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