MPLab 32bit Multiplication

32-bit Binary Multiplication

1. Define the model and architecture for the assembly code

.model small 
.stack 100H 
.386          ; specifies that the program will use instructions of the 80386 processor (32-bit instructions).

2. Define your data in the data section ( here you can define various data items: variables, constants, strings, arrays)

.data          ; .data directive indicates the beginning of data section 

; DATA1 and DATA2 are 32-bit variables initialized to 0. 
DATA1 dd 00000000H         
DATA2 dd 00000000H

; PROD1 and PROD2 32-bit uninitialized variables.
PROD1 dd ?           
PROD2 dd ?

;define strings that will be displayed as messages to the user while executing the program.
msg db 10,13,"Enter the First Number: $"
msg1 db 10,13,"Enter the Second Number: $"
msg2 db 10,13,"The Product (in Hexadecimal) is: $"

3. Define your instruction in the code section (here you write the assembly instructions that perform computations, control program flow, and interact with data variables and memory)

.code        ; .code directive indicates the beginning of code section 
.startup     ;  the entry point of the program

; Using DOS interrupt 21H to display the message.
MOV AH,09         
MOV DX,OFFSET msg        
INT 21H  

; Read the First Number and store it into DATA1
MOV EBX, 0     ;sets the value of the EBX register to zero
MOV CX, 8       ;sets the value of the CX register to 8  (CX is used as a loop counter)
; Using Loop
AGAIN: 
             ; Using DOS interrupt 21H to read a character
             MOV AH, 01 
             INT 21H
             
             ;compares the value in the AL register (the character just read) with the character 'A'.
             CMP AL,'A'      
            
             ; If the character in AL is greater than or equal to 'A', program jumps to Label L5. 
             JGE L5      

             ; Otherwise it subtracts 30H from the ASCII value of the character.  To convert numeric character('0' to '9') to the numerical value (0 to 9) 
             SUB AL,30H      
             
             ;After subtraction program jumps to Label L6
             JMP L6

             ;If the character in AL is an upper case letter, it subtracts 37H from the ASCII value of the character
             L5: SUB AL,37H

             ;shift the contents of the EBX register left by 4 bits.
             L6: SHL EBX,4

             ;Add the value in AL (which has been adjusted to be a numerical value) to the low byte (BL) of the EBX register.
             ADD BL,AL
             
             ;This is a loop instruction that decrements the CX register and jumps back to the label AGAIN if CX is not zero. 
             LOOP AGAIN

; EBX contents are moved to DATA1
MOV DATA1, EBX

;Read the Second Number and store it into DATA2
MOV AH,09
MOV DX, OFFSET msg1
INT 21H

MOV EBX, 0
MOV CX, 8

AGAIN1: 
             MOV AH,01 
             INT 21H
             CMP AL,'A'
             JGE L7
             SUB AL, 30H
             JMP L8
             L7: SUB AL, 37H
             L8: SHL EBX, 4
             ADD BL,AL
             LOOP AGAIN1
MOV DATA2, EBX