How is std::is_function implemented?
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?
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 ] [...]
There are only two categories of types in the language that cannot have const-qualification: reference types, and function types. So, if const T
fails to be a const-qualified type, it means T
is either a function type or a reference type. If you can rule out reference types, then you are left with only function types.
Note that a function type that carries a cv-qualifier, such as int(int) const
, is not a const-qualified type. It's an example of an "abominable function type", whose only real use is to compose or decompose pointer-to-member-function types. The type int(int) const
cannot be obtained by adding const-qualification on top of int(int)
. Rather, the const
applies to the implied object parameter.