Hey
marc fleury wrote:
> did you know that
>
> byte[] byte1 = {1,2,3};
> byte[] byte2 = {1,2,3};
>
> yields
> byte1.equals(byte2) = false;
> and
> byte1.hashCode() != byte2.hashCode();
Of course.. since you are comparing the arrays, and not their contents.
Both equals and hashCode will be the Object defaults, which work on
memory adress of the actual object.
> blows my mind man....
>
> what is the best way to compute a byte[].hashCode()?
Iterate and add the elements together. Also depends on the distribution
of the contents. Why do you need a byte array BTW? Is this for the Xid?
Why not use an enumerated long or something?
/Rickard
--
Rickard �berg
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