>
> >Rotating both bytes together like that is a common technique, and I've seen
> >it used frequently, but it's almost always better to use a routine to shift
> >out eight bits at a time
>
> I won't presume to argue with you (my degree is in biology and I'm sure you
> know more about what you're talking about than I do), but I would like to
> know why you make this argument.  What difference does it make?

I'm sorry if I sounded too pedantic; in cases where execution speed is not
an issue, a multi-byte shift seems fine.  Unfortunately, many people who get
used to doing multi-byte shifts in non-time-critical things sometimes end up
doing them in more time-sensitive things (such as multi-precision multipli-
cation or division) and I wanted to ensure that people were well aware that
multi-byte shifts are often not the best idea.  (Actually, I must [sheepishly]
admit that if one is complying with the port timing requirement of leaving a
cycle between operations on different bits, doing a double-shift imposes only
a little extra cost if done in a once-unrolled loop:

ShiftEm:
        bsf     NumBits,3       ; Assumes NumBits was zero
Loopy:
        bcf     DATA
        btfss   Byte1,7
         bsf    DATA
        rlf     Byte0
        bsf     CLOCK
        bcf     CLOCK
        rlf     Byte1
        bcf     DATA
        btfss   Byte1,7
         bsf    DATA
        rlf     Byte0
        rlf     Byte1
        bsf     CLOCK
        bcf     CLOCK
        decfsz  NumBits
         goto   Loopy