Assembly: LC3 Division

Question

I'm very new to the assembly language and in class we're working on a division program using the LC3 simulator. Below is my code for my division algorithm.

DIVISION:

AND     R3, R3, 0   ; Zero out R3 /This is the remainder
AND R4, R4, 0   ; Zero out R4 /This is the quotient

NOT     R3, R2      ; Takes the inverse of 2nd input ->R3
ADD     R3, R3 #1   ; Add one to the inverse (for 2s comp)

LOOPD
ADD     R4, R4, #1  ; Add 1 to R4 repeatedly
ADD     R1, R1, R3  ; Subtract input2 from R1
BRN     NEGATIVE
BRZ     ZERO
BRP     LOOPD

NEGATIVE
ADD     R4, R4, #-1
ADD     R3, R1, R2

; Done with divison algorithm.
ZERO
LD  R0, DECCONV     ; Load Decimal converter
ADD     R3, R3, R0  ; Convert back to ASCII
ADD     R4, R4, R0  ; Convert back to ASCII

ST  R3, REMRESULT   ; Store the remainder result
ST  R4, DIVRESULT   ; Store the division result.

LD  R0, DIVRESULT   ; Load Division result into R0
PUTC            ; Print it.
LEA     R0, DIVSTRING   ; Load the string for division.
PUTS            ; Print the string.

LD  R0, REMRESULT   ; Load Remainder result into R0
PUTC            ; Print it.
LEA     R0, REMSTRING   ; Load the string for remainder
PUTS            ; Print the string.

When I enter two inputs, for example: 4 and 2. I get 2 for the quotient and 1 for the remainder. When I enter 9,3 I get 6 remainder 1..

Any ideas?

Answer 1

Use the simulator to single step the code and find where it goes wrong. That said, it's obvious that the ZERO case still has the remainder in R1 so using R3 is wrong. I recommend you keep the remainder in R1 for the NEGATIVE block too and adjust the final part of the code for this.

The quotient should have been fine, I don't know why you got 6 for the 9/3 (if you really did).