移动语义和原始类型

Question

示例代码:

int main()
{
    std::vector<int> v1{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    std::cout << "Printing v1" << std::endl;
    print(v1);
    std::vector<int> v2(std::make_move_iterator(v1.begin()),
                         std::make_move_iterator(v1.end()));
    std::cout << "Printing v1" << std::endl;
    print(v1);
    std::cout << "Printing v2" << std::endl;
    print(v2);

    std::vector<std::string> v3{"some", "stuff", "to",
                        "put", "in", "the", "strings"};
    std::cout << "Printing v3" << std::endl;
    print(v3);
    std::vector<std::string> v4(std::make_move_iterator(v3.begin()),
                                 std::make_move_iterator(v3.end()));
    std::cout << "Printing v3" << std::endl;
    print(v3);
    std::cout << "Printing v4" << std::endl;
    print(v4);
}

输出:

Printing v1
1 2 3 4 5 6 7 8 9 10
Printing v1
1 2 3 4 5 6 7 8 9 10
Printing v2
1 2 3 4 5 6 7 8 9 10
Printing v3
some stuff to put in the strings
Printing v3

Printing v4
some stuff to put in the strings

问题

1. 由于对原始类型的移动操作只是一个副本,我可以假设v1它将保持不变或者即使使用原始类型也未指定状态吗？

2. 我假设原始类型没有移动语义的原因是因为复制速度一样快,甚至更快,这是正确的吗？

Answer 1

1. 不,如果你想能够假设你应该复制,而不是移动.

2. 移动使源对象处于有效但未指定的状态.原始类型确实表现出移动语义.源对象保持不变的事实表明复制是实现移动的最快方式.