 '15 at 16:33

score 1 · Answer 1 · edited Aug 02 '15 at 16:32

1

It should be possible with trailing return type and decltype, like

template<typename T>
auto CallableWrapper(T callableObj) -> decltype(std::declval<T>()())
{
    ...
}

edited Aug 02 '15 at 16:32

Quentin

answered Aug 02 '15 at 16:30

Thanks! I know this is possible, I was just wondering what I'm missing in the code above. – Mr. Anderson Aug 02 '15 at 16:31

1 Answers1