SHA1 Hash on Hex String

Question

I'm trying to hash a number, represented by hex string with Java security library. Meaning, If I have the String "AABBCCDD" I want to hash it not as this is an ascii input, which is 0x65, 0x65, 0x66, 0x66, 0x67, 0x67, 0x68, 0x68, but as four bytes - 0xAA, 0xBB, 0xCC, 0xDD. I managed to do it with low values such as "112233445566" (since bytes are signed in Java) but failed with high values.

Does someone know how to implement such thing?

Thanks, Binyamin

These isn't a method that take byte[] as input? I'm not quite sure I understand what the problem is. — user166390
– user166390, Commented Dec 2, 2010 at 7:39
Yes it is. I assume I made some mistakes in my implementation before. — MByD
– MByD, Commented Dec 2, 2010 at 8:10

Community · Accepted Answer · 2017-05-23 12:15:42Z

6

First convert your hex into byte[] using for example this: Convert a string representation of a hex dump to a byte array using Java?

After it use

byte[] data = hexStringToByteArray(hexData);
MessageDigest md = MessageDigest.getInstance("SHA-1");
md.update(data, 0, data.length);
byte[] sha1hash = md.digest();

edited May 23, 2017 at 12:15

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answered Dec 2, 2010 at 7:43

gleber

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4 Comments

MByD Over a year ago

I actually already implemented hexStringToByteArray() before for RC4 implementation, but for some reason I forgot it exists and tried to solve it in a different (way more complicated and inefficient way). Thanks.

Jon Skeet Over a year ago

Why are you creating a byte array and then effectively discarding it? Just declare sha1Hash on the final line when you call md.digest().

Jon Skeet Over a year ago

(Fixed the odd use of text.length() too.)

gleber Over a year ago

Indded, fixed sha1hash blunder

Community · Accepted Answer · 2017-05-23 11:44:30Z

2

Basically you just need to find a hex parser - there are plenty around, with one example here, or Apache Commons Codec for this and other conversions. While Java bytes are indeed signed, you'll get the same bit pattern as if they were unsigned, so they'll have the same way. Unless you're performing your own arithmetic/bit-shifting on byte values, you can usually ignore the fact that bytes are signed in Java.

edited May 23, 2017 at 11:44

CommunityBot

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answered Dec 2, 2010 at 7:39

Jon Skeet

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1 Comment

MByD Over a year ago

Thanks, as I wrote to gleber(above), I already implemented your first option.

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