How to convert short array to BitSet with 2 bit precision in Java?-CodePudding

I have a short array {0,2,3,1,…} which I would like to convert to a BitSet.

Expected bits in bitset: 00 10 11 01 …

Every two bit in the bitset should represent a short. (2-bit precision) This should work fine for short values (0,1,2,3).

I know that I can use ByteBuffer and BitSet to access the bits of the numbers but those are formatted into 2 bytes (16 bit). I assume I need to bitshift the values to access the correct bits but I don't know how.

int nBit = 0;
BitSet result = new BitSet();
for (int i = 0; i < numbers.length; i  ) {
  ByteBuffer buffer = ByteBuffer.allocate(2);
  buffer.putShort(number);
  BitSet bits = BitSet.valueOf(buffer.array());

  result.set(nBit  , bits.get(?));
  result.set(nBit  , bits.get(?));
}

It there maybe an easier way?

CodePudding user response：

My current pragmatic solution which may be okay since only 4 values need to be handled:

public static BitSet transformToBitSet(short[] numbers) {
        BitSet data = new BitSet();
        int nBit = 0;
        for (int i = 0; i < numbers.length; i  ) {
            short number = numbers[i];
            switch (number) {
            case 0:
                data.set(nBit  , false);
                data.set(nBit  , false);
                break;
            case 1:
                data.set(nBit  , true);
                data.set(nBit  , false);
                break;
            case 2:
                data.set(nBit  , false);
                data.set(nBit  , true);
                break;
            case 3:
                data.set(nBit  , true);
                data.set(nBit  , true);
                break;
            default:
                throw new RuntimeException("Invalid number encountered. Can't map values >4");
            }
        }
        return data;
    }

CodePudding user response：

Here is one way to do it.

set the desired bitLength (used for padding on the left with zero bits);
initialize the BitSet index to 0.
The iterate over the values.
convert each to a bit string.
pad on the left to achieve the desired bit length
then set the bits based on the numeric value of the bit

BitSet b = new BitSet();
short[] vals = { 0, 2, 3, 1 };


int bitLength = 2;
int idx = 0;
for (short v : vals) {
    String bits = Integer.toBinaryString(v);
    bits = "0".repeat(bitLength - bits.length()) bits;
    for (char c : bits.toCharArray()) {
        b.set(idx  , c - '0' == 1);
    }
}
System.out.println(b)

prints

2,4,5,7