Function using alias

Question

What exactly does the following code declare;

using f1 = void(int);

I know that the following;

using f2 = void(*)(int);
using f3 = void(&)(int);

f2 is a pointer to a function and f3 would be the reference.

Answer 1

What is it?

It's a function type. When you declare a function, such as:

void func(int);

its type is not a pointer nor a reference. The above function's type is void(int).

We can "prove" it by using type traits as follows:

void func(int) {}

int main() {
    std::cout << std::is_same<decltype(func), void(int)>::value << '\n';
    std::cout << std::is_same<decltype(func), void(*)(int)>::value << '\n';
    std::cout << std::is_same<decltype(func), void(&)(int)>::value << '\n';
}

Live demo

The above code will return true only for the first row.

Is it the same as a pointer or a reference?

No, but a function lvalue can be implicitly converted to a function pointer as per:

§4.3/1 Function-to-pointer conversion [conv.func]

An lvalue of function type T can be converted to a prvalue of type “pointer to T.” The result is a pointer to the function.

The relationship between a function type A(Args...) and its reference (namely A(&)(Args...)) is basically the same as the relationship between any type T and its reference (namely T&).

Where's it used?

It's often used as a template parameter.

For example std::function takes a function type to be stored inside the std::function object and you can declare such an object with:

std::function<void(int)> fn;