BitSet to and from integer/long

Question

If I have an integer that I'd like to perform bit manipulation on, how can I load it into a java.util.BitSet? How can I convert it back to an int or long? I'm not so concerned about the size of the BitSet -- it will always be 32 or 64 bits long. I'd just like to use the set(), clear(), nextSetBit(), and nextClearBit() methods rather than bitwise operators, but I can't find an easy way to initialize a bit set with a numeric type.

Ferri

Personally, I'd say raw bit manipulation is the way to go here. It really isn't that complicated, and as you say I don't see a simple way to get an int or long into a BitSet. – Salesperson

Answer 1

The following code creates a bit set from a long value and vice versa:

public class Bits {

  public static BitSet convert(long value) {
    BitSet bits = new BitSet();
    int index = 0;
    while (value != 0L) {
      if (value % 2L != 0) {
        bits.set(index);
      }
      ++index;
      value = value >>> 1;
    }
    return bits;
  }

  public static long convert(BitSet bits) {
    long value = 0L;
    for (int i = 0; i < bits.length(); ++i) {
      value += bits.get(i) ? (1L << i) : 0L;
    }
    return value;
  }
}

EDITED: Now both directions, @leftbrain: of cause, you are right

Nine

FYI, this is creating a bitset in little-endian order – Titled

You should really use BitSet.nextSetBit() for BitSet -> long. – Astera

Answer 2

Add to finnw answer: there are also BitSet.valueOf(long[]) and BitSet.toLongArray(). So:

int n = 12345;
BitSet bs = BitSet.valueOf(new long[]{n});
long l = bs.toLongArray()[0];

Africander

If the bitset has no "1", bs.toLongArray()[0] will throw ArrayIndexOutOfBoundsException exception! – Lianneliao

Answer 3

Java 7 has BitSet.valueOf(byte[]) and BitSet.toByteArray()

If you are stuck with Java 6 or earlier, you can use BigInteger if it is not likely to be a performance bottleneck - it has getLowestSetBit, setBit and clearBit methods (the last two will create a new BigInteger instead of modifying in-place.)

Stout

Answer 4

To get a long back from a small BitSet in a 'streamy' way:

long l = bitSet.stream()
        .takeWhile(i -> i < Long.SIZE)
        .mapToLong(i -> 1L << i)
        .reduce(0, (a, b) -> a | b);

Vice-versa:

BitSet bitSet = IntStream.range(0, Long.SIZE - 1)
        .filter(i -> 0 != (l & 1L << i))
        .collect(BitSet::new, BitSet::set, BitSet::or);

N.B.: Using BitSet::valueOf and BitSet::toLongArray is of course easier.

Astera

Answer 5

Pretty much straight from the documentation of nextSetBit

value=0;
for (int i = bs.nextSetBit(0); i >= 0; i = bs.nextSetBit(i+1)) {
 value += (1 << i)
 }

Sleuth

This will fail for a BitSet larger than 32 or 64 bits, in such a case you'll need to handle an int[] or long[] at the output. But OP explicitly doesn't care, so fair enough. Just a few minor glitches: in case of a long you should 1L << i, to prevent overflow, and an OR like value |= 1L << i is enough. – Astera

Answer 6

For Kotlin those extensions can be used.

BitSet to Int extension:

fun BitSet.toInt() =
    toLongArray()
        .getOrNull(0)
        ?.toInt()
        ?: 0

Int to BitSet extension:

fun Int.toBitSet(size: Int): BitSet {
    val byteArray = byteArrayOf(toByte())

    return BitSet.valueOf(
        if (byteArray.size == size) byteArray
        else byteArray.copyOf(size))
}

Usage example:

val intValue = 1

val bitSetValue = intValue.toBitSet(size = 2)
println("bitSetValue: $bitSetValue")

val firstValue = bitSetValue.get(0)
println("firstValue: $firstValue")

val secondValue = bitSetValue.get(1)
println("secondValue: $secondValue")

bitSetValue.set(0, true)
bitSetValue.set(1, false)
println("bitSetValue: $bitSetValue")

val newIntValue = bitSetValue.toInt()
println("newIntValue: $newIntValue")

Laddy

Answer 7

Isn't the public void set(int bit) method what your looking for?

Syzran

That sets one single bit with the index you provide. I'd like to set each bit that's set in the integer. – Ferri