do pure virtual functions prevent implicitly generated move constructors?

Question

#include <type_traits>

struct test {
    virtual void foo() noexcept = 0;
};

struct test2 : test {
    void foo() noexcept override final {}   
};

// fails
static_assert(std::is_move_constructible<test>::value, "test not move constructible");
// succeeds
static_assert(std::is_move_constructible<test2>::value, "test2 not move constructible");

(Live)
According to cppreference.com (and as far as I understand), test should have an implicitly-generated move constructor:

The implicitly-declared or defaulted move constructor for class T is defined as deleted in any of the following is true:

T [= test] has non-static data members that cannot be moved (have deleted, inaccessible, or ambiguous move constructors)

T has direct or virtual base class that cannot be moved (has deleted, inaccessible, or ambiguous move constructors)

T has direct or virtual base class with a deleted or inaccessible destructor

T is a union and has a variant member with non-trivial copy constructor

(until C++14) T has a non-static data member or a direct or virtual base without a move constructor that is not trivially copyable.

Why does the compiler not generate an implicit move constructor for test? Any why does it do so for test2?

Barry Barry · Accepted Answer · 2015-04-22T21:57:28

std::is_move_constructible checks whether the type can be constructed from an rvalue argument. test is an abstract class type. It cannot be constructed at all, regardless of argument.

do pure virtual functions prevent implicitly generated move constructors?

1 Answers