At 1999-11-04 17:58, Ben Yenko-Martinka wrote:
>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?
The char gets promoted to int before being shifted, so I suppose it
doesn't matter which one you use.
-- Ashley Yakeley, Seattle WA Almost empty page: <http://semantic.org/>
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