Does Stream#findFirst() (or findAny) shortcircuit a flatmapped stream?

Question

Java 8 Stream#findFirst() is a short-circuiting terminal operation. Meaning it will stop generating the stream once an element is found (usually used with a filter() operation).

However, I want to know if it will short circuit on a stream created by flatmap. I.e. will it short circuit the streams created within the flatmap operation.

In other words, which code is more efficient (in terms of generating fewer elements in the stream):

Example 1 (with flatmap):

    // produce range 0..99 (10 elements at a time)
    return IntStream.rangeClosed(0, 9)
        .flatMap(i -> IntStream.rangeClosed(i * 10, i * 10 + 9))
        .filter(i -> i == 55)
        .findFirst()
        .getAsInt();

Example 2 (no flatmap):

    // produce range 0..99 (all 100 elements at once)
    return IntStream.rangeClosed(0, 99)
        .filter(i -> i == 55)
        .findFirst()
        .getAsInt();

Use a `peek` with `System.out.print` to get to see the result. Elaborate *efficient* otherwise. — Naman, Apr 20 '19 at 02:10
@Naman thank you. Never heard of ```peek```, the answer is yes. They both only generated 55 elements. — wilmol, Apr 20 '19 at 02:12

wilmol · Accepted Answer · 2019-04-24T09:48:59.513

Answer is yes (using Java 11)

using peek:

Example 1 (with flatmap):

    // produce range 0..99 (10 elements at a time)
    return IntStream.rangeClosed(0, 9)
        .flatMap(i -> IntStream.rangeClosed(i * 10, i * 10 + 9))
        .peek(System.out::println)
        .filter(i -> i == 55)
        .findFirst()
        .getAsInt();

Example 2 (no flatmap):

    // produce range 0..99 (all 100 elements at once)
    return IntStream.rangeClosed(0, 99)
        .peek(System.out::println)
        .filter(i -> i == 55)
        .findFirst()
        .getAsInt();

Both generated 56 elements. (Originally thought it would be 60 with flatmap.)

Does Stream#findFirst() (or findAny) shortcircuit a flatmapped stream?

1 Answers1