Why is the move constructor neither declared nor deleted with clang?

Question

Consider the following classes.

struct with_copy {
    with_copy() = default;
    with_copy(with_copy const&) {}
    with_copy& operator=(with_copy c         


        

Answer 1

I'm not quite sure what you tested but it foo is surely both move assignable and move constructible. Admittedly, this doesn't say anything about a move constructor or a move assignment being accessible, just that construction or assignment from an rvalue works. Both clang (clang version 3.5 (trunk 196718)) and gcc (gcc version 4.9.0 20131031 (experimental) (GCC)) agree with this assessment. This is the complete source I tried:

#include <iostream>
#include <type_traits>
#include <memory>

struct with_copy {
    with_copy() = default;
    with_copy(with_copy const&) {}
    with_copy& operator=(with_copy const&) { return *this; }
};

struct foo {
    with_copy c;
    std::unique_ptr<int> p;
};

int main()
{
    std::cout << "move constructible: "
              << std::is_move_constructible<foo>::value << '\n';
    std::cout << "move assignable: "
              << std::is_move_assignable<foo>::value << '\n';
    foo f0;
    foo f1 = std::move(f0);
    f0 = std::move(f1);
}

Answer 2

C++11, or rather n3485, [class.copy]/9:

If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

• X does not have a user-declared copy constructor,
• X does not have a user-declared copy assignment operator,
• X does not have a user-declared move assignment operator,
• X does not have a user-declared destructor, and
• the move constructor would not be implicitly defined as deleted.

and /11:

An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/ move constructor for a class X is defined as deleted (8.4.3) if X has:

• [...]
• for the copy constructor, a non-static data member of rvalue reference type, or
• for the move constructor, a non-static data member or direct or virtual base class with a type that does not have a move constructor and is not trivially copyable.

As with_copy is not trivially copyable, foo will have no move-constructor (it would be defined as deleted, therefore it won't be implicitly declared).

---

C++1y, or rather github repo commit e31867c0 from 2013-11-12; incorporating DR1402:

/9:

If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

• X does not have a user-declared copy constructor,
• X does not have a user-declared copy assignment operator,
• X does not have a user-declared move assignment operator, and
• X does not have a user-declared destructor.

and /11:

An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/ move constructor for a class X is defined as deleted (8.4.3) if X has:

• [...]
• for the copy constructor, a non-static data member of rvalue reference type.

A defaulted move constructor that is defined as deleted is ignored by overload resolution (13.3, 13.4).

Here, foo will have a move-constructor.