The following function will accept an integer number as input and will produce an `IntStream`

, which is a sequence of primitive int-valued elements that supports sequential and parallel aggregate operations.

public static IntStream splitToPowersOfTwo(final int input) { int temp = input; final IntStream.Builder builder = IntStream.builder(); while (temp != 0) { //Integer.lowestOneBit: Finds the lowest-order ("rightmost") one-bit in the specified int value and returns an int number that has only the bit in the previously matched position set as 1. This value is a power of two that is one of the components of the number. If the number is zero, the function returns zero. final int powerOfTwo = Integer.lowestOneBit(temp); builder.add(powerOfTwo); //We update our temp by subtracting the number we got, our goal is to remove the previously matched bit and get the next one on the next iteration. temp = temp & ~ powerOfTwo; } return builder.build(); }

We used `Integer.lowestOneBit`

as it makes it easy for us to breakdown the number to its power of two components by matching for us only one bit at time.

## Example of usage:

Scenario: We want to breakdown an integer to the powers of two that their sum produces the number and create a String with those values. Using our function, we can do the following:

final String output = splitToPowersOfTwo(flags).mapToObj(Integer::toString).collect(Collectors.joining(", "));

Note: an `IntStream`

is not the same as a `Stream<Integer>`

, we used `.mapToObj(Integer::toString)`

which calls the static method `Integer.toString(int)`

. This is different to calling `.map(Integer::toString)`

on a `Stream<Integer>`

as the latter won’t compile because it is ambiguous.

This post is also available in: Greek

This code will NOT convert the number to Base 2. It will return a stream of base 10 numbers that if you sum them up, will return the original value.

For the number 25 (base 10) it will create a stream of numbers (also base 10). The stream will contain the values [1, 8, 16].