Misunderstanding with move constructor

Question

I have such class, where I create a move constructor

class Test
{
private:
    int m_a;

public:
    Test(int val) { m_a = val; }

    Test (const Test &) {}
    
    // move constructor
    Test (Test && d)
    {
        std::cout << &m_a << std::endl;   // Line X
        std::cout << &d.m_a << std::endl;
    }

    void print()
    {
        std::cout << m_a << std::endl;
    }
};

I also create a function to test move constructor

void fun(Test a)
{ return ; }

Than in main function I create 2 objects of class above and call function to test move constructor

int main()
{
    Test a {50};
    Test b {100};

    fun(a);
    fun(std::move(a));

    fun(b);
    fun(std::move(b));

    return 0;
}

When I looked at the output, I was surprised, because the address of o m_a variable in line X has same address for both objects.

0x7ffc40d37bb4   // look at this
0x7ffc40d37bac
0x7ffc40d37bb4   // look at this
0x7ffc40d37bb0

How is it possible ? It's not static member, what is going on ? Compiler optimization or what ?

Answer 1

How is it possible ?

The parameters of the separate invocations of fun don't have overlapping storage lifetimes. Hence, they can use the same storage.