[Gmecm] 68HC11 assembler help needed

[Rob Handley]

Aargh. My brain hurts. Been bashing it against the $58 code at $F650 - which 
appears to be a routine to multiply a 16 bit number by an 8 bit number. But 
my worked example doesn't work - so my understanding of the code must be 
wrong.

In my worked example, I set the 16 bit number (pointed to by X register) to: 
571 (MSB = 2, LSB = 59), and the 8 bit (stored in A register) to: 45.

Here's the code, with my worked example in [square brackets]:

; start of routine

F650:  PSHX
F651:  PSHA

F652:  LDAB 1,X     ; (8 bit) B = contents of (X+1) = LSB of incoming 16 bit 
quantity [B = 59]
F654:  MUL            ; (16 bit) D = A * B [D = 45 * 59 = 2655 = $0A5F so 
MSB(A) = 10 & LSB(B) = 95]
F655:  ADCA #$00  ; A = A + 0 + carry [A = 10 + 0 + 0 = 10]

In D, we now have the 1st of the 2 part calculation - the result of the LSB 
of the 16 bit quantity multiplied by the 8 bit quantity. The MSB (A) is now 
saved (on the stack), but the LSB (B) seems to be lost (this can't be 
right):

F657:  PSHA          ; Copy (8 bit) A to stack [10]

F658:  LDAA 0,X     ; (8 bit) A = contents of (X+0) = MSB of incoming 16 bit 
quantity [A = 2]
F65A:  TSX             ; (16 bit) X = stack pointer+1 (so X is address of 
last byte on stack)
F65B:  LDAB 1,X     ; B = contents of (X+1) i.e. last but one byte on stack 
= incoming value of A [B = 45]
F65D:  MUL            ; (16 bit) D = A * B [D = 2 * 45 = 90 so MSB(A) = 0 & 
LSB(B) = 90]
F65E:  ADDB 0,X     ; B = B + contents of (X+0) i.e. relative stack address 
(-3) = earlier value of A [B = 90 + 10 = 100]
F660:  ADCA #$00   ; A = A + 0 + carry [A = 0 + 0 + 0 = 0]

At this point we should have 100 in A and 95 in B, to make D = 25695 = 571 * 
45, but I make the above: 0 in A and 100 in B. Help!

F662:  PULX
F663:  PULX
F664:  RTS

; end of routine

Can someone please tell me where I'm going wrong!

Robin