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Simple md5 question - leading zeros

Posted on 2004-09-19
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Last Modified: 2008-01-16
hi,

I have come across this code at many places on net

StringBuffer hexString = new StringBuffer();
for (int i=0; i < digest.length; i++) {
    String hex = Integer.toHexString(0xFF & digest[i]);
    if (hex.length() == 1) {
        hexString.append("0" + hex);
    } else {
        hexString.append(hex);
    }
}

my question is -- what's the use of doing
0xFF & digest[i]

Thnx...
0
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Question by:kumvjuec
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8 Comments
 
LVL 86

Expert Comment

by:CEHJ
ID: 12096622
That would be to 'unsign' the byte value, which of course is a signed type
0
 
LVL 3

Author Comment

by:kumvjuec
ID: 12096633
thnx cehj, can u explain li'l more...

how can a byte value be signed.. and how this operation will make difference
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12096856
It's actually not only to unsign it - it's to ensure the value does not exceed 0xFF after being passed to Integer.toHexString. Let's assume that digest[i] is itself 0xFF (-1)

Integer.toHexString(0xFF)

would produce

"ffffffff"

because of 'sign extension'. ANDing 0xFF with 0xFF would produce as a hex int

000000FF, and when that int is passed to Integer.toHexString, you'd get

"ff", which is what you want
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LVL 3

Author Comment

by:kumvjuec
ID: 12096928
cehj,
can u also give me a code to demonstrate this. I tried
"System.out" ing the Integer.toHexString(0xff) and Integer.toHexString(0xff & 0xff)
but both are printing out the same thing
i.e 0xff

Thanks,
0
 
LVL 3

Author Comment

by:kumvjuec
ID: 12096934
sorry, they are printing out ff instead.
0
 
LVL 86

Accepted Solution

by:
CEHJ earned 125 total points
ID: 12097008
That's because you're using an integer literal instead of a byte and those values will not be sign extended. Try this

byte b = (byte)0x80; // (-128)
System.out.println(Integer.toHexString(b));
0
 
LVL 3

Author Comment

by:kumvjuec
ID: 12097020
thnx cehj
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12097062
8-)

Rather than calling Integer.toHexString, you may find the following more efficient:

            System.out.println(byteArrayToHexString(new byte[] { (byte)0xca, (byte)0xfe, (byte)0xba, (byte)0xbe}));

...........

            public static String byteArrayToHexString(byte[] rawBytes) {
                  String charIndex = "0123456789abcdef";
                  StringBuffer sb = new StringBuffer(rawBytes.length * 2);
                  for (int i = 0; i < rawBytes.length; i++) {
                        int ix = (rawBytes[i] >> 4) & 0xF;
                        sb.append(charIndex.charAt(ix));
                        ix = rawBytes[i] & 0xF;
                        sb.append(charIndex.charAt(ix));
                  }
                  return sb.toString();
            }
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