Name: Objective Create a LC-3 program that will perform addition (or potentially

Question

Name: Objective Create a LC-3 program that will perform addition (or potentially more operations) on numbers with 4 digits or less. Problem Statement We are now familiar with the LC-3 simulator and can debug and develop asm code. For this assignment, you are to develop a LC-3 assembly program to perform math operations and display the result on the screen. A skeleton code is NOT provided to you. You are to develop your own program to accomplish this task. Feel free to refer to previous labs for ideas. The requirement for the lab is to ask the user to enter two numbers of up to four digits in length and display their sum.The expected program output is shown below. Please enter SRC1: 1234 Please enter SRC2 4321 The sum is: 5555 Two references you might find useful for this lab are the ASCII table and the LC-3 instructions guide (on Blackboard). Bonus You can gather more points by adding extra operations and building a calculator. Each additional operation is worth extra 15 points. Multiplication (Extra 15 points) You can modify your code to offer multiplication as an option. Recall that we discussed how we can multiply numbers by adding them repeatedly. You are only responsible for results with up to 4 digits. Subtraction (Extra 15 points) Modify your code to allow subtraction of two 4-digit numbers. You should allow for negative results as well. Division (Extra 15 points) Modify your code to allow for division. Recall that we can divide numbers by repeated subtraction or by using bit-shifting.

Explanation / Answer

I could do the first part of your code
Add and sub
.MODEL SMALL
.STACK 100h
.DATA
choice_msg db 13,10,'Add or Sub?',13,10,'$'
first_msg db 13,10,'Enter the first number:',13,10,'$'
second_msg db 13,10,'Enter the second number:',13,10,'$'
result_msg db 13,10,'The result is:',13,10,'$'
new_line db 13,10,'$'
val1 db ?
num1 db ?
num2 db ?
num3 db ? ; purely a buffer variable
ten db 10   
t1 db 0
t2 db 0   
result db 0
.CODE ;where the code is written
start:
mov ax, @data ;Moves the address of the variables under .DATA into ax
mov ds,ax ;moves ax into ds. the two lines allow you to display string using the 21h interrupt sequence 9
mov ah,09
mov dx, offset choice_msg
int 21h ;displays the string in choice_msg
mov ah,01
int 21h ;copies a value into the al, using subfunction 01
cmp al,'+' ;compares the entered value in with "+"
jne subtraction ;if the enterd value is "+" then it jumps to addition else it jumps to subtraction
addition:
call read ;Read the input
call endl ;output new line
mov bl, num2 ;move the value of num 2 into bl
add num1,bl ;adds num2 and num1 to form the sum1
mov al, num1 ;mov num1 to al
mov result, al ;store the result of the sum in result
call write ;write the output
jmp exit
subtraction:
call read ;Read the input
call endl ;output new line
mov bl, num2 ;move value of num2 to bl
sub num1,bl ;subtracts the value in num2 from the value in num1
mov al, num1 ;move result to a register
mov result, al ;move the result of the subtraction to result
call write ;display result with write procedure
jmp exit

;-----------------------
;procedure declarations:
proc endl
mov ah,09
mov dx, offset new_line
int 21h ;goes to next line, i.e. "enter"
ret
endp

proc read

mov ah,09
mov dx, offset first_msg
int 21h ;displays the string in first_msg
mov ah,01 ;read char
int 21h ;copies a value into the al, using subfunction 01
sub al,48 ;subtracts 48 from the vaule in the al
mov num1,al ;moves the value in the al to the variable num1
mov ah,01 ;read second char
int 21h ;copies a value into the al, using subfunction 01
sub al,48 ;subtracts 48 from the value in the al
mov num2,al ;moves the value in the al to the variable num2
mov al,num1 ;moves the value in num1 into the al
mul ten ;multiplies the value in the al by ten
add al,num2 ;adds the value in num2 to the al, to get the two-digit number
mov num1,al ;moves the two digit value into
call endl
mov ah,09
mov dx, offset second_msg ;displays the string in second_msg
int 21h   
mov ah,01
int 21h ;copies a value into the al, using subfunction 01
sub al,48 ;subtracts 48 from the value in the al
mov num2,al ;moves the new value in the al into the variable num2
mov ah,01
int 21h ;copies a value into the al, using subfunction 01
sub al,48 ;subtracts 48 from the value in the al
mov num3,al ;moves the new value in the al into the variable num3
mov al,num2 ;moves the value in num2 into the al
mul ten ;multiplies the value in the al by ten
add al,num3 ;adds the value in num3 to the al, to get a two-digit number
mov num2,al ;moves the value in the al into the variable num2
ret ;first number in num1, second in num2
endp

;The write procedure writes the decimal stored in result.
;by dividing by ten it seperates the two digits as quotient
;and remainder. Then it outputs the quotient and remainder
;in ascii form.
proc write
mov dx,offset result_msg
mov ah,09h
int 21h ;display the result_msg string
mov al,result ;move the result from add/sub to al
mov ah,00 ;initialize ah
div ten ;div al by ten, quotient is in al
;remainder is stored in ah.
mov dl,ah ;move the remainder to dl
mov t2,dl ;store the remainder in t2
mov dl,al ;move quotient into dl
add dl,48 ;add 48 to dl, to convert it to ascii
mov ah,02h ;char display interupt code
int 21h ;display char in dl register
mov dl,t2 ;move remainder to t2
add dl,48 ;convert it to ascii by adding 48
mov ah,02h ;display character in dl interupt code
int 21h ;diplays contents of dl
call endl ;output a new line
ret   

endp

exit:
mov ax, 4c00h ;This is just a failsafe exit
int 21h

END

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