Java integer to byte array to short

Question

I am trying to convert an Integer to a byte array (array of 4 bytes) and then I will extract the last 4 bits of each byte, then I will create a short from those last 4 digits of each bytes.

I have the following code but it always prints zero.


    private static short extractShort(byte[] byteArray) {
        short s = 1;
        ByteBuffer byteBuffer = ByteBuffer.allocate(4);
        for(int i = 0; i < byteArray.length; i++) {
            byte lastFourBytes = extractLast4Bits(byteArray[i]);
            // Uses a bit mask to extract the bits we are interested in
            byteBuffer.put(lastFourBytes);
        }
       return  ByteBuffer.wrap(byteBuffer.array()).getShort();
    }


    private static byte extractLast4Bits(byte byteParam) {
        return  (byte) ( byteParam & 0b00001111);
    }

    private static byte[] intToByteArray(int i) {
        return new byte[] {
            (byte)((i >> 24) & 0xff),
            (byte)((i >> 16) & 0xff),
            (byte)((i >> 8) & 0xff),
            (byte)((i >> 0) & 0xff),
         };
    }

}

Any help will be sincerely appreciated

Is there a reason why you don't just manipulate the integer and cast it to a short? — Jannik
– Jannik, Commented Oct 7, 2019 at 16:17
You should shift input to right on each iteration of the loop — Palash Goel
– Palash Goel, Commented Oct 7, 2019 at 16:18
Jannik I am learning by building, casting covers up the magic. — Edward Banfa
– Edward Banfa, Commented Oct 7, 2019 at 17:27

WJS · Accepted Answer · 2019-10-07 17:33:55Z

1

Once you get the byte array, try this..


    short val = 0;
    for (byte b : bytes) {
       val <<= 4;
       val |= (b&0xf)
    }

If you want to do it starting from an int, you can do it like this.

      int v = 0b1110_1010_1111_1001_1010_1000_1010_0111;
      short verify = (short) 0b1010_1001_1000_0111;
      // initialize a short value
      short val = 0;

      // increment from 24 to 0 by 8's  the loop will
      // repeat 4 times.
      for (int i = 24; i >= 0; i -= 8) {
         // start by shifting the result left 4 bits
         // first time thru this does nothing.
         val <<= 4;

         // Shift the integer right by i bits (first time is
         // 24, next time is 16, etc
         // then mask off the lower order 4 bits of the right
         // most byte and OR it to the result(val).
         // then when the loop continues, val will be
         // shifted left 4 bits to allow for the next "nibble"

         val |= ((v >> i) & 0xf);
      }
      System.out.println(val);
      System.out.println(verify);

For more information on bitwise operators, check out this Wikipedia link.

edited Oct 7, 2019 at 17:33

answered Oct 7, 2019 at 16:20

WJS

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4 Comments

Edward Banfa Over a year ago

Could you please explain what's going on in the for loop ? It works very well.

WJS Over a year ago

I added some comments. If you have more questions let me know. I presume you know and & and | work with bit manipulation.

Edward Banfa Over a year ago

Yes thank you very much. How do I accept this answer ?

Gus Over a year ago

@EdwardBanfa To mark an answer as accepted, click on the check mark beside the answer to toggle it from greyed out to filled in. For more detailed information about managing answers on your question, see stackoverflow.com/help/someone-answers

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