java replace list contents or to replace reference value itself

Question

I have a list that used for adapter for ListView in Android.

Most time, the List is used for read but sometimes replace of all List contents is needed.

which type of code below (replaceListA, replaceListB) is preferred? or another way to replace the list?

thank you in advance.

private ArrayList<Model> list = new ArrayList<Model>();

public synchronized void replaceListA(List newList) {
            list.clear();
            list.addAll(newList);
}

public synchronized void replaceListB(List newList) {
    list = newList;
}

Is it possible that `newList` would be edited after calling this method? — Flight Odyssey, Jul 31 '14 at 03:12
If `newList` is updated after calling the method, it could change the value of `list` as well (if you use `replaceListB`). Depending on the situation, you might want that to happen or you might not. — Flight Odyssey, Jul 31 '14 at 03:21

score 1 · Answer 1 · edited May 23 '17 at 12:20

1

See this answer.

It's hard to know without a benchmark, but if you have lots of items in your ArrayList and the average size is lower, it might be faster to make a new ArrayList.

edited May 23 '17 at 12:20

Community

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1

answered Jul 31 '14 at 03:11

mattbdean

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You should just vote to close as a duplicate. – Sotirios Delimanolis Jul 31 '14 at 03:20

score 1 · Accepted Answer · answered Jul 31 '14 at 03:15

In general, replaceListB will probably run faster, but if newList is later modified, list will also change. For example:

ArrayList<Model> newList = new ArrayList<Model>();
replaceListB(newList);
newList.add( /*something*/ ); //This will update both newList and list

By contrast

ArrayList<Model> newList = new ArrayList<Model>();
replaceListA(newList);
newList.add( /*something*/ ); //This will only update newList

To be more specific, replaceListB will run in a constant amount of time, since it is only changing a pointer, where replaceListA has to copy over each element of the list individually, so it will take a long time if the list is large. However, depending on the details of the garbage collecter, it's possible that replaceListA could free up memory sooner. Either way, it is unlikely to make a major difference unless the lists are very large, so I would recommend choosing based on whether you want to make a separate copy of the list (replaceListA) or allow editing (replaceListB).

score 0 · Answer 3 · answered Jul 31 '14 at 03:09

0

The second is preferred. It runs in O(1) time, the other runs in O(n) time. Just be sure to call notifyDataSetChanged afterwards.

answered Jul 31 '14 at 03:09

Gabe Sechan

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java replace list contents or to replace reference value itself

3 Answers3