Strange algorithm to convert binary SHA-1 digest to hex string

Question

in the Internet I found this code to generate a SHA1 hash:

        public static String hash(String str) {
            try {
                    MessageDigest mg = MessageDigest.getInstance("SHA-1");
                    byte[] result = mg.digest(str.getBytes());
                    StringBuffer sb = new StringBuffer();
                    for (int i = 0; i < result.length; i++) {
                            sb.append(Integer.toString((result[i] & 0xff) + 0x100, 16).substring(1));
                    }
                    return sb.toString();
            } catch (NoSuchAlgorithmException e) {
                    System.err.println("SHA-1 not found.");
                    return "";
            }
    }

But why is there (result[i] & 0xff) + 0x100?

@pino copying code and asking here. Can't he refer official documents first. and show hhat he studied.is efforts t — Nikhil Agrawal
– Nikhil Agrawal, Commented Apr 27, 2013 at 16:33
@Nikhil: I don't agree. What official document could answer his question? Copying someone else's code is not discouraged by the FAQ, on the contrary FAQ says "We feel the best Stack Overflow questions have a bit of source code in them". He has a problem with a software algorithm he cannot understand: it's perfectly in-topic. — Pino
– Pino, Commented Apr 27, 2013 at 17:04

Pino · Accepted Answer · 2013-04-27 17:07:52Z

6

Bytes are signed: they could be negative. When a negative byte is handled by Integer.toString() generates a string beginning with "FFFFFF", but this doesn't happen with positive bytes, so the length of the resulting string is not fixed. The & 0xff converts the byte to an unsigned integer. Then 0x100 is added to ensure that the hex string is 3 chars long; this is needed because we want a string with 2 hex digits for each byte but a byte between 0 and 15 would produce 1 char only. Finally the third digit is discarded with substring(1).

I suggest to substitute StringBuffer with StringBuilder because it is slightly more efficient and also to specify the initial buffer length:

StringBuilder sb = new StringBuilder(result.length * 2);

edited Apr 27, 2013 at 17:07

answered Apr 27, 2013 at 16:40

Pino

9,5737 gold badges61 silver badges79 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Edward Falk · Accepted Answer · 2013-04-28 15:27:52Z

1

The & 0xff is there just in case byte gets promoted to something larger than 8 bits and sign-extended.

Sign-extension is a very real problem here, so the 0xff is needed for that reason at the very least.

edited Apr 28, 2013 at 15:27

answered Apr 27, 2013 at 16:16

Edward Falk

10.2k11 gold badges84 silver badges124 bronze badges

2 Comments

Pino Over a year ago

"just in case byte is more than 8 bits" sounds very bad.

Edward Falk Over a year ago

True. I never like to make assumptions about the underlying hardware, but in this case, the Java spec seems to say that a byte can never be more than 8 bits, so I've simplified my answer.

Collectives™ on Stack Overflow

Strange algorithm to convert binary SHA-1 digest to hex string

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related