如何实现 std: : is_function？

下面是如何实现 std::is_function的？

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

(来自 参考资料)

在我看来，int应该是这个定义下的一个函数。我遗漏了什么？

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最佳答案

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 ] [...]

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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.


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