> data in of the second, etc...  tie all the clock and load lines together,
> and rotate all three bytes out on the pin.  Something like this:
>
>   mov COUNTER, #24               ;we're going to rotate 24 bits
> LOOP
>   rl  BYTE3                      ;rotate byte 3.  MSB goes into carry bit.
>   rl  BYTE2                      ;carry goes into LSB, MSB goes into carry
>   rl  BYTE1                      ;same idea
>   movb C,DATAOUTPIN              ;move carry bit to the output pin
>   setb CLOCKPIN                  ;strobe the clock pin
>   clrb CLOCKPIN
>   djnz COUNTER,LOOP              ;repeat for all 24 bits
>   setb LOADPIN                   ;strobe the load pin
>   clrb LOADPIN
>
> This is written in Parallax's version of the instruction set but shouldn't
> be too hard to follow.  the 'djnz' instruction is decrement and jump if not
> zero, and like many of the mnemonics compiles into several opcodes.

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 (note: the seemingly-odd placement of the "rlf" is
to ensure that there is at least one instruction between the writes to CLOCK
and DATA.  Without such a delay, capacitive loading on the line could cause
the program to malfunction--especially at higher crystal speeds.

ShiftW:
        movwf   Byte1
ShiftOne:
        bsf     Counter,3
ShiftLp:
        bcf     DATA
        btfsc   Byte1,7
         bsf    DATA
        rlf     Byte1,f
        bsf     CLOCK
        bcf     CLOCK
        decfsz  Counter
         goto   ShiftLp
        return

ShiftAll:
        clrf    Counter
        call    ShiftOne
        movf    Byte2,w
        call    ShiftW
        movf    Byte3,w
        call    ShiftW
        bsf     LATCH
        bcf     LATCH
        return