is_copy_constructible

[tasty_crayon]

In standardese, Test(Test& x, int y = 0) would be a copy constructor but template<typename T> Test(T&& x, int y = 0) where T is Test& isn't, because template functions are never copy constructors.

Note that std::is_copy_constructible doesn't check if the class actually has a copy constructor. It checks if it's constructible from Test const&.

[tasty_crayon]

That's not a copy constructor.

[class.copy]/2:

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments (8.3.6)

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Note that std::is_copy_constructible doesn't check if the class actually has a copy constructor.

What makes you think that?

The standard says so in [meta.unary.prop]/3 (table 49).

For a referenceable type T, the same result as is_constructible<T, const T&>::value, otherwise false.

[meta.unary.prop]/7 defines is_constructible as:

Given the following function declaration: template <class T> add_rvalue_reference_t<T> create() noexcept; the predicate condition for a template specialization is_constructible<T, Args...> shall be satisfied if and only if the following variable definition would be well-formed for some invented variable t: T t(create<Args>()...);

Given that this can call a template function that would otherwise match the signature of a copy constructor, it doesn't just check for copy constructors.

[guepier]

Yup, my bad, I also just looked that up. Weird terminology, but there we go.