Java 8 stream for each and peek

Question

I stumbled upon a task with a stream that I couldn't solve. I had to modify the code to print abcABC instead of aAbBcC.

I understand why it's printed this way.

// prints 'a b c' then prints 'A B C'
List<String> strings = List.of("a", "b", "c");
    strings.stream()
           .peek(str -> System.out.println(str))
           .map(str -> str.toUpperCase())
           .forEach(str -> System.out.println(str));

I appreciate any help.

score 3 · Accepted Answer · answered Nov 18 '21 at 09:51

3

Here's one way to do it:

List<String> strings = List.of("a", "b", "c");
    strings.stream()
           .peek(str -> System.out.println(str))
           .sorted()
           .map(str -> str.toUpperCase())
           .forEach(str -> System.out.println(str));

The sorted() operation must process all the elements of the Stream before the terminal operation can start outputting the elements (since you can't sort a Stream without going over all its elements).

Therefore, this causes peek() on all the elements to be executed before the first upper case element is outputted by forEach().

answered Nov 18 '21 at 09:51

Eran

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Is this actually guaranteed, or could the sorting be theoretically omitted if the implementation can determine that the input is already sorted? – Hulk Nov 18 '21 at 10:24
I guess such an optimization is not possible here, as [java.util.stream](https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/stream/package-summary.html#StreamOps) explicitly uses it as an example for an intermediate operation that needs to process the entire input, but I was thinking along the lines of how `.count()` can be optimized if the size is known. – Hulk Nov 18 '21 at 10:31
3

@Hulk this optimization does already exist, try `Set strings = new TreeSet<>( List.of("a", "b", "c")); strings .stream() .peek(System.out::println) .sorted() .map(String::toUpperCase) .forEach(System.out::println);` – Holger Nov 18 '21 at 12:18
@Holger thanks, interesting - that is exactly what I had in mind. So the trick/hack used in this answer only works for streams based on Spliterators that do not report [Spliterator.SORTED](https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/Spliterator.html#SORTED). – Hulk Nov 18 '21 at 12:26
1

@Holger that;s good to know – Eran Nov 18 '21 at 12:51
2

@Hulk in fact, even the answer’s example is not guaranteed to work, formally. `forEach` is an *unordered* operation, so it’s not required to do the preceding `sorted()` operation. Or, well, whether the programmer should be burdened with such a surprising behavior, is debatable. As Brian Goetz says in [this comment](https://stackoverflow.com/questions/30843279/30916033#comment49971789_30916033), the case of chaining an unordered operation after `sorted` is special and has been exempted from such optimization deliberately, to simplify the implementation (and solve related bugs). – Holger Nov 18 '21 at 13:20
Oh yes, I missed that - strictly speaking, [`forEachOrdered()`](https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/stream/Stream.html#forEachOrdered(java.util.function.Consumer)) would be needed, also discussed here: https://stackoverflow.com/a/34253279/2513200 – Hulk Nov 18 '21 at 13:28

Java 8 stream for each and peek

1 Answers1