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.
movfw bcdh
movwf bin ;bin = bcdh
clrc
rlf bin, F ;bin = bcdh * 2
rlf bin, F ;bin = bcdh * 4
addwf bin, F ;bin = bcdh * 4 + bcdh = bcdh * 5
rlf bin, F ;Now bin = bcdh * 5 * 2 = bcdh * 10
movfw bcdl
addwf bin, F ;Finally bin = bcdh*10 + bcdl