Question

我想实现一个“take”方法。 “take”方法类似于get方法，但是从其所有者窃取getted对象：所有者将该对象保留为空状态。当然，这涉及C ++ 11移动语义。但是以下哪一项是实施它的最佳方式？

class B
{
public:

    A take_a_1()
    {
        return std::move(m_a);
    }

    // this can be compiled but it looks weird to me that std::move 
    // does not complain that I'm passing a constant object
    A take_a_2() const
    {
        return std::move(m_a);
    }

    A&& take_a_3()
    {
        return std::move(m_a);
    }

    // this can't be compiled (which is perfectly fine)
    //
    // A&& take_a_4() const
    // {
    //     return std::move(m_a);
    // }

private:

    A m_a;
};

Answer 1

他们都不是。我会用这个：

Username for 'https://github.com': jgoldstick
Password for 'https://jgoldstick@github.com': 
Counting objects: 29, done.
Delta compression using up to 2 threads.
Compressing objects: 100% (14/14), done.
Writing objects: 100% (16/16), 2.61 KiB | 0 bytes/s, done.
Total 16 (delta 12), reused 0 (delta 0)
To https://github.com/jgoldstick/baseball.git
   7be5c2d..d324acb  master -> master
fatal: Unable to create '/home/jcg/code/python/venvs/baseball/.git/refs/remotes/origin/master.lock': Permission denied
Unexpected end of command stream
(baseball)jcg@jcg:~/code/python/venvs/baseball$

这样你只能从struct B { A && get() && { return std::move(m_a); } };：

的右值“取”

使用std :: move进行“take”方法实现

1 个答案: