Move ctor 没有被调用

Question

下面是我的代码

#include<iostream>
#include<string.h>
using namespace std;

class mystring {
    private:
        char * ptr;
    public:
        mystring() {
            ptr = new char[1];
            ptr[0] = '[=10=]';
        }
        mystring(const char * obj) {
            cout<<"param called "<<endl;
            int len = strlen(obj);
            ptr = new char[len+1];
            strcpy(ptr,obj);
        }
        mystring(mystring& obj) {
            cout<<"copy called "<<endl;
            ptr = new char[strlen(obj.ptr)+1];
            strcpy(ptr, obj.ptr);
        }
        mystring(mystring&& obj) noexcept{
            cout<<"shallow copy created "<<endl;
            ptr = obj.ptr;
            obj.ptr = NULL;
        }
        friend ostream& operator<<(ostream& out, mystring& obj) {
            out<<obj.ptr<<endl;
            return out;
        }
};

int main() {
    mystring s1 = move("Hello World");
    mystring s2 = s1;
    return 0;
}

当我说 mystring s1 = move("Hello World"); 时，我的理解是应该调用 move ctor，但由于某种原因，上述代码的输出是 param called copy called。我不确定为什么在尝试进行浅拷贝时会为此调用 param ctor。有人可以帮我理解输出吗？谢谢！

Answer 1

在行

mystring s1 = move("Hello World");

您正在 move 字符串文字 "Hellow Word"，而不是 mystring.

类型的对象

int main() {
    mystring s1 = move("Hello World"); // moving a c-string, not very useful
    mystring s2 = s1; // calls copy constructor
    mystring s3 = std::move(s1); // new line, calls move constructor
    return 0;
}

将打印

param called 
copy called 
shallow copy created

https://godbolt.org/z/cxh7qrzcd

最后，请注意“浅拷贝”根本不能很好地描述移动的目的。

Move ctor 没有被调用

Move ctor is not getting invoked

c++

oop

move-constructor

move-semantics