From: Mike Keitz
;You have to have the 2 BCD digits in the low 4 bits of
;bcdh and bcdl, with the high 4 bits all zero.
;The result is place in bin.
;Note that it could be modified to just reuse one of the
;bcd locations to store the binary result.
mov W, bcdh
mov bin, W ;bin = bcdh
clrb C
rl bin ;bin = bcdh * 2
rl bin ;bin = bcdh * 4
add bin, W ;bin = bcdh * 4 + bcdh = bcdh * 5
rl bin ;Now bin = bcdh * 5 * 2 = bcdh * 10
mov W, bcdl
add bin, W ;Finally bin = bcdh*10 + bcdl