Convert Big Integer value to eight bit bytes(2s complement big endian) sequence which is multiple of 8 in Java

Question

How to convert big integer to the following byte array form in Java:

Big Integers are encoded as a sequence of eight-bit bytes, in two's complement notation, transmitted big-endian. If the length of the sequence is not a multiple of eight bytes, then Big Integers SHALL be padded with the minimal number of leading sign-extended bytes to make the length a multiple of eight bytes.

This is to be in terms with KMIP protocol, section 9.1.1.4 Item Value

Answer 1

As far as I know, there is no padding functionality provided by the BigInteger API, so you have to do the padding yourself:

For a BigInteger bigInt, use

byte[] array = bigInt.toByteArray();
int len = array.length, len8 = len+7 & ~7;
if(len != len8) {
    int pad = len8 - len;
    byte[] nArray = new byte[len8];
    if(bigInt.signum() < 0) Arrays.fill(nArray, 0, pad, (byte)-1);
    System.arraycopy(array, 0, nArray, pad, len);
    array = nArray;
}

• First, use toByteArray() to get a byte array
• calculate the next multiple of eight for the array length
• If this number is not identical to the length, you need padding
  • allocate an array of the required size
  • fill the padding with -1 (sign extension) when negative (it already has the required zeros in the other case)
  • copy the original bytes

Note that a sign extended padded array still is compatible to the BigInteger(byte[]) constructor, so an assert bigInt.equals(new BigInteger(array)); after the operation should never fail.