What does & 0xff do And MD5 Structure?

Question

import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;

public class JavaMD5 {

    public static void main(String[] args) {
        Stri         


        

Answer 1

It's a really messy way of translating to a hexadecimal string.

• & 0xFF performs a binary AND, causing the returning value to be between 0 and 255 (which a byte always is anyway)
• + 0x100 adds 256 to the result to ensure the result is always 3 digits
• Integer.toString(src, 16) converts the integer to a string with helix 16 (hexadecimal)
• Finally .substring(1) strips the first character (the 1 from step 2)

So, this is a very elaborate and obfuscated way to convert a byte to an always 2-character hexadecimal string.

Answer 2

Presumably most of the code is clear and the only mystery for you here is this expression:

(bytes[i] & 0xff) + 0x100

The first part:

bytes[i] & 0xff

widens the byte at position i to an int value with zeros in bit positions 8-31. In Java, the byte data type is a signed integer value, so the widening sign-extends the value. Without the & 0xff, values greater than 0x7f would end up as negative int values. The rest is then fairly obvious: it adds 0x100, which simply turns on the bit at index 8 (since it is guaranteed to be 0 in (bytes[i] & 0xff). It is then converted to a hex String value by the call to Integer.toString(..., 16).

The reason for first adding 0x100 and then stripping off the 1 (done by the substring(1) call, which takes the substring starting at position 1 through the end) is to guarantee two hex digits in the end result. Otherwise, byte values below 0x10 would end up as one-character strings when converted to hex.

It's debatable whether all that has better performance (it certainly isn't clearer) than:

sb.append(String.format("%02x", bytes[i]));