CODEWITHSUNDEEP
javalistjava-8java-stream
Lahiru Chandima
Lahiru Chandima

Reputation: 24068

Java stream map and collect - order of resulting container

List<MyObject> myList = new ArrayList<>(); 
//populate myList here

List<String> nameList = myList.stream()
        .map(MyObject::getName)
        .collect(Collectors.toList());

In above code, can I expect that order of MyObject names in nameList is always the same as the order of myList?

Upvotes: 68

Views: 42209

Answers (1)

Tagir Valeev
Tagir Valeev

Reputation: 100159

Yes, you can expect this even if you are using parallel stream as long as you did not explicitly convert it into unordered() mode.

The ordering never changes in sequential mode, but may change in parallel mode. The stream becomes unordered either:

  • If you explicitly turn it into unordered mode via unordered() call
  • If the stream source reports that it's unordered (for example, HashSet stream is unordered as order is implementation dependent and you cannot rely on it)
  • If you are using unordered terminal operation (for example, forEach() operation or collecting to unordered collector like toSet())

In your case none of these conditions met, thus your stream is ordered.

Upvotes: 92

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