Is there a typical use case that must use `is_detected_v`?

Question

Below is excerpted from cppref:

#include <experimental/type_traits>

template<class T>
using copy_assign_t = decltype(std::declval<T&>() = std::declval<const T&>());

struct Meow { };

using namespace std::experimental;

int main()
{
    static_assert(is_detected_v<copy_assign_t, Meow>,
        "Meow should be copy assignable!"); // version 1

    static_assert(is_copy_assignable_v<Meow>,
        "Meow should be copy assignable!"); // version 2
}

Is there any difference between version 1 and version 2?

Is there a typical use case that must use is_detected_v?

Answer 1

std::is_detected can be used as a building block for std::is_copy_assignable. If you want to check for copy-assignability, you should use std::is_copy_assignable. If you need to check for the existence of a custom operation/member function, std::is_detected gives you an easy way of doing that.

template<class T>
using foo_detector = decltype(std::declval<T&>().foo());

static_assert(!is_detected_v<foo_detector, Meow>, "Meow shouldn't have `.foo()`!");

---

A realistic example use case is unification of different APIs:

template<class T>
using clean_detector = decltype(std::declval<T&>().clean());

template<class T>
using clear_detector = decltype(std::declval<T&>().clear());

template <typename T>
auto clear(const T& x) -> std::enable_if_t<is_detected_v<has_clean, T>>
{
    x.clean();
}

template <typename T>
auto clear(const T& x) -> std::enable_if_t<is_detected_v<has_clear, T>>
{
    x.clear();
}

Usage example:

struct Foo { void clean(); };
struct Bar { void clear(); };

int main()
{
    Foo f; Bar b;
    clear(f);
    clear(b);
}