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Home/ Questions/Q 259557
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T22:16:26+00:00 2026-05-11T22:16:26+00:00

In the code below, why it is that when I take the address of

  • 0

In the code below, why it is that when I take the address of a map index (which contains a list) and I take the address of the list itself, they both have different values.

See the code below for clarification.

#include <iostream>
#include <list>
#include <map>

using namespace std;

int main()
{
    list<char> listA;   //list of chars

    map<int,list<char> > mapper;    //int to char map

    mapper[1] = listA;

    cout << &(mapper[1]) << endl;
    cout << &listA << endl;
}
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  1. Editorial Team
    Editorial Team
    2026-05-11T22:16:26+00:00Added an answer on May 11, 2026 at 10:16 pm

    You get different addresses because you create a copy of the original list and assing it to the map structure.

    Consider using pointers (map< int, list<char>* >).

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