Is There a declval for Function Pointers?

问题

I have a function and I need to test whether I can pass an argument of a given type to it. For example:

template<typename T, auto F>
decltype(F(declval<T>{})) foo();

Calling foo<int, bar>() does 2 things:

1. Sets the return type of foo would have the same return type as bar
2. Ensures that bar is a function that accepts an argument of type T

Unfortunately I don't have access to auto template types, but I still want to accomplish both of these. What I need is a decltype for function pointers, which would allow me to do something like this:

template <typename T, typename F>
decltype(declval<F>(declval<T>{})) foo();

So I could still call foo<int, bar>() and get the same result. Of course there isn't a declval for function pointers. But is there another way I could accomplish this?

回答1:

Of course there isn't a declval for function pointers.

What do you mean? std::declval works perfectly with function pointer types:

template<typename F, typename... Args>
using call_t = decltype(std::declval<F>()(std::declval<Args>()...));

In this example, F can be a function pointer type, a lambda type or any callable types.

Here's an example of usage:

template<typename T, typename F>
auto foo() -> call_t<F, T>;

Another example using the detection idiom (implementable in C++11):

template<typename F, typename... Args>
using is_callable = is_detected<call_t, F, Args...>;

static_assert(is_callable<void(*)(int), int>::value, "callable")

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Note that all this can be replaced by std::invoke_result_t and std::is_invocable in C++17. I'd suggest mimicking those to have the most seamless upgrade.

来源：https://stackoverflow.com/questions/54947367/is-there-a-declval-for-function-pointers