移动构造函数没有被调用，而是复制构造函数被调用

Question

我正在尝试调用移动构造函数，但正在调用复制构造函数。 我做错了什么？

#include <iostream>
#include <string.h>

class X
{
    char* name;
public:
    X(const char* p)
    {
        name = new char[strlen(p) + 1];
        strcpy(name, p);
        std::cout<<"Constructor: "<<name<<"; Address: "<<this<<std::endl;
    }

    ~X()
    {
        std::cout<<"Destructor: "<<name<<"; Address: "<<this<<std::endl;
        delete [] name;
    }

    X(const X& a)
    {
        name = new char[strlen(a.name) + 1];
        strcpy(name, a.name);
        std::cout<<"Copy Constructor: "<<name<<"; Address: "<<this<<"; made from: "<<&a<<std::endl;
    }

    X& operator=(const X& a)
    {
        if (name)
        {
            delete [] name;
        }

        name = new char[strlen(a.name) + 1];
        strcpy(name, a.name);
        std::cout<<"Assignment operator: "<<name<<"; Address: "<<this<<"; made from: "<<&a<<std::endl;

        return *this;
    }

    X operator+(const X& a)
    {
        std::string s1 = name;
        std::string s2 = a.name;
        s1.append(" ");
        s1.append(s2);

        X x(s1.c_str());

        std::cout<<"operator+: "<<s1.c_str()<<"; Address: "<<&x<<"; Created from"
            <<this<<" and "<<&a<<std::endl;

        return x;
    }

    // move copy constructor
    X(X&& a)
    {
        name = a.name;
        a.name = nullptr;
        std::cout<<"Move Copy Constructor: "<<name<<"; Address: "<<this<<std::endl;
    }

    friend X fun(const X& a);
};

X fun(const X& a)
{
    std::cout<<"Inside fun()"<<std::endl;
    X p = a;
    return p;
}

int main()
{
    X h("Harry");
    X r("Ron");

    std::cout<<"Trying to create a temporary object"<<std::endl;
    X a = fun(h + r);
    std::cout<<"Check above if a temporay object was created"<<std::endl;

    return 0;
}

上述程序的O/P：

Constructor: Harry; Address: 0x79315dbc31b0
Constructor: Ron; Address: 0x79315dbc31c0
Trying to create a temporary object
Constructor: Harry Ron; Address: 0x79315dbc31e0
operator+: Harry Ron; Address: 0x79315dbc31e0; Created from0x79315dbc31b0 and 0x79315dbc31c0
Inside fun()
Copy Constructor: Harry Ron; Address: 0x79315dbc31d0; made from: 0x79315dbc31e0
Destructor: Harry Ron; Address: 0x79315dbc31e0
Check above if a temporay object was created
Destructor: Harry Ron; Address: 0x79315dbc31d0
Destructor: Ron; Address: 0x79315dbc31c0
Destructor: Harry; Address: 0x79315dbc31b0

这是否意味着程序无法创建临时对象？ 我以为地址为 0x79315dbc31e0 的对象是临时对象。

注意：由于我的 gcc 很旧，我在支持 C++11 和 C++14 的在线 C++ 编译器之一上尝试了此代码。

Answer 1

在“Inside fun()”之后对“复制构造函数”的调用实际上是在fun() 中的以下行中对复制构造函数的调用：

X p = a;

这是对复制构造函数的调用。您显然希望调用移动构造函数，因为：

X a = fun(h + r);

这个临时的被完全“忽略”掉了。整个序列：

X p = a;
return p;

编译器实际上将 p 构造为函数的返回值，因此“return p”什么也不做。

然后，因为函数的返回值用于在 main() 中实例化 a 对象，所以调用者本质上将 main() 的“a”传递给 fun()，作为要构造的对象fun().

您可以自己查看这些事件的顺序，只需使用调试器单步执行您的代码。