I'm trying to modify AN2101 for 16 bit divide.
The 8 bit version works fine but the 16 bit version just give trash.
this is the code, it is in assembly only
;We have [x+0], [x+1], The result will be in dividend+0 A1h ,dividend+1 A2h, ;Lets load the data
div_s_8::
mov [dividend+0],e0h ;we want to div this
mov [dividend+1],01h ;we want to div this
;
mov [divisor+0],02h ;Div by
mov [divisor+1],00h ;Div by 2
;div8:
mov [remainder+0],00h ;initialize remainder to 0
mov [remainder+1],00h ;initialize remainder to 0 New
;New
and F,fbh ;clear carry bit in flags
mov [lcount],16 ;set loop counter to 16 New changed from 8
d8u_1:
rlc [dividend+0] ;shift MSB of dividend to LSB of remainder
rlc [remainder+0] ; continued
rlc [dividend+1] ;shift MSB of dividend to LSB of remainder
rlc [remainder+1] ; continued
;
mov [temp+0],[remainder+0] ;store remainder
mov [temp+1],[remainder+1] ;store remainder New
;
mov a,[remainder+0] ;subtract divisor from remainder
sub a,[divisor+0] ; continued
mov [remainder+0],a ; continued
;New
mov a,[remainder+1] ;subtract divisor from remainder
sub a,[divisor+1] ; continued
mov [remainder+1],a ; continued
jnc d8u_2 ;jump if result was positive
;
mov [remainder+0],[temp+0] ;restore remainder
and [dividend+0],feh ;clear LSB of dividend
;New
mov [remainder+1],[temp+1] ;restore remainder
and [dividend+1],feh ;clear LSB of dividend
;
jmp chkLcount8 ;jump to loop counter decrement
d8u_2:
or [dividend+0],01h ;set dividend LSB to 1
;New
or [dividend+1],01h ;set dividend LSB to 1
;
chkLcount8:
dec [lcount] ;decrement loop counter
jnz d8u_1 ;repeat steps if loop counter not zero
;
ret ;back to caller
;******************************************************************************
Does anyone see the error ?
Thanks
Gord
gkearns@luxcom.com
