纯虚函数会阻止隐式生成的移动构造函数吗？

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)
根据 cppreference.com（据我所知），test 应该有一个隐式生成的移动构造函数：

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.

为什么编译器不为 test 生成隐式移动构造函数？ test2 为什么要这样做？

Answer 1

std::is_move_constructible 检查类型是否可以从右值参数构造。 test 是抽象的 class 类型。无论参数如何，它都无法构造。