Double move on same object is copying from left to right?

Question

I am just beginner in move operation in c++11, so playing with it. But found something which i am not able to understand.

#include <iostream>
using namespace std;

class A{
    public:
        A(){cout << "default ctor" << endl;}
        A(const string& str):_str{str}{cout << "parameter ctor" << endl;}
        A(A&& obj):_str{std::move(obj._str)}{cout << "move ctor" << endl;}
        A& operator =(A&& rhs){_str = std::move(rhs._str);cout << "move assignment operation" << endl; return *this;}
        void print(){cout << _str << endl;}
    private:
        string _str;
};

int main(){
    A a("rupesh yadav"); // parameter ctor
    A b(std::move(a));   // move ctor

    cout << "print a: ";
    a.print();           // NOT printing  --> CORRECT!!
    cout << "print b: ";
    b.print();           // printing      --> CORRECT!!

    b = std::move(a);    // i don't know may be silly but still lets do it WHY NOT!!!, could be just mistake??

    cout << "print a: "; 
    a.print();           // printing      --> WRONG!! 
    cout << "print b: "; 
    b.print();           // NOT printing  --> WRONG!!
}

I was expecting that b = std::move(a) operation would behave something different because i am applying move on object a second time but it is copying left side object b to right hand side object a, this part i don't understand.

Or i have done something wrong in programming. Please help if i am doing something wrong in move operation.

EDIT: I know this is undefined behavior. My doubt is if i will do it again then it is copying from object a to object b, and if again i will do the same thing then will copy object b to object a?

Hence it is copying form left to right and right to left why?

Answer 1

You can't move from the same object twice.

After you first moved a into b, a had a "valid but unspecified state" (can't remember the exact terminology). Then, you attempted to move a into b again! Now all heck has broken loose. (I suspect that, internally, the data pointers have just been swapped around.)

Simply don't do this. I see no reason to want to.