Note: I am aware of the Iterator#remove() method.
In the following code sample, I don’t understand why the List.remove in main method throws ConcurrentModificationException, but not in the remove method.
public class RemoveListElementDemo {
private static final List<Integer> integerList;
static {
integerList = new ArrayList<Integer>();
integerList.add(1);
integerList.add(2);
integerList.add(3);
}
public static void remove(Integer toRemove) {
for(Integer integer : integerList) {
if(integer.equals(toRemove)) {
integerList.remove(integer);
}
}
}
public static void main(String... args) {
remove(Integer.valueOf(2));
Integer toRemove = Integer.valueOf(3);
for(Integer integer : integerList) {
if(integer.equals(toRemove)) {
integerList.remove(integer);
}
}
}
}
Here’s why:
As it is says in the Javadoc:
This check is done in the
next()method of the iterator (as you can see by the stacktrace). But we will reach thenext()method only ifhasNext()delivered true, which is what is called by the for each to check if the boundary is met. In your remove method, whenhasNext()checks if it needs to return another element, it will see that it returned two elements, and now after one element was removed the list only contains two elements. So all is peachy and we are done with iterating. The check for concurrent modifications does not occur, as this is done in thenext()method which is never called.Next we get to the second loop. After we remove the second number the hasNext method will check again if can return more values. It has returned two values already, but the list now only contains one. But the code here is:
1 != 2, so we continue to the
next()method, which now realizes that someone has been messing with the list and fires the exception.Hope that clears your question up.
Summary
List.remove()will not throwConcurrentModificationExceptionwhen it removes the second last element from the list.