Change of address after re-allocation

Question

In the code below, whenever a new member is inserted to the std::vector<int>A, due to memory re-allocations, std::vector<reference_wrapper<int>>B points to a wrong address. Is it possible to make the referencing vector to track the re-allocations and store the correct address always?

#include <iostream>
#include <functional>
#include <vector>
using namespace std;

int main()
{
    vector<int> A (0,3);
    vector<reference_wrapper<int>> B;
    B.push_back ( ref(A[0]) );
    B.push_back ( ref(A[1]) );
    B.push_back ( ref(A[2]) );

    A.push_back (0);

    cout << &A[0] << endl;
    cout << &B[0].get() << endl;

    return 0;
}

what is the usecase for this? can't you just save the iterator instead of reference? — cerkiewny, Sep 01 '14 at 10:17
@cerkiewny: How would that help? Iterators are generally invalidated at least as often as references. — Benjamin Lindley, Sep 01 '14 at 10:18
ok guys yes you are right, sorry my bad. the iterators will be invalidated as well. — cerkiewny, Sep 01 '14 at 10:21
Do you really need the address? Can't you just store the index? — molbdnilo, Sep 01 '14 at 10:27
You could use `list` or `deque` if you don't want the elements to move. — Mike Seymour, Sep 01 '14 at 10:41

Vlad from Moscow · Answer 1 · 2014-09-01T10:25:04.623

The only way I see is to reserve enough memory for the vector that it would not be reallocated when a new element is pushed back provided that you do not remove or insert elements in the middle of the vector.

Take into account that this statement

vector<int> A (0,3);

is wrong.

There has to be

vector<int> A (3,0);

or simply

vector<int> A (3);

You could write

#include <iostream>
#include <functional>
#include <vector>
using namespace std;

int main()
{
    vector<int> A( 3 );
    A.reserve( 4 );

    vector<reference_wrapper<int>> B;

    B.push_back ( ref(A[0]) );
    B.push_back ( ref(A[1]) );
    B.push_back ( ref(A[2]) );

    A.push_back (0);

    cout << &A[0] << endl;
    cout << &B[0].get() << endl;

    return 0;
}

The output can look like

0x9cbd018
0x9cbd018

score 0 · Answer 2 · answered Sep 01 '14 at 10:58

You didn't specify the actual usage for this pattern. So here is some possible solutions:

Store index rather than reference/iterator.
Change A to a list or deque which does not invalidate iterators other than the ones actually changed.
Dynamically allocated all the integers (e.g. std::shared_ptr<int>) and let A and B both store the (shared) pointer.

Change of address after re-allocation

2 Answers2