Printing double digit numbers in LC-3

Question

Been learning about LC-3 lately and was wondering how do I go about printing a number that's bigger then 9? In this program I made it asks for width and length and multiplies the two to get the area of the shape. My problem is any output bigger then 9 it starts printing a letter or a number not close to what I wanted. How should I go about doing this? my code:

.ORIG x3000

; Reset Registers
AND R0, R0, #0
AND R1, R1, #0
AND R2, R2, #0
AND R3, R3, #0
AND R4, R4, #0
AND R5, R5, #0
AND R6, R6, #0
AND R7, R7, #0

LEA R0, numberone
PUTS
GETC
OUT
LD R3, HEXN30 
ADD R0, R0, R3
ADD R1, R0, #0


LEA R0, numbertwo
PUTS
GETC
OUT
ADD R0, R0, R3
ADD R6, R0, #0
LOOP
      ADD R2, R2, R1
      ADD R6, R6, #-1
BRp LOOP

LEA R0, MESG 
PUTS
ADD R0, R2, x0

LD R2, NEG_TEN  
ADD R2, R2, R0  
BRn JUMP    
AND R4, R4, #0  
ADD R4, R4, R2  
LD R0, ASCII_1 
OUT    
AND R0, R0, #0  
ADD R0, R0, R4  

JUMP

LD R3, HEX30 ;add 30 to integer to get integer character
ADD R0, R0, R3
OUT 

HALT ;{TRAP 25}


numberone  .stringz "\nPlease enter the length: "
numbertwo  .stringz "\nPlease enter the width: "
MESG .STRINGZ  "\n\nThe Area of the Rectangle is: "
HEXN30 .FILL xFFD0 ; -30 HEX
HEX30 .FILL x0030 ; 30 HEX
NEG_TEN .FILL #-10
ASCII_1 .FILL x0031 ; ASCII char '1'
.END

Example Output:

Please enter the length: 4
Please enter the width: 5
The area of the object is: 20

Answer 1

This code will always print 1 :

LD R0, ASCII_1 
OUT 

There's no chance for it to print a 2 like you'd want from 4 times 5.

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The next character prints as : because you have 10 in R4 , as you haven't subtracted 10 enough times in the division by repetitive subtraction. You've only subtracted 10 once and 20 needs to have 10 subtracted twice (to get 0 as the remainder).

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You should be able to see the first problem by simply reading the code there. How could loading an ascii 1 and printing it do anything other than print a 1 ? So there is some missing code there.

The latter problem is a form of off-by-one error (looping one to few iterations), which is something you should be looking for during single-step debugging. You want your division by repetitive subtraction to leave you with a remainder between 0 and 9 — definitely not 10!

Boundary conditions are very error prone. Using < when it should be <= (or vice vesa), for example, can result in an off by one error. What we do is try to use idioms to avoid these problems (like for(int i = 0; i < n; i++) ), but when an idiom isn't applicable, then you want to be suspicious of and test those boundary conditions: at the boundary, at one less, at one more.