XOR Hex String in JAVA of different length

Question

I have two strings

String s1="426F62";
String s2="457665";

The strings are in hex representation. I want to XOR them. XORing normally character by character gives the correct result for others except F XOR 6.(It gives 112, the answer should be 9)

Please tell me the correct way to implement it in JAVA

EDIT: Converting to int and xoring works. But how to xor when two strings are of different length.

Answer 1

Rather than XORing the Unicode representations, just convert each character into the number it represents in hex, XOR those, then convert it back to hex. You can still do that one character at a time:

public String xorHex(String a, String b) {
    // TODO: Validation
    char[] chars = new char[a.length()];
    for (int i = 0; i < chars.length; i++) {
        chars[i] = toHex(fromHex(a.charAt(i)) ^ fromHex(b.charAt(i)));
    }
    return new String(chars);
}

private static int fromHex(char c) {
    if (c >= '0' && c <= '9') {
        return c - '0';
    }
    if (c >= 'A' && c <= 'F') {
        return c - 'A' + 10;
    }
    if (c >= 'a' && c <= 'f') {
        return c - 'a' + 10;
    }
    throw new IllegalArgumentException();
}

private char toHex(int nybble) {
    if (nybble < 0 || nybble > 15) {
        throw new IllegalArgumentException();
    }
    return "0123456789ABCDEF".charAt(nybble);
}

Note that this should work however long the strings are (so long as they're the same length) and never needs to worry about negative values - you'll always just get the result of XORing each pair of characters.

Answer 2

Try this:

String s1 = "426F62";
String s2 = "457665";
int n1 = Integer.parseInt(s1, 16);
int n2 = Integer.parseInt(s2, 16);
int n3 = n1 ^ n2;
String s3 = String.format("%06x", n3);

Why are you storing hex values as strings? it'd be a much better idea to represent hex numbers as hex integers or longs.