On Sun, 29 Nov 1998 13:29:44 +1300 Seth Fischer <aureo@xtra.co.nz>
writes:
>What does it mean to rotate a file _through the carry bit_ (RLF and
>RRF)?

The value of the carry bit before the instruction goes into one end of
the byte being rotated.  The bit that is rotated out of the byte goes
into the carry bit.  In somewhat graphical form, we have

rlf :  C <- dddddddd <- C
rrf :  C -> dddddddd -> C

A single rrf is merely a shift.  But repeated rrf or rlf will rotate 9
bits of data, 8 being in the data byte and one in the carry bit.

>
>When executing RLF on portB (8 LEDs) I get this:
>
>portB = 00000001
>        RLF     portb
>portB = 00000011        ; I want only one bit high and 'walking' left
>        RLF     portB
>portB = 00000110
>        RLF     portB
>portB = 00001100
>
>What is happening?

The result you see is exactly what should happen if the C bit were 1 at
the start of the process.  To keep from getting more than one 1 bit in
the result, you should clear the carry bit before each rotate, except for
the case of 10000000 going back to 00000001.

This code will approximatley "work."  However I don't recommend that you
use it for anything other than an exercise.  It implements a true 8 bit
rotate by setting the C bit before shifting.  You would need to
initialize PORTB to 00000001 first.

        clrc            ;Always clear C
        btfsc   PORTB,7 ;except if a 1 is coming out
        setc            ; in which case set C.
        rlf     PORTB,f

Now this code does "work" but it has a couple of bad practices in it.
First, it is possible for patterns other than the desired ones containing
just one 1 to continually rotate around.  If a glitch causes the PORTB
pins to read back improperly (which can happen because of heavy loads on
the pins for example), some bits may get set when they aren't supposed
to.  Or, an all-zero pattern could persist.  We should use a better test
to be sure that the routine will "count out of" undesired states.  A
better method is to shift in zeros unless the value is X0000000.  Here's
a test for X0000000:

        movfw   PORTB
        andlw   b'01111111'     ;Test if 7 low bits are all zero.
        clrc                    ;Prepare to have C clear if they
aren't.
        skpnz
        setc
        rlf     PORTB

The other bad practice is reading back the output pins.  Sure it works
most of the time, but unless you just can't avoid it it is best to read
ports only to determine the state of the input pins.  This makes the
program less likely to malfunction due to transients from the external
circuitry.  Set up the output value in RAM instead and write it to the
port.  So we'd define a RAM varaible "shadowb" and do all the rotating on
it instead.  Combining that with the enhanced test above:

        movfw   shadowb
        andlw   b'00000001'
        clrc
        skpnz
        setc
        rlf     shadowb,f
        movfw   shadowb         ;Get bit pattern out of RAM
        movwf   PORTB           ;Write to PORTB.

This code can be optimized some, though it becomes more obscure.

        rlf     shadowb,w       ;W= dddddddX
        andlw   b'11111110'     ;See if need to put in a new 1.
        skpnz                   ;No, value OK.
        movlw   b'00000001'     ;Restart at 00000001
        movwf   shadowb         ;Update RAM variable
        movwf   PORTB           ;Update port value.

With these routines there isn't a need to initialize the shadowb/PORTB
values unless it is important that the output be correct for the first 8
shifts.


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