On Wed, 26 Sep 2001, Regulus Berdin wrote:

Reggie, welcome back.

> Hi Scott,
>
> I'll try (untested):
>
> ;x <<= y;
>
>  clrc
>  btfsc  y,0
>   rlf   x,f         ;x = x*2
>
>  clrc
>  rlf    x,w         ;w = x*2
>  addwf  x,w         ;w = w+x = x*2+x = x*3
>  btfsc  y,1
>   addwf x,f         ;x = w+x = x*3+x = x*4
>
>  swapf  x,w
>  andlw  0xF0
>  btfsc  y,2
>   movwf x           ;x=x*8



Interesting. Mine is slightly different and is suitable for this type of
operation:


  unsigned char x;
  unsigned char y;
  unsigned char z;


  z = x << y;


  swapf  x,w
  andlw  0xf0
  btfss  y,2
   movf  x,w

  movwf  z
  btfss  y,0
   addwf z,f

  rlf    z,w
  andlw  0xfe
  addwf  z,w
  btfsc  y,1
   addwf x,f

-----------------

Dmitry, The "unsigned char" implies an 8-bit register (at least on all of
the C compilers that target a PIC that I know of). Only the lower three
bits of the shift count are examined. It looks like an additional 4
instructions would be required to verify that the upper bits are not set.


------
Now, to do the same thing in a looping snippet:

  ; if x and z are different registers
  movf  x,w
  movwf z

  comf  y,w   ; the shift count
              ; note w= 0-y+1
  rrf   z,f
loop:
  rlf   z,f
  addlw 1
  skpc
   goto loop



Scott

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