指定派生类的析构函数时,std :: move不起作用

Question

我试图将一个实例移动到另一个实例,如下所示:

#include <iostream>

class student {
public:
    student() = default;
    student(const student& student) {
        std::cout << "copy student" << std::endl;
    }
    student(student&& student) {
        std::cout << "move student" << std::endl;
    }
    virtual ~student() = default;
};

class undergraduate: public student {
public:
    undergraduate(): student() {}
    ~undergraduate() override = default; // removing this line will make std::move work
};

int main() {
    undergraduate student;
    undergraduate student1 = std::move(student);
}

这是输出:

copy student

正如你所看到的,std::move没有工作,student被复制而不是被移动,但是,如果我删除了undergraduate析构函数,即以下行:

~undergraduate() override = default; // removing this line will make std::move work

输出将变为move student,这意味着std::move有效.为什么？为什么在std::move指定派生类的析构函数时不起作用？

Answer 1

指定类的析构函数会禁止自动生成移动构造函数并移动赋值.您可以使用= default以下命令恢复它们:

class undergraduate: public student {
public:
    undergraduate(): student() {}
    ~undergraduate() override = default;

    undergraduate(const undergraduate&) = default;
    undergraduate& operator=(const undergraduate&) = default;

    undergraduate(undergraduate&&) = default;
    undergraduate& operator=(undergraduate&&) = default;
};

Howard Hinnant为他的"你想要了解的关于移动语义的东西(以及其他一些)"创建了一个很好的表格"演示: