Stream Spliterator error after concat Java

Question

When attempting to create an iterator from a Stream that was concatenated from two previous streams, I get a NoSuchElementException, as the iterator doesn't recognise that the Stream has elements due to some problems with the Spliterator. The concatenated Stream seems to have two spliterators, despite the previous streams being of the same type. I also get this error when I tried to convert the concatenated stream into an array to try get round the problem.

Stream<Node> nodeStream = Stream.of(firstNode);
while (!goal) {
    Iterator<Node> iterator = nodeStream.iterator();
    Node head = iterator.next();
    Stream<Node> tail = Stream.generate(iterator::next).filter(n -> n != head);
    Node[] newNodes = head.expand(end);
    if (newNodes.length == 1) {
        goal = true;
        endNode = newNodes[0];
    }
    nodeStream = Stream.concat(Arrays.stream(newNodes), tail);
    nodeStream = nodeStream.sorted(Comparator.comparing(n -> n.routeCost(end)));
}

The error is as follows:

Exception in thread "main" java.util.NoSuchElementException
    at java.base/java.util.Spliterators$1Adapter.next(Spliterators.java:688)
    at java.base/java.util.stream.StreamSpliterators$InfiniteSupplyingSpliterator$OfRef.tryAdvance(StreamSpliterators.java:1358)
    at java.base/java.util.stream.StreamSpliterators$WrappingSpliterator.lambda$initPartialTraversalState$0(StreamSpliterators.java:292)
    at java.base/java.util.stream.StreamSpliterators$AbstractWrappingSpliterator.fillBuffer(StreamSpliterators.java:206)
    at java.base/java.util.stream.StreamSpliterators$AbstractWrappingSpliterator.doAdvance(StreamSpliterators.java:161)
    at java.base/java.util.stream.StreamSpliterators$WrappingSpliterator.tryAdvance(StreamSpliterators.java:298)
    at java.base/java.util.stream.Streams$ConcatSpliterator.tryAdvance(Streams.java:723)
    at java.base/java.util.stream.StreamSpliterators$WrappingSpliterator.lambda$initPartialTraversalState$0(StreamSpliterators.java:292)
    at java.base/java.util.stream.StreamSpliterators$AbstractWrappingSpliterator.fillBuffer(StreamSpliterators.java:206)
    at java.base/java.util.stream.StreamSpliterators$AbstractWrappingSpliterator.doAdvance(StreamSpliterators.java:161)
    at java.base/java.util.stream.StreamSpliterators$WrappingSpliterator.tryAdvance(StreamSpliterators.java:298)
    at java.base/java.util.Spliterators$1Adapter.hasNext(Spliterators.java:681)
    at java.base/java.util.Spliterators$1Adapter.next(Spliterators.java:687)
    at inf.ed.ac.uk.Route.generateRoute(Route.java:35)

I am trying to expand the first node (returns 16 new nodes) add them to the stream, sort it, and repeat, this is part of an implementation of the A* algorithm on gps coordinates

Code for Node is

public class Node {
    boolean goal = false;
    Node parent;
    final LngLat coords;
    LngLat.Compass direction;
    double cost;
    private Route route;

    public Node[] expand(LngLat end) {
        ArrayList<Node> nodes = new ArrayList<>();
        for (LngLat.Compass direction: LngLat.Compass.values()) {
            Node node = new Node(coords.nextPosition(direction), this, direction);

            if (noFlyClear(node)) {
                if (!route.contains(node)) {
                    if (node.coords.closeTo(end)) {
                        node.goal = true;
                        return new Node[]{node};
                    }
                    nodes.add(node);
                    route.visited.add(node);
                }
            }

        }
        return nodes.toArray(Node[]::new);
    }

But that's since I changed to ArrayLists I'm still unsure what the issue was originally

And please explain what you're trying to do. Why would you try to call `next()` twice on a singleton stream? — shmosel, Nov 28 '22 at 21:48
@AlistairTait I'm guessing but I think `Stream.generate(iterator::next)` is going to call it again. — markspace, Nov 28 '22 at 22:01
@AlexanderIvanchenko Node is an object, I am working with arrays of Nodes, I am trying to expand the node, and add the new nodes to a stream — Alistair Tait, Nov 28 '22 at 22:03
@AlexanderIvanchenko Node has a geographic coordinate, and it generates new nodes by taking a step in 16 cardinal directions, I don't think that's relevant to what I am having a problem with — Alistair Tait, Nov 28 '22 at 22:06
@shmosel it should be a new stream, when I debug the code, it adds the streams together, it just gives it two spliterators — Alistair Tait, Nov 28 '22 at 22:07
I don't know what that means, but I doubt streams is the right tool here. — shmosel, Nov 28 '22 at 22:12
After executing `iterator.next()` for a stream with one element, if `iterator::next` in `Stream.generate(iterator::next)` is evaluated, `NoSuchElementException` will be thrown. — ﾓｷｬﾃﾞ, Nov 28 '22 at 22:37
@ﾓｷｬﾃﾞWould it not create a new iterator each time the loop is run? — Alistair Tait, Nov 28 '22 at 22:40
@ﾓｷｬﾃﾞIt fails on the second pass through the loop — Alistair Tait, Nov 28 '22 at 22:41
Is that true? I think it's happening in the first loop. Because `.sorted()` tries to pick out all the elements. — ﾓｷｬﾃﾞ, Nov 28 '22 at 22:56
Can add the code of the `Node` class (only fields and `expand()`) method? — Alexander Ivanchenko, Nov 28 '22 at 23:40
Also, what's the line `35` in your code? I bet `Node head = iterator.next();` — Alexander Ivanchenko, Nov 28 '22 at 23:41
Basically, you're trying to use a Stream as a replacement of a sorted Collection. — Alexander Ivanchenko, Nov 28 '22 at 23:45
Streams are definitely the wrong tool for this task, but most obviously, Stream.generate expects its supplier to work forever. — Louis Wasserman, Nov 29 '22 at 05:27
You have to use the right way [to create a Stream from an iterator](https://stackoverflow.com/questions/23439780/how-to-create-a-java-8-stream-from-an-iterator) but using an `ArrayDeque` instead will be straight-forward and more efficient. — Holger, Nov 29 '22 at 11:21
(or use an `ArrayList` and always append to the end, since you’re sorting the result anyway) — Holger, Nov 29 '22 at 11:27
@AlexanderIvanchenko line 35 is ```nodeStream = nodeStream.sorted(Comparator.comparing(n -> n.routeCost(end)));``` — Alistair Tait, Nov 30 '22 at 23:54

Stream Spliterator error after concat Java

0 Answers0