比较Java中的两个十六进制字符串

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.

Answer 1

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.

Answer 2

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

Answer 3

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
}

Answer 4

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