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javaarraylist
i3wangyi
i3wangyi

Reputation: 2409

Java ArrayList.clear() function

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?

Upvotes: 1

Views: 522

Answers (3)

Pshemo
Pshemo

Reputation: 124215

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

Upvotes: 5

Peter Lawrey
Peter Lawrey

Reputation: 533492

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.

Upvotes: 1

Elliott Frisch
Elliott Frisch

Reputation: 201429

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.

Upvotes: 1

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