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Problem While Removing Element From List

Gowher Naik
Ranch Hand

Joined: Feb 07, 2005
Posts: 643

output is-> 1 3
i am not able to understand why 1 and 3 is not removed from List.
wise owen
Ranch Hand

Joined: Feb 02, 2006
Posts: 2023

Because i add 1 in each loop.
1) i=0, remove "0"; The "1" element move to index 0 and l.size()=3
2) i=1, remove "2"; The "3" element move to index 1 and l.size()=2
3) i=2, exit loop because the condition of "i<l.size();"
[ August 04, 2006: Message edited by: wise owen ]
Paul Michael
Ranch Hand

Joined: Jul 02, 2001
Posts: 697
wise is correct. try to remove the elements from the list backwards.
using highest indeces first. that way, it won't mess up your for loop.

SCJP 1.2 (89%), SCWCD 1.3 (94%), IBM 486 (90%), SCJA Beta (96%), SCEA (91% / 77%), SCEA 5 P1 (77%), SCBCD 5 (85%)
Gautam Malik

Joined: Mar 17, 2006
Posts: 1
try out this:

import java.util.*;
class test
public static void main(String args[]){
List<String> l=new ArrayList<String>();
int size = l.size();
for(int i=0; i < size ; i++){
System.out.println("Removing "+l.get(0));
System.out.println("Size is now: "+l.size());
for(int i=0;i<l.size();i++){
Yatendra Varshney

Joined: Jan 30, 2006
Posts: 5

See, your code is running in follwoing manner:
1. Go to first for loop: size of list is 4 and value of i =0.
2. First Attampt: In for loop i < l.size() means 0<4, condition is true go to inside loop and remove "0" from the list and "1" moved to 0th location of the list, now value of list is 3 and i is 1.
Second Attampt: Again go to inside loop and remove "2" because "2" on the 1st position, and then "3" moved to 1th location and then i is 2 and size of list is 2 then condition is false.
dennis du
Ranch Hand

Joined: Dec 31, 2004
Posts: 59
You should use iterator to remove them.

Iterator iterator=list.iterator();

So you can scan and remove all of them. It is more reasonable.

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subject: Problem While Removing Element From List
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