Pointer to class data member

Question

In this little test program, can someone explain me:

Why is every addresses printed in both loops the same?

Why is it different from one loop to another?

struct A {
    A(){ std::cout << &v << "\n"; };
    int v;
};

int main()
{
    std::vector<A> vec;

    int i = 10;
    while (i--)
        vec.push_back(A());

    for (A b : vec)
        std::cout << &(b.v) << "\n";

    while (true);
    return 0;
}

I actually expected to see ten different addresses repeated 2 times

Because `b` is a *copy* of the vector element, not the vector element itself. — Kerrek SB, Nov 19 '16 at 11:48
Try `for (A& b : vec)` to see the address within the `vector`. — πάντα ῥεῖ, Nov 19 '16 at 11:49

score 1 · Answer 1 · answered Nov 19 '16 at 11:49

In this range based for loop :

    for (A b : vec)
        std::cout << &(b.v) << "\n";

b is a copy of the element in the vector. Thats why you see 10 times the same adress. It is different for two different loops, because that copy happens to be in a different memory location. You probably wanted this:

    for (A& b : vec)
        std::cout << &(b.v) << "\n";

here b is a reference to the element in the vector.

score 1 · Accepted Answer · edited May 23 '17 at 10:32

If you want to see the same addresses you need to look at the addresses inside the vector and not the addresses of temporary copies.

With vector::emplace_back, you can see the addresses inside the vector during the construction (see emplace_back). A reference access gives you the address inside the vector during the iterations.

int main()
{
    std::vector<A> vec;

    int i = 10;
    while (i--)
        vec.emplace_back();

    for (const A& b : vec)
        std::cout << &(b.v) << "\n";
    ...
}

Pointer to class data member

2 Answers2