IndexOutOfBoundsException异常

Question

Hi this code will return indexoutofboundsException and really I don't know why? I want to remove those objects from pointlist which are as the same as an object in the list .

    public void listOfExternalPoints(List<Point> list) {
    System.out.println(list.size());
    System.out.println(pointList.size());
    int n = pointList.size();
    for (int i = pointList.size() - 1; i >= 0; i--) {
        for (int j = 0; j < list.size(); j++) {
            if (pointList.get(i)==(list.get(j))) {
                pointList.remove(i);
                n--;
            }
        }
    }

}

Also the out put of println will be :

54
62

Also the exception:

Exception in thread "AWT-EventQueue-0" java.lang.IndexOutOfBoundsException: Index: 60, Size: 60
    at java.util.ArrayList.RangeCheck(ArrayList.java:547)
    at java.util.ArrayList.get(ArrayList.java:322)
    at ConvexHull.BlindVersion.listOfExternalPoints(BlindVersion.java:83)

thanks.

Answer 1

hey, you removed some elements from the list. So the list is smaller than it was at the beginning of the loop.

I suggest you to use:

pointList.removeAll(list)

Or an iterator.

Answer 2

When you do pointList.remove(i) , you should break from the inner loop. Otherwise, it will try to index pointList.get(i) , which you just removed, again on the next iteration of the loop, which is why are you getting the exception.

When arrayLists remove elements, that element is taken out, and all elements after it gets shifted down. So if you remove index 3 and there are only 4 elements, the new arrayList only has size 3 and you try to get index 3, which is out of bounds.

EDIT: A better way to do this is:

for(Point p : list) {
    pointList.remove(p);
}

It will have the same efficiency, but more correct, I think. Remember that == compares references for the same object. Where as the remove method uses .equals to check for equality, which is what I assume you want.

Answer 3

pointList.remove(i);

This will decrease your pointList size. Hope this helps.

Answer 4

Removing the object from pointList will reduce its size. Therefore, in one iteration of the "for j" block, you may be removing two elements of PointList, shifting all other elements to the left. However, in the next iteration "i" will refer to an out of bounds location (60).

Answer 5

The problem is with the order of loop evaluation. You are only evaluating the length of pointList once, and never checking it again.

If you remove the last item from pointList , and you have not reached the end of list , then you will attempt to get() same item from pointList again and you will be reading off the end of the list, causing the exception. This is what shoebox639 noticed; if you break the inner loop after removing something, the decrement in the outer loop will fix the issue.

Because of the order of iteration, it is impossible to remove two elements from the same run through the loop- you're only considering the end of the list, so nothing beyond the current point in pointList is candidate for removal. If you were to remove two elements at once, it would be possible to shorten the list to a degree that the next i is off the list, but that can't happen here. Watch out for it in other, similar loops, though.

Answer 6

You may remove more then one element of the pointList in every run of the first loop (over the pointList). Put a breackpoint in the line pointList.remove(i); and step through your code with the debugger and you will see it.

Answer 7

Instead of iterating over your list with an integer, use the for each construct supported by the Collections interface.

public void listOfExternalPoints(List<Point> list) { 
    for (Point pointListEntry : pointList) { 
        for (Point listEntry : list) { 
            if (pointListEntry == listEntry) { 
                pointList.remove(pointListEntry); 
            } 
        } 
    } 
}

I haven't debugged this, but it looks right to me.