# 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.

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

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

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.)

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.

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)
}
```

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