why the objects in the list giving a different address

[pradeeppani]

#include<iostream>
#include<list>
using namespace std;
class a
{
private:
    int x;
public:
    void setx(int xx)
    {
        x=xx;
    }
    void showx()
    {
        cout<<" x is "<<x;
    }
    void display(){
        cout<<"i am in display"<<endl;
        cout<<" x is "<<x <<endl;
    }
    void display2(){
        cout<<"i am in  display2"<<endl;
    }
    ~a()
    {
        cout<<"destructor of a"<<endl;
    }
};
 
list<a> lst;
    int main()
    {
 
 
        a *a1;  
 
        a1=new a();
        a1->setx (3);
 
        cout<<"Address1...."<<&(*a1)<<endl;
        cout<<"Address1.1..."<<a1<<endl;
 
        lst.push_back(*a1);
 
        list<a>::iterator p=lst.begin();
 
         cout<<"Address2......"<<&(*p)<<endl;
        p->display();
 
        cout<<lst.size()<<endl;
 
        return 0;
    }
 

[sicarie]

The objects in the list are giving different addresses because they are all stored at different locations - the list is what has the references to tie them together. If you used a vector, you would find that they were all contiguous in memory (if you returned many of them, there would be a pattern as the containers of the individual objects in the list are all of the same size - so there is the same amount of memory between them...). I don't believe (please anyone correct me if I'm wrong!) Lists need to have the contiguous memory space (though they can...). But each element in the list, vector, or STL container will have a different memory location because each element takes up space, and the list is ready to grow/shrink as you want it to, it just references those places the containers are at.