How to get the bytes of a long?

Hi,
I need an 8-byte number for my application. First I create an int and then cast it to long. After that this long number must be represented by 8 bytes, or 64 bits, in memory. As far as I know the leading zeros in its binary representation are present, even though they mean nothing. However, I need to get those 8 bytes and put them in a byte array. And I need all 8 bytes, no matter what their values are. Can anyone tell me what the easiest way to that is? Thanks.

legriv wrote:
Thanks for the info about the negative values. One quick question though - is this final masking for printing only or do I need to mask the byte before I put it in the array? Because later this array will be used to reconstruct the number (for a SecretKey if you remember my last posting) and I'm not sure whether the right number will be produced.As I'm pretty sure I said in your thread concerning the DH generation of a SecretKey, an integer is too small to be used as DES key. You will only have at most 2^32 keys using a key based on an int and I can crack that in about 5 minutes even with my slow old dog of a computer.
I don't understand why you could only produce an int using the DH key exchange. There is nothing in the algorithm that restricts one to an int. In fact, because of overflow problems I would have thought it much more difficult to use an int than a BigInteger.
Edit: On second thoughts, using your approach there are only 2^28 keys since the DES key ignores the least significant bit of the 8 bytes of the key! My old Apple II could break that in 5 minutes!
Edited by: sabre150 on Oct 31, 2007 10:59 AM

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...The following code does not work. Work?! It does not even compile! Please consider posting code in the form of an SSCCE in future.
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