CODEWITHSUNDEEP
javabit-manipulationbitset
ataylor
ataylor

Reputation: 66059

BitSet to and from integer/long

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.

Upvotes: 62

Views: 75984

Answers (7)

Eugene Babich
Eugene Babich

Reputation: 1689

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")

Upvotes: -1

charlie
charlie

Reputation: 1527

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.

Upvotes: 5

Grigory Kislin
Grigory Kislin

Reputation: 17980

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];

Upvotes: 50

user3799584
user3799584

Reputation: 927

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)
 }

Upvotes: 0

finnw
finnw

Reputation: 48619

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.)

Upvotes: 19

Arne Burmeister
Arne Burmeister

Reputation: 20594

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

Upvotes: 67

kukudas
kukudas

Reputation: 4924

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

Upvotes: -3

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