On Mon, 17 Jul 2000, Chris Eddy wrote:

> Folks, I have a very long data string that I want to shift in, to the
> tune of 90 or more bits.  I want to shift the data in (or if possible
> use a pointer to a bit with increment) in C.  (You assy folks sit back
> down, I know it is fairly easy).  I can implement longs and use the left
> shift operator, but how can I avoid carrying the one bit with a pack of
> code just for that purpose?  If I just do a bunch of rotates, I think C
> will start with a 0 in carry for the roll operation every time.
>
> Any grand ideas?
>
> Chris Eddy

The assembly version is just too easy. For 90 bits, you can do it in 12
instructions without resorting to optimization from Dmitry.


The C version is harder. But here is something for bit indexing that is more
generic.


void set_bit(unsigned char *bit_array, char bit_position, char new_state)
{


  if(new_state)
    bit_array[bit_position >> 3] |= ( 1 << (bit_position & 7) );
  else
    bit_array[bit_position >> 3] ^= ~( 1 << (bit_position & 7) );

}

I bet the compiler generates more than 12 instructions for this!


but I reckon you can do it in assembly like so:

set_bit:

  ;Get a pointer to the byte containing the bit we want:

    rrf   bit_position,w
    movwf fsr

    rrf   fsr,f
    rrf   fsr,w
    andlw BIT_ARRAY_MASK
    addlw bit_array
    movwf fsr

  ; find 1<<(bit_position&7) [note, this could be made 2-cycles
  ;                           shorter]

    movlw  0101b
    btfsc  bit_position,0
     movlw 1010b

    movwf  mask

    movlw  0011b
    btfsc  bit_position,1
     movlw 1100b

    andwf  mask,f

    btfsc  bit_position,2
     swapf mask,f

  ; now set or clear the bit

    movf   mask,w
    iorwf  indf,f         ; Assume we're setting.
    btfss  new_state,0
     xorwf mask,w         ; assumption was wrong


Scott

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