a simple array question...

[zknyy wolf]

I have a String array like this:
String[] arrStr = {"123",null,null,"345",null,"fg",null}
if I turn it to like arrStr2 neither use loop to remove the null nor rebuild a arrstr array,how can I do :
arrStr2 = {"123","345","fg"}

[Lance Finney]

There's probably a more efficient way, but you could do this:

    List list = Arrays.asList(arrStr);
    List newList = new ArrayList();
    for (int i = 0; i < arrStr.length; i++) {
        if (list.get(i) != null) {
            newList.add(list.get(i));
        }
    }
    String[] arrStr2 = (String[]) newList.toArray(new String[0]);

[Layne Lund]

I can't think of a way off the top of my head to do it without a loop. Why don't you want to use a loop? Or did I just misunderstand the question?

[zknyy wolf]

tahnk you.Lance Finney and Layne Lund
I do it as Lance do.but I want to find a simple way as use API like
Object[] removeObject(Object[],Object)
now I think it's impossible.

[Anonymous]

It's quite some overkill, but not using (visible) loops can be done as follows:
String[] arr1= { ... };
List list= Arrays.asList(arr1);
List temp= Arrays.asList( new Object[] { null });
Object[] arr2= list.removeAll(temp).toArray();
The trouble is, you'll end up with an Object[], not a String[]. I think a simple (explicit) loop would be way more efficient.
kind regards

[Jim Yingst]

You can create a String[] rather than Object[] by precreating the array with the correct size:

    String[] arr2 = (String[]) list.toArray(new String[list.size()]);

The final line of Lance Finney's solution works as well - I just hate creating that extra zero-length array when I don't have to.
Unfortunately Arrays.asList() creates a fixed-length list - which means that removeAll() throws an UnsupportedOperationException. . We can work around this by creating another List:

    String[] arr1 = {"123", null, null, "345", null, "fg", null};
    List list = new LinkedList(Arrays.asList(arr1));
    list.removeAll(Arrays.asList(new Object[] { null }));
    String[] arr2 = (String[]) list.toArray(new String[list.size()]);

Of course, in the original question, "if I turn it to like arrStr2 neither use loop to remove the null nor rebuild a arrstr array" doesn't make much sense - there's no way to avoid "rebuilding" an array here, since the new array has a different size from the original. So I don't know if this answer is considered acceptable or not - but I'm not going to worry too much about it.
[ February 01, 2003: Message edited by: Jim Yingst ]