Compare two hex strings in Java?

Question

I am implementing a simple DHT using the Chord protocol in Java. The details are not important but the thing I'm stuck on is I need to hash strings and then see if one hashed string is "less than" another.

I have some code to compute hashes using SHA1 which returns a 40 digit long hex string (of type String in Java) such as:

69342c5c39e5ae5f0077aecc32c0f81811fb8193

However I need to be able to compare two of these so to tell, for example that:

0000000000000000000000000000000000000000

is less than:

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

This is the complete range of values as the 40 digit string is actually representing 40 hex numbers in the range 0123456789ABCDEF

Does anyone know how to do this?

Thanks in advance.

James Cronen · Accepted Answer · 2010-11-23 18:59:21Z

12

The values 0..9 and A..F are in hex-digit order in the ASCII character set, so

string1.compareTo(string2)

should do the trick. Unless I'm missing something.

answered Nov 23, 2010 at 18:59

James Cronen

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3 Comments

Chad Okere Over a year ago

As long as the strings are always going to be the same length, and case.

James Cronen Over a year ago

@Chad: I'm assuming that's true since he's using a canned SHA1 algorithm.

Poindexter Over a year ago

@Chad and Tenner: Even if it's not, it's rather easy to pad length and unify the cases.

Adam · Accepted Answer · 2010-11-23 19:03:08Z

6

BigInteger one = new BigInteger("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF",16);
BigInteger two = new BigInteger("0000000000000000000000000000000000000000",16);
System.out.println(one.compareTo(two));
System.out.println(two.compareTo(one));

Output:
1
-1

1 indicates greater than -1 indicates less than 0 would indicate equal values

answered Nov 23, 2010 at 19:03

Adam

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Comments

Adrian Pronk · Accepted Answer · 2010-11-23 19:00:12Z

1

Since hex characters are in ascending ascii order (as @Tenner indicated), you can directly compare the strings:

String hash1 = ...;
String hash2 = ...;

int comparisonResult = hash1.compareTo(hash2);
if (comparisonResult < 0) {
    // hash1 is less
}
else if (comparisonResult > 0) {
    // hash1 is greater
}
else {
    // comparisonResult == 0: hash1 compares equal to hash2
}

answered Nov 23, 2010 at 19:00

Adrian Pronk

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Comments

vitaut · Accepted Answer · 2010-11-23 19:06:05Z

0

Since the strings are fixed length and '0' < '1' < ... < 'A' < ... < 'Z' you can use compareTo. If you use mixed case hex digits use compareToIgnoreCase.

answered Nov 23, 2010 at 19:06

vitaut

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