c++移动构造函数什么时候会被调用?

Question

class x{
public:
    x( moveClass&& item) {
        std::cout << 123 << std::endl;
    }
    
    x() = default;
};


int main(){
    x(x());
}

如上的代码, 不应该输出123吗, 为什么什么都没有输出... 这种情况到底算调用了哪种构造方法.

Answer 1

Your code should not output 123. That's a function declaration, it doesn't call x(moveClass&& item). My understanding is that the move constructor is used on rvalues ​​to reduce calls to temporary constructors and destructors. When you use rvalues ​​to call the constructor, the move constructor can be called.

#include <iostream>

using namespace std;

class x
{
public:
    x(int&& val)
    {
        cout << 123 << endl;
    }
    x() = default;
};

int main()
{
    x(567);

    return 0;
}

In the above code, because I am passing an rvalue in, the constructor will be called. http://stackoverflow.com/ques... This discusses when to use the move constructor.
Title title! ! ! A major discovery. I checked the information and then did my own experiments and found that x(x()) is a function declaration! If you declare another x(x()) below your x a, an error should be reported when compiling and running; the following is my test code, you can also try it, so your line does not call the move constructor!

#include <iostream>
#include <utility>

using namespace std;

class x
{
public:
    x(int&& val)
    {
        cout << "sdfs";
    }
    x()=default;
};

int func()
{
    return 123;
}
class AA
{
public:
    AA(int&&){cout << "no" << endl;}
};

//x(func())
//{
//    cout << "function called" << endl;
//}
//AA(func());
AA(funcall())
{
    cout << "here\n";
}

//x(x());
//AA(x());

int main()
{
    
    //(x)func();
    x(123);
    x(func());
    //AA(123);
    //funcall();
    return 0;
}

Answer 2

#include <iostream>
#include <utility>

class X {
public:
    X( X&& item) {
        std::cout << 123 << std::endl;
    }

    X() = default;
};


int main() {
    X(X()); // X(std::move(X())); 这个会输出 123
}

copy assignment operator should be written like this:

class X {
    X operator=( const X &rhs ) {
        // ...
    }
}

Answer 3

Try x{x{}}