Why memory is new when doing std::move

In this code:

int main()
{
    std::vector<int> src{1, 2, 3};

    std::cout << "src: ";
    for (std::vector<int>::const_iterator x = src.begin(); x != src.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;
    }   

    std::vector<int> dest(std::move(src));

    std::cout << "src: ";                                                       
    for (std::vector<int>::const_iterator x = src.begin(); x != src.end(); ++ x)
    {
         std::cout << *x << ' ' << &(*x) << std::endl ;;
    }   
    std::cout << "\ndest: ";
    for (std::vector<int>::const_iterator x = dest.begin(); x != dest.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;;
    }   
    std::cout << '\n';
}

src: 1 0x43ea7d0
2 0x43ea7d4
3 0x43ea7d8
src: 
dest: 1 0x43ea7d0
2 0x43ea7d4
3 0x43ea7d8

This make sense because the memory address of dest is now what previously src was.

But when I do:

int main()
{
    std::vector<int> src{1, 2, 3};
    std::vector<int> dest(src.size());

    std::cout << "src: " << std::endl;
    for (std::vector<int>::const_iterator x = src.begin(); x != src.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;
    }   
    std::cout << "\ndest: " << std::endl;
    for (std::vector<int>::const_iterator x = dest.begin(); x != dest.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;
    }    
    std::cout << '\n';

    std::move_backward(src.begin() , src.end(), dest.end());

    std::cout << "src: " << std::endl;
    for (std::vector<int>::const_iterator x = src.begin(); x != src.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;
    }   
    std::cout << "\ndest: " << std::endl;
    for (std::vector<int>::const_iterator x = dest.begin(); x != dest.end(); ++ x)
    {
        std::cout << *x << ' ' << &(*x) << std::endl ;
    }    
    std::cout << '\n';
}

   src: 
1 0x41e0140
2 0x41e0144
3 0x41e0148

dest: 
0 0x41e0160
0 0x41e0164
0 0x41e0168

src: 
1 0x41e0140
2 0x41e0144
3 0x41e0148

dest: 
1 0x41e0160
2 0x41e0164
3 0x41e0168

Why is the second case the address being different? I thought std::move only changing the pointer, without touching the memory of the original object