Answer:
;nasm 2.13.02
section .data
section .bss
number: resb 1;
section .text
global _start
_start:
;;; main
mov al,5;
mov [number],al
; encode number
mov ecx,number
call Encode
; print number
mov ecx,number
call Print
; decode number
mov ecx,number
call Decode
; add 1 to number
inc byte [number]
; encode number
mov ecx,number
call Encode
; print number
mov ecx,number
call Print
jmp end
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; encode adds '0' to the variable in ECX
Encode:
ADD BYTE [ECX],'0'
ret
Decode:
SUB BYTE [ECX],'0'
ret
Read:
mov eax,3
mov ebx,0
int 80h
ret
Print:
mov eax,4
mov ebx,1
mov ecx,number
mov edx,1
int 80h
ret
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
end:
mov eax,1
mov ebx,0
int 80h;
Step-by-step explanation:
In the main program, we first set the value of number to 5 using mov al,5 and mov [number],al. Then, we call the Encode function to add the character '0' to number. Next, we call the Print function to print the value of number on the console. After that, we call the Decode function to remove the character '0' from number. Then, we increment the value of number by 1 using inc byte [number]. Finally, we call the Encode function again to add the character '0' to number, and then call the Print function to print the updated value of number on the console.
The Encode function takes the variable address as an argument in ECX, and adds the character '0' to the variable using ADD BYTE [ECX],'0'.
The Decode function also takes the variable address as an argument in ECX, and subtracts the character '0' from the variable using SUB BYTE [ECX],'0'.
The Print function uses the mov ecx,number instruction to load the address of number into ECX, and then prints the byte value at that address using the int 80h instruction.