Why can you implement std::is_function in terms of std::is_const and std::is_reference?

Question

How is the following an implementation for std::is_function?

template<class T>
struct is_function : std::integral_constant<
    bool,
    !std::is_const<const T>::value && !std::is_reference<T>::value
> {};

(from CPP Reference)

Seems to me, an int would be a function under this definition. What am I missing?

Answer 1

Let's go over the conditions as they appear:
If const T isn't const (const doesn't really apply to function types since functions aren't objects), and T isn't a reference (const doesn't apply to references either for the same reason), it's a function type. int (or any other non-function-non-reference type) wouldn't fit in because is_const<const int>::value is true.

According to C++17 Standard §11.3.5 Functions / section 7: (Emphasis mine)

The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type. In the latter case, the cv-qualifiers are ignored. [ Note: A function type that has a cv-qualifier-seq is not a cv-qualified type; there are no cv-qualified function types. — end note ] [...]