c++ 移动语义不执行移动

Question

想法是使用移动语义来避免不必要的复制。给定以下代码：

#include <iostream>
#include <string>
#include <utility>



class Address {
private:
    const std::string m_street;
    const std::string m_city;
    const int m_suite;

public:
    Address(const std::string &street, const std::string &city, int suite) : m_street {std::move(street)},
                                                                            m_city {std::move(city)},
                                                                            m_suite {suite}
    {
    }

    friend std::ostream& operator<<(std::ostream &out, const Address &address)
    {
        out << "Address: (street: " << address.m_street << ", city: " << address.m_city << ", suite: " << address.m_suite << ")\n";

        return out;
    }

    Address(const Address &other) = delete;

    Address& operator=(const Address &other) = delete;

    Address(Address &&other) : m_street {std::move(other.m_street)},
                               m_city {std::move(other.m_city)},
                               m_suite {std::move(other.m_suite)}
    {
    }
};


class Contact {
    const std::string m_name;
    const Address m_address;

public:
    Contact(const std::string &name, Address &address) : m_name {std::move(name)},
                                                        m_address {std::move(address)}
    {
    }

    friend std::ostream& operator<<(std::ostream& out, const Contact &contact)
    {
        out << "Contact: " << contact.m_name << "\n" << contact.m_address << "\n";

        return out;
    }
};


int main()
{
    Address address1 {"123 East Dr", "London", 123};
    Contact john {"John Doe", address1};

    std::cout << john;
    std::cout << address1;

    return 0;
}

我得到了：

Contact: John Doe
Address: (street: 123 East Dr, city: London, suite: 123)

Address: (street: 123 East Dr, city: London, suite: 123)

为什么 address1 变量的内容没有移动？打印输出不应该是

Address: (street: , city: , suite: <whatever>)

此外，为什么对主要是代码的 post 有限制？一切都在代码中给出。我对移动语义很感兴趣，所以我创建了 address1 变量来保存一些地址。我使用相同的变量来初始化 Contact 类型的对象（使用移动语义），但是没有执行移动语义并且 address1 变量仍然保持相同的值。

Answer 1

数据成员m_street和m_city声明为const；然后在 Address 的移动构造函数的成员初始化列表中，如 m_street {std::move(other.m_street)}，使用 std::string 的复制构造函数（但不是移动构造函数）。

您可能想要删除 const 限定符，然后您将得到

Contact: John Doe
Address: (street: 123 East Dr, city: London, suite: 123)

Address: (street: , city: , suite: 123)

LIVE

顺便说一句：对于像int这样的内置类型，移动的效果与复制相同。这就是为什么 suite 的值仍然是 123.

BTW2：对于std::string、

的移动构造函数

Move constructor. Constructs the string with the contents of other using move semantics. other is left in valid, but unspecified state.

移动操作后移动的对象不保证被修改为空。

c++ 移动语义不执行移动

c++ move semantics does not perform move

c++

move-semantics

c++11