Why userdefined move constructor is not called in below example?

Question

I am expecting that after using std::move(any_object) we are converting that object in rvalue reference and due to this it should call my userdefined move constructor. but it is not working like this.Could anyone tell whats happening?

#include <iostream>
#include <memory>

struct Demo {
    int Value;
    Demo():Value(10) { std::cout << "Demo's default constructor called\n";}
    Demo(Demo& tempObj) { std::cout << "Demo non-const copy constructor called\n";}
    Demo(const Demo& tempObj) { std::cout << "Demo const copy constructor called\n"; }
    Demo(Demo&& tempObj) { std::cout << "Demo move constructor called\n"; }
    Demo(const Demo&& tempObj) { std::cout << "Demo const move constructor called\n"; }
    Demo& operator= (const Demo& tempObj){ std::cout << "Demo copy assignment operator called ";}
    Demo& operator= (Demo&& tempObj){ std::cout << "Demo move assignment operator called";}
    ~Demo(){ std::cout << "Demo's destructor called\n";}
};

void fun(Demo&& tempObj)
{
    tempObj.Value = 20;
}

int main()
{
    Demo demo;
    fun(std::move(demo));
}

//output:

Demo's default constructor called
Demo's destructor called

Answer 1

Why userdefined move constructor is not called in below example?

Because you're only binding an rvalue reference to an xvalue and not actually constructing an instance of type Demo . That is, a move constructor will be called when you actually construct an instance of type Demo using a rvalue as shown below for example:

//-----------v-------->removed && from here
void fun(Demo tempObj)
{
    tempObj.Value = 20;
}
fun(std::move(demo)); //now this uses move constructor

---

Additionally, you missing return statement in your assignement operators.