Convert Hex string to Byte Array for RSA Decryption

Question

I have a hex string after RSA encryption. When I convert it to a byte[], the RSA decryption gives javax.crypto.BadPaddingException: Blocktype mismatch: 0

I am using this method for conversion (got it on Stack overflow itself)

public static byte[] hexStringToByteArray(String data) {
    int k = 0;
    byte[] results = new byte[data.length() / 2];
    for (int i = 0; i < data.length();) {
        results[k] = (byte) (Character.digit(data.charAt(i++), 16) << 4);
        results[k] += (byte) (Character.digit(data.charAt(i++), 16));
        k++;
    }
    return results;
}

Any suggestions please.

Your conversion looks good. I suppose your mistake is somewhere else. (By the way, instead of "got it on Stack Overflow itself", the proper way would be to link to the relevant question/answer.) — Paŭlo Ebermann
– Paŭlo Ebermann, Commented Oct 14, 2011 at 12:41
Can you post the relevant decryption section of the code? Perhaps you're using String.getBytes() somewhere instead of this routine? — maerics
– maerics, Commented Oct 14, 2011 at 13:58

trojanfoe · Accepted Answer · 2011-10-14 15:07:29Z

1

The encryption method requires the input to be a fixed-length; you're going to have to add padding to the required length in order to avoid this exception. This size will depend on the key size.

EDIT: There is also a potential bug in your iteration of data: If its length is not divisible by two then the second i++ will cause an IndexOutOfBoundsException. You are better off incrementing i by 2 in the for loop and using [i] and [i+1] when accessing the data:

for (int i = 0; i + 1 < data.length(); i += 2, k++)
{
    results[k] = (byte) (Character.digit(data.charAt(i), 16) << 4);
    results[k] += (byte) (Character.digit(data.charAt(i + 1), 16));
}

edited Oct 14, 2011 at 15:07

answered Oct 14, 2011 at 13:43

trojanfoe

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2 Comments

maerics Over a year ago

This is true; however, typically when you instantiate the cipher it chooses a padded transform by default, so there's no need to play with madding manually. But it depends entirely on OPs code...

Paŭlo Ebermann Over a year ago

Your version of the code will give exactly the same IndexOutOfBoundsException. (It is more clear, though.)

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