CODEWITHSUNDEEP
c++stl
Tanay
Tanay

Reputation: 181

Does erase() in C++'s STL actually delete the element?

For the following code:

 list<int> P;
int i,n,x;
cin>>n>>x;
for(i=0;i<n;i++)
    P.push_back(i+1);
list<int>::iterator t2,t1=P.begin();
advance(t1,x-1);
t2=t1;
t1++;
P.erase(t1);
t2++;
cout<<*t2<<" ";
cout<<*(P.end())<<endl;

when the input I give is: 5 3,

the ouput comes out to be 5 4.

I am actually deleting the element 4 from the list. But it is going to the end of list. How do I avoid this? I want to completely remove 4 from the list so that the last element in the list is 5.

Upvotes: 0

Views: 187

Answers (2)

Jake Freeman
Jake Freeman

Reputation: 1698

When you call the erase() it deletes the element in the internal storage.

A test you can perform to test it is this:

int main()
{
    vector<int> x = {3, 4, 5, 6};
    int * temp  = &x[1];
    cout<<*temp<<endl;
    x.erase(x.begin() + 1);
    cout<<*temp<<endl;

}

The output shows:

4
5

Which shows the the erase removes the element and shifts all the elements in its spot.

Hope this helps.

Upvotes: 0

eerorika
eerorika

Reputation: 238351

Does erase() in C++'s STL actually delete the element?

Yes. erase does destroy the element.

But it is going to the end of list.

*(P.end())

end returns past the end iterator. Dereferencing it has undefined behaviour. You can use back to get the last element.

Upvotes: 6

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