On Sun, 2 Dec 2001, Dwayne Reid wrote:

> Good day to all.
>
> Second attempt:
>
> F1_TIME_1, 2, 3 contain the number to be divided /2
> R1_TIME_1, 2, 3 contains the result
>
>      rrf         F1_TIME_1,W     ;/2
>      andlw       b'01110111'     ;
>      btfsc       F1_TIME_1,4     ;is nxt nyb odd?
>        addlw     5               ;add 5 to lower nyb
>      btfsc       F2_TIME_1,0     ;is nxt nyb odd?
>        addlw     0x50            ;add 5 to upper nyb
>      movwf       R1_TIME_1       ;
>
> Any thoughts?

Yeah. You haven't read my response to your first message :).

If you wish to try the approach in that snippet above, you don't need to
bother with the ANDing and adding 5. Instead, you can skip the AND and
add a -3.

For example, suppose

  F1 = 0x52

If we divide this BCD number by two, we should get 0x26. A shift right by
itself will produce 0x29. But if we subtract 3 then we end up with 0x26.
The reason this works is simple. When an odd BCD digit is binary shifted
right it will add 8 to the next BCD digit to its right. When an odd
decimal digit is divided by two, 5 is added to the digit to its right. The
difference between 8 and 5 is 3.

      clrc
      rrf         F1_TIME_1,W     ;/2
      movwf       R1_TIME_1       ;
      rrf         F1_TIME_2,W     ;/2
      movwf       R1_TIME_2       ;
      rrf         F1_TIME_3,W     ;/2

      btfsc       F1_TIME_3,4     ;is nxt nyb odd?
        addlw     -3              ;subtract 3 from lower nyb
      btfsc       F2_TIME_2,0     ;is nxt nyb odd?
        addlw     -0x30           ;subtract 3 from upper nyb

      movwf       R1_TIME_2       ;

   etc.

But this doesn't lead to any shorter code than what was posted earlier.

BTW, there was an error in the constants in the earlier post.

Scott

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