What does (b & 0xff) mean?

Question

With Java, I am converting a String value to a hash using SHA1 on a MessageDigest instance. I am at the point where I have created a hash object:

MessageDigest md = MessageDigest.getInstance("SHA1");
byte[] hash = md.digest(password.getBytes("UTF-8"));

The part I do not understand is what b & 0xff means in the following code:

StringBuilder sb = new StringBuilder(2*hash.length);
for(byte b : hash) {
   sb.append(String.format("%02x", b & 0xff));
}

I know that %02x means to specify a format where there are two characters using hexadecimal, but I've no idea what the second parameter is, what it does to each byte or what it means. A simple explanation would be great! :-)

Random side note: if you can use third-party libraries, this whole thing is the one line `Hashing.sha1().hashString(password, Charsets.UTF_8).toString()` with [Guava](https://code.google.com/p/guava-libraries/). — Louis Wasserman, Aug 13 '13 at 17:36

score 2 · Answer 1 · edited Aug 13 '13 at 17:23

2

That's the bitwise AND operator. b & 0xff gets the last byte of b. Since b is already a byte, I don't see any point to this: the result of the String.format is the same.

edited Aug 13 '13 at 17:23

Tim S.

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answered Aug 13 '13 at 17:21

Hovercraft Full Of Eels

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1

See the duplicate answer for why that bitwise-AND is useful in the OP's context. Also, b is a byte. – Andy Thomas Aug 13 '13 at 17:24
1

One utility of "b & 0xff" here is that it works independently of whether the format symbol is currently %02x or %03d. – Andy Thomas Aug 13 '13 at 17:54

What does (b & 0xff) mean?

1 Answers1