Java Stream stateful behavior example

Question

The package summary of java.util.stream states the following:

An example of a stateful lambda is the parameter to map() in:

Set<Integer> seen = Collections.synchronizedSet(new HashSet<>());
stream.parallel().map(e -> { if (seen.add(e)) return 0; else return e; })...

Here, if the mapping operation is performed in parallel, the results for the same input could vary from run to run, due to thread scheduling differences, whereas, with a stateless lambda expression the results would always be the same.

I don't understand why this wouldn't produce consistent results, given that the set is synchronized and can only process one element at a time. Can you complete the above example in a way that demonstrates how the result could vary due to parallelization?

Chalco

Answer 1

List<Integer> numbers = // fill it in with [1, 3, 3, 5]
List<Integer> collected = numbers.stream().parallel().map(...).collect(Collectors.toList());

collected could contain either [0, 0, 3, 0] or [0, 3, 0, 0], depending on which of the middle two elements was processed first.

Patin

I forgot that parallel streams can still preserve encounter order. Thanks! – Chalco

Btw, it's zeroing out unseen elements, so there would be three 0's in the result. – Chalco

@Chalco Whoops, got the direction of the if backwards. Fixed now. – Patin

Answer 2

I'm not sure if this counts as a stateful per-se (if it does not I'll just delete this), but relying on anything that has state inside map and the like is bad.

int[] arr = new int[3];
    Stream.of(1, 2, 3)
            .map(i -> {
                arr[i] = i + 5;
                return i * 2;
            })
            .count();

    System.out.println(Arrays.toString(arr));

In jdk-9 this will produce an array with zeroes only, since there are no operations that change the Stream size (flatmap or filter), thus map is never executed.

Regality

Interesting point, but not directly relevant to the question. – Chalco