当Iterator不支持remove（）时如何动态修剪Java List？

Question

I have the following code:

Widget[] widgetArray = widgetService.getAllWidgets();
List<Widget> widgets = Arrays.asList(widgetArray);

// Prune out any Widgets named "Melvin".
Iterator<Widget> iter = widgets.iterator();
while(iter.hasNext()) {
    Widget w = iter.next();

    if("Melvin".equals(w.getName()))
        iter.remove();
}

When I run this code I get a runtime java.lang.UnsupportedOperationExceptionError with a vague exception message of null that gets thrown on the iter.remove() line. It seems that some Java Iterators don't support the remove method and will throw this exception.

I can't change the widgetService.getAllWidgets() method to return a List<Widget> and am stuck with the Widget[] array return value.

So I ask: what can I do to loop through my widgets array and dynamically prune out ones that are named " Melvin "?

Answer 1

If you can afford it, just make a mutable copy of the list. Replace

List<Widget> widgets = Arrays.asList(widgetArray);

with

List<Widget> widgets = new ArrayList<Widget>(Arrays.asList(widgetArray));

Answer 2

Just defer removal until the iterator is done:

Widget[] widgetArray = widgetService.getAllWidgets();
List<Widget> widgets = Arrays.asList(widgetArray);

// Prune out any Widgets named "Melvin".
List<Widget> toRemove = new ArrayList<Widget>();
Iterator<Widget> iter = widgets.iterator();
while(iter.hasNext()) {
    Widget w = iter.next();

    if("Melvin".equals(w.getName()))
        toRemove.add(w);
}
widgets.removeAll(toRemove);

Alternatively, just build the list from eligible widgets using the inverse logic:

List<Widget> widgets = new ArrayList<Widget>();
// Add all Widgets not named "Melvin"
for (Widget w : widgetService.getAllWidgets()) {
    if(!"Melvin".equals(w.getName()))
        widgets.add(w);
}

Answer 3

The asList() list is still backed by the array.

You may want to loop through each element of the array and add it to a brand new list. This would take two loops.

Or better yet, compare the string value and then add to the list. This way, you have one loop and a brand new list.

Answer 4

With guava you can get rid of all that code and let a fast, well tested library take care of it for you.

Collection<Widget> noMelvins = Collections2.filter( Arrays.asList(widgetArray), new Predicate<Widget>() {
    @Override public boolean apply( final Widget arg) {
        return !"Melvin".equals(arg.getName());
    }
});