Can't seem to generate hex-encoded string from SHA256 digest in Java

Question

Can't seem to figure out where Im going wrong here:

 private static String generateHashFromFile(String filePath) {
  try {
   final int BUFSZ = 32768;
   MessageDigest sha = MessageDigest.getInstance("SHA-256");
   FileInputStream in = new FileInputStream(filePath);
   BufferedInputStream is = new BufferedInputStream(in, BUFSZ);
   byte[] buffer = new byte[BUFSZ];
   int num = -1;
   while((num = is.read(buffer)) != -1) {
    sha.update(buffer, 0, num);
   }
   is.close();
   byte[] hash = sha.digest();
   return byteArrayToHex(hash);
  } catch (NoSuchAlgorithmException e) {
   // TODO Auto-generated catch block
   e.printStackTrace();
  } catch (FileNotFoundException e) {
   // TODO Auto-generated catch block
   e.printStackTrace();
  } catch (IOException e) {
   // TODO Auto-generated catch block
   e.printStackTrace();
  }
  return null;
 }

 private static String byteArrayToHex(byte[] barray)
 {
     char[] c = new char[barray.length * 2];
     byte b;
     for (int i = 0; i < barray.length; ++i)
     {
         b = ((byte)(barray[i] >> 4));
         c[i * 2] = (char)(b > 9 ? b + 0x37 : b + 0x30);
         b = ((byte)(barray[i] & 0xF));
         c[i * 2 + 1] = (char)(b > 9 ? b + 0x37 : b + 0x30);
     }
     return new String(c);
 }

Im getting strings like:

")469.76F5941+31E25)6,9,C26)978)4*917180A4C(B7C,E,D+6,7133C705167"

Clearly not hexadecimal!

Questions:

1. Is the hash generation code correct?
2. Is the hex encoding method correct?
3. Am I missing something with regards to encoding?

Answer 1

A byte value is signed, and you are using the signed-preserving right-shift. This will result in negative values for b when computing the high-order nybble.

For example, consider what your code does with a byte value -112 (0x90). When right-shifting, it is first promoted to an int value, 0xFFFFFF90. Then it is shifted right 4 bits, preserving the sign, and becomes 0xFFFFFFF9. This is then cast back to a byte, which simply discards the high-order 24 bits, and 0xF9 (-7 decimal) is assigned to b. b is not greater than 9, so the resulting character is (-7 + 48), or ')'.

Do this instead:

int hi = (barray[i] & 0xF0) >>> 4, lo = barray[i] & 0xF;

The use of a byte as a local variable doesn't do any good on a 32- or 64-bit machine. In fact, the cast to byte is a wasted instruction.

Answer 2

Both your hash generation and hex encoding code works. I would take a closer look at the contents of the file you're reading.

You can also write your hex encoding method like so:

public static String byteArrayToHex(byte[] barray) {
 StringBuffer sb = new StringBuffer();
 for (int i = 0; i < barray.length; i++) {
     String hex = Integer.toHexString(0xff & barray[i]);
     if (hex.length() == 1) sb.append('0');
     sb.append(hex);
 }
 return sb.toString();
}

Answer 3

Your hash generation code is correct. To convert byte[] barray into a hex string, you can simply do:

String c = new String();
for(short i = 0; i < barray.length; i++) {
    c += Integer.toString((barray[i] & 255) + 256, 16).substring(1).toUpperCase();
}