Question

我是移动构造函数的新手，我从一些网站调查并尝试使用Visual Studio 11 Express Beta ..

以下是我的测试代码......

#include <iostream>

using namespace std;

class Foo
{
public:
    Foo()
      : Memory(nullptr)
    {
        cout<<"Foo Constructor"<<endl;
    }
    ~Foo()
    {
        cout<<"~Foo Destructor"<<endl;
        if(Memory != nullptr)
            delete []Memory;
    }
    Foo(Foo& rhs)
      : Memory(nullptr)
    {
        cout<<"Copy Constructor"<<endl;

        //allocate
        //this->Memory = new ....
        //copy 
        //memcpy(this->Memory, rhs.Memory...);
    }

    Foo& operator=(Foo& rhs)
    {
        cout<<"="<<endl;
    }
    void* Memory;

    Foo(int nBytes) { Memory = new char[nBytes]; }

    Foo(Foo&& rhs)
    {
        cout<<"Foo Move Constructor"<<endl;
        Memory = rhs.Memory;
        rhs.Memory = nullptr;
    }
};

Foo Get()
{
    Foo f;
    return f; 
    //return Foo();
}
void Set(Foo rhs)
{
    Foo obj(rhs);
}
int main()
{
    Set(Get());
    return 0;
}

我不知道为什么它不会进入移动构造函数。

它实际上是来自Get（）;

的Rvalue

如果我从const构造函数修改了非常量复制构造函数，

它将进入移动构造函数。行为改变了......

有人可以解释为什么会这样吗？

Answer 1

#include <iostream>

using namespace std;

class Foo
{
public:
    Foo():
        Memory(nullptr)
    {
        cout<< this << "Foo Constructor"<<endl;
    }

    ~Foo()
    {
        cout<< this << "~Foo Destructor"<<endl;
        if(Memory != nullptr)
            delete []Memory;
    }

    Foo(Foo& rhs)
        :Memory(nullptr)
    {
        cout<<this << "Copy Constructor"<<endl;

        //allocate
        //this->Memory = new ....
        //copy 
        //memcpy(this->Memory, rhs.Memory...);
    }

    Foo& operator=(Foo& rhs)
    {
        cout<<"="<<endl;
    }
    void* Memory;

    Foo(int nBytes) { Memory = new char[nBytes]; }

    Foo(Foo&& rhs)
        {
        cout<<this << "Foo Move Constructor"<<endl;

                 Memory = rhs.Memory;


                 rhs.Memory = nullptr;
        }

};

Foo Get()
{
    Foo f;
    cout << &f << "f" <<endl;
    return f; 
}

void Set(Foo rhs)
{
    Foo obj(rhs);
    cout << &obj << "obj"<<endl;
}

int main()
{
    Set(Get());
    return 0;
}

...输出

0x7fffe38fa0a0 Foo Constructor
0x7fffe38fa0a0 f
0x7fffe38fa070 Copy Constructor
0x7fffe38fa070 obj
0x7fffe38fa070 ~Foo Destructor
0x7fffe38fa0a0 ~Foo Destructor

答案：由于命名返回值优化，参数rhs在本地构建为局部变量f的别名。（也就是说rhs和f是相同的实例）。

由于rhs是左值，复制构造函数用于从rhs复制构造obj。

移动构造函数和非const复制构造函数

1 个答案: