Java ArrayList.clear() function

Question

List<List<Integer>> result = new ArrayList<List<Integer>>();
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(5);temp.add(6);
result.add(temp);
temp.clear();

I wrote codes like something above, what puzzles me is when I debug the code, I found the result contains size of 1 but the values (5,6,...) are lost after I applying the clear function, can any one explain why?

You didn't clear the result list, so it will still have one element (an empty list). — Keppil, Jul 20 '14 at 20:35
@StefanoSanfilippo it seems the person just wants to delete one item of the 2D list? — Kick Buttowski, Jul 20 '14 at 20:36
Sorry my intention is to clear the temp not the result list. I mean after clearing the members in the temp, even the result's list is affected — i3wangyi, Jul 20 '14 at 20:37
Can you tell us what result were you expecting exactly? It is not very clear in the question — gpeche, Jul 20 '14 at 21:16

Pshemo · Answer 1 · 2014-07-20T20:56:44.173

You have list of lists. After this code

List<List<Integer>> result = new ArrayList<List<Integer>>();
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(5);temp.add(6);
result.add(temp);

situation looks like this

                     ┌───> 5
result ─────> tmp ───┤
                     └───> 6

result list contains one element which is tmp list
tmp list contains two elements 5 and 6

after

temp.clear();

situation changes to

           // ↓↓↓ `temp.clear()` affects only this list
result ─────> tmp

so now

tmp list is empty
but result still contains tmp list that is why its size is 1

+1 for the graphical representation of the situation :) – GameDroids Jul 20 '14 at 20:42 — GameDroids, Jul 20 '14 at 20:42

score 1 · Answer 2 · answered Jul 20 '14 at 20:38

1

This line of code

result.add(temp);

Adds a reference to temp to result, the next line

temp.clear(); // <-- here

clears temp. I think you wanted a copy of temp (so that you could then clear temp without altering result) so,

result.add(new ArrayList<Integer>(temp)); // <-- copy temp.

Then clearing temp will not change the values in result.

answered Jul 20 '14 at 20:38

Elliott Frisch

198,278
20
158
249

I've no idea what's the difference between deep and normal/ – i3wangyi Jul 20 '14 at 20:40
@ElliottFrisch yeah, this explains my confusion well. I don't even know the reference thing would happen in java – i3wangyi Jul 20 '14 at 20:41
1

@i3wangyi normal copies one level. deep copies all levels. – Peter Lawrey Jul 20 '14 at 20:41
@i3wangyi: A deep copy copies all the elements too. This creates a new `List` that still contains the same `Integer` references. – Keppil Jul 20 '14 at 20:41
@Elliott: Yep, I think this answers the question well. +1 – Keppil Jul 20 '14 at 20:42

Peter Lawrey · Answer 3 · 2014-07-20T20:46:29.517

temp is a reference to an ArrayList object.

ArrayList<Integer> temp;

This add the reference to the result List.

result.add(temp);  // adds a copy of the reference, not a copy of the list.

This clear the original and only list (apart from the result list)

temp.clear();

Note: Java only has references and primitives, there is no other types.

How could I do to avoid this situation? copy the temp list?

for every new list you want, create a new one. Instead of temp.clear() call

temp = new ArrayList<>();

Ideally, you shouldn't even reuse the local variable unless it has the same purpose.

// don't use temp again.
List<Integer> temp2 = new ArrayList<>();

BTW I advocate that you re-use mutable objects to maximise performance. You should only do this after you have measured this a problem for your allocation rate and you know what you are doing.

thanks @peter How could I do to avoid this situation? copy the temp list? — i3wangyi, Jul 20 '14 at 20:42

Java ArrayList.clear() function

3 Answers3