CODEWITHSUNDEEP
javaparallel-processingsynchronizationjava-stream
shmosel
shmosel

Reputation: 50716

Java Stream stateful behavior example

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?

Upvotes: 5

Views: 361

Answers (2)

Eugene
Eugene

Reputation: 120848

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.

Upvotes: 3

Douglas
Douglas

Reputation: 5087

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.

Upvotes: 5

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