Thread: The proper way to erase a vector element

ok, i know i want to erase element number n, however, i do not "know the iterator" for element n.

Code:

vector<int *> a;   
.... //'a' filled with 20 values.
a.erase(15); //gives syntax error
//so does a.erase(a[15]);

?????

What if you use:

a.erase(a.begin() + 15);

Originally Posted by laserlight

What if you use:

a.erase(a.begin() + 15);

that would make too much sense


how about
a.erase((vector<int *>::iterator) &a[15])

Code:

a.erase( a.begin() + 15 );

[edit] My post came reaaally late.

how about
a.erase((vector<int *>::iterator) &a[15])

That shouldn't work.

Originally Posted by Dante Shamest

Code:

a.erase( a.begin() + 15 );

[edit] My post came reaaally late.

how about
a.erase((vector<int *>::iterator) &a[15])

That shouldn't work.

i don't see why not. it works fine (so it seems). an iterator is nothing but a pointer. so i grab the & of a[n], tell the erase function that it's an iterator, and BAM! - it compiles.

if i remember correctly 'iterator' is declared as

Code:

template<type T, ......>
class vector
{
     typedef  T * iterator;
}

but this is of course why i ask.

i don't see why not. it works fine (so it seems). an iterator is nothing but a pointer. so i grab the & of a[n], tell the erase function that it's an iterator, and BAM! - it compiles.

I apologise. I thought it wouldn't work because I mistakenly assumed you were working with a vector of ints, instead of a vector of int*s.

i don't see why not. it works fine (so it seems). an iterator is nothing but a pointer. so i grab the & of a[n], tell the erase function that it's an iterator, and BAM! - it compiles.

But it's something of a kluge, in my opinion.

You're going from location to pointer then cast to iterator where you can just directly simulate pointer arithmetic with the iterator.

works just the same

Code:

#include <iostream>
#include <stdlib.h>
#include <vector>
using namespace std;

int main(int argc, char *argv[])
{
  vector<int> a;
  int i;
  
  for(i = 0; i < 10; i++)
    a.push_back(i);
  
  for(i = 0; i < 10; i++)
    cout << a[i] << ' ';
  cout << endl;
  
  a.erase((vector<int>::iterator) &a[5]);
  
  for(i = 0; i < 9; i++)
    cout << a[i] << ' ';
  cout << endl;
  system("PAUSE");	
  return 0;
}

Originally Posted by anonytmouse

It works, but is it guaranteed to work? Does the standard guarantee that a vector iterator is a pointer or is this an implementation detail? Calling erase on an item in the middle or beginning of a vector is an expensive operation, as all subsequent elements must be moved. If you need to add and remove elements from the beginning or middle of a container, you should consider another container type.

i'm not arguing that my ingenius, super hackish way of erasing an element is the correct way. hell, look at the thread title. i am saying that it DOES work. i would not go into a job interview and do it that way, that's for sure.

a vector is fine. i only need to erase a pointer here and there. it won't happen more than, on average, about 0.5 times per program execution and it only will contain about 6 (pointer) elements tops. i'm not to concerned with it's efficiency.