How to ensure order of processing in java8 streams?

Question

I want to process lists inside an XML java object. I have to ensure processing all elements in order I received them.

Should I therefore call sequ

Answer 1

In a nutshell:

Ordering depends on the source data structure and intermediate stream operations. Assuming you are using a List the processing should be ordered (since filter won't change the sequence here).

More details:

Sequential vs Parallel vs Unordered:

Javadocs

S sequential()
Returns an equivalent stream that is sequential. May return itself, either because the stream was already sequential, or because the underlying stream state was modified to be sequential.
This is an intermediate operation.

S parallel()
Returns an equivalent stream that is parallel. May return itself, either because the stream was already parallel, or because the underlying stream state was modified to be parallel.
This is an intermediate operation.

S unordered()
Returns an equivalent stream that is unordered. May return itself, either because the stream was already unordered, or because the underlying stream state was modified to be unordered.
This is an intermediate operation.

Stream Ordering:

Javadocs

Streams may or may not have a defined encounter order. Whether or not a stream has an encounter order depends on the source and the intermediate operations. Certain stream sources (such as List or arrays) are intrinsically ordered, whereas others (such as HashSet) are not. Some intermediate operations, such as sorted(), may impose an encounter order on an otherwise unordered stream, and others may render an ordered stream unordered, such as BaseStream.unordered(). Further, some terminal operations may ignore encounter order, such as forEach().

If a stream is ordered, most operations are constrained to operate on the elements in their encounter order; if the source of a stream is a List containing [1, 2, 3], then the result of executing map(x -> x*2) must be [2, 4, 6]. However, if the source has no defined encounter order, then any permutation of the values [2, 4, 6] would be a valid result.

For sequential streams, the presence or absence of an encounter order does not affect performance, only determinism. If a stream is ordered, repeated execution of identical stream pipelines on an identical source will produce an identical result; if it is not ordered, repeated execution might produce different results.

For parallel streams, relaxing the ordering constraint can sometimes enable more efficient execution. Certain aggregate operations, such as filtering duplicates (distinct()) or grouped reductions (Collectors.groupingBy()) can be implemented more efficiently if ordering of elements is not relevant. Similarly, operations that are intrinsically tied to encounter order, such as limit(), may require buffering to ensure proper ordering, undermining the benefit of parallelism. In cases where the stream has an encounter order, but the user does not particularly care about that encounter order, explicitly de-ordering the stream with unordered() may improve parallel performance for some stateful or terminal operations. However, most stream pipelines, such as the "sum of weight of blocks" example above, still parallelize efficiently even under ordering constraints.

Answer 2

You are asking the wrong question. You are asking about sequential vs. parallel whereas you want to process items in order, so you have to ask about ordering. If you have an ordered stream and perform operations which guarantee to maintain the order, it doesn’t matter whether the stream is processed in parallel or sequential; the implementation will maintain the order.

The ordered property is distinct from parallel vs sequential. E.g. if you call stream() on a HashSet the stream will be unordered while calling stream() on a List returns an ordered stream. Note that you can call unordered() to release the ordering contract and potentially increase performance. Once the stream has no ordering there is no way to reestablish the ordering. (The only way to turn an unordered stream into an ordered is to call sorted, however, the resulting order is not necessarily the original order).

See also the “Ordering” section of the java.util.stream package documentation.

In order to ensure maintenance of ordering throughout an entire stream operation, you have to study the documentation of the stream’s source, all intermediate operations and the terminal operation for whether they maintain the order or not (or whether the source has an ordering in the first place).

This can be very subtle, e.g. Stream.iterate(T,UnaryOperator) creates an ordered stream while Stream.generate(Supplier) creates an unordered stream. Note that you also made a common mistake in your question as forEach does not maintain the ordering. You have to use forEachOrdered if you want to process the stream’s elements in a guaranteed order.

So if your list in your question is indeed a java.util.List, its stream() method will return an ordered stream and filter will not change the ordering. So if you call list.stream().filter() .forEachOrdered(), all elements will be processed sequentially in order, whereas for list.parallelStream().filter().forEachOrdered() the elements might be processed in parallel (e.g. by the filter) but the terminal action will still be called in order (which obviously will reduce the benefit of parallel execution).

If you, for example, use an operation like

List<…> result=inputList.parallelStream().map(…).filter(…).collect(Collectors.toList());

the entire operation might benefit from parallel execution but the resulting list will always be in the right order, regardless of whether you use a parallel or sequential stream.