Converting to lambda expression with ForEach for a breaking for loop

Question

Have the following codes with breaking behavior in a for loop:

package test;

import java.util.Arrays;
import java.util.List;

public class Test {

    private static List<Integer> integerList = Arrays.asList(1, 2, 3, 4);

    public static void main(String[] args) {
        countTo2(integerList);
    }

    public static void countTo2(List<Integer> integerList) {

        for (Integer integer : integerList) {
            System.out.println("counting " + integer);
            if (integer >= 2) {
                System.out.println("returning!");
                return;
            }
        }
    }
}

trying to express it with Lambda using forEach() will change the behavior as the for loop is not breaking anymore:

public static void countTo2(List<Integer> integerList) {

    integerList.forEach(integer -> {
        System.out.println("counting " + integer);
        if (integer >= 2) {
            System.out.println("returning!");
            return;
        }
    });
}

This actually makes sense as the return; statements are only enforced within the lambda expression itself (within the internal iteration) and not for the whole execution sequence, so is there a way to get the desired behavior (breaking the for loop) using the lambda expression?

Jacob G. · Accepted Answer · 2017-09-11 18:47:48Z

3

The following code is logically equivalent to yours:

public static void countTo2(List<Integer> integerList) {
    integerList.stream()
               .peek(i -> System.out.println("counting " + i))
               .filter(i -> i >= 2)
               .findFirst()
               .ifPresent(i -> System.out.println("returning!"));
}

If you're confused about anything, please let me know!

answered Sep 11, 2017 at 18:47

Jacob G.

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Comments

Eugene · Accepted Answer · 2017-09-12 07:34:04Z

2

What you are looking for is a short-circuit terminal operation and while this is the way to do it:

integerList.stream()
            .peek(x -> System.out.println("counting = " + x))
            .filter(x -> x >= 2)
            .findFirst()
            .ifPresent(x -> System.out.println("retunrning"));

That's an equivalent only when dealing with sequential stream. As soon as you add parallel that peek might show elements that you would not expect, because there is no defined processing order, but there is encounter order - meaning that elements will be correctly fed to the terminal operation.

answered Sep 12, 2017 at 7:34

Eugene

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1 Comment

user892960 Over a year ago

Sure, it has to be sequential as the logic is short-circuiting when the condition matches on the smallest value.

Naman · Accepted Answer · 2017-09-11 18:38:23Z

1

One way I could think of doing that would be using anyMatch and the inverse:

if (integerList.stream().noneMatch(val -> val >= 2)) {
    System.out.println("counting " + val);
}

if (integerList.stream().anyMatch(val -> val >= 2)) {
    System.out.println("returning!");
}

but internally that would iterate over the list twice and wouldn't be very optimal I believe.

answered Sep 11, 2017 at 18:38

Naman

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