In which case it still doesn't matter because as Markus pointed out, char is
an unsigned type so the high bit isn't treated as a sign thus there is none
to get shifted with either operator except as data so & 0xF isn't necessary
for the high nibble, correct? Or is the type transparent to the shift
operator?
Quibbles 'n' Bits ...
Ashley Yakeley wrote:
> Whoops... as Ben Yenko-Martinka pointed out to me, the right-shift
> operator doesn't matter if you're doing &0xF, unless of course you wanted
> to do:
>
> hexBuffer[2] = hex[c >>> 12];
>
> --
> Ashley Yakeley, Seattle WA
> Almost empty page: <http://semantic.org/>
This archive was generated by hypermail 2.1.2 : Tue Jul 10 2001 - 17:20:54 EDT