Cryptographic Applications - Introduction to Cryptography with Java Applets - page 329

public byte[] digestOf(byte[] msg) {
//Convert message to BigInteger-easier to work with
//Ensure the BigInteger is positive using BigInteger(int signum,byte[] b)
//constructor
BigInteger msgInt=new BigInteger(1,msg);
//b is bitlength of msg
BigInteger b=BigInteger.valueOf(msgInt.bitLength());
//n is largest multiple of 16 not exceeding bitlength of modulus
int n=modulus.bitLength()/16*16;
//Check that msg is not too large for use with MASH2
if (b.compareTo(two.pow(n/2))>0) throw new IllegalArgumentException
(“Message is too large”);
//Pad msg with enough zeros to make it a multiple of n/2
int amountToShift=msgInt.bitLength()%(n/2)==0?0:(n/2)-
msgInt.bitLength()%(n/2);
msgInt=msgInt.shiftLeft(amountToShift);
//Define variable for 2 raised to n power
BigInteger twon=two.pow(n);
//Define initialization vector H
BigInteger H=BigInteger.valueOf(0);
//Define n bit binary numeric constant A=11110000...0000
BigInteger A=BigInteger.valueOf(15).multiply(two.pow(n-4));
//Process the first t blocks
int t=msgInt.bitLength()/(n/2);
BigInteger prevH;
for (int i=0;i<t;i++) {
prevH=H;
H=BigInteger.valueOf(0);
//Process the 4 bit nybbles-there are n/8 of them
BigInteger rem;
for (int j=n/2-4;j>=0;j-=4) {
//Each byte begins with 1111B
H=H.shiftLeft(4).or(fifteen);
//Shift msg to right and keep last 4 bits
rem=msgInt.shiftRight(j+n/2*(t-1-i)).mod(sixteen);
//Append this remainder to H
H=H.shiftLeft(4).or(rem);
}