One small "bug". You're printing a null terminator. You have two options in "correcting" it. You can either remove all those null terminators which will make the sys_write contain the correct length, or you can subtract 1 from the length to ignore that extra null byte. I like to keep the null byte, if you ever decide to change sys_write to printf, it'll already be zero terminated.
section .data
; Messages
msg1 db 10,'-Calculator-',10,0
lmsg1 equ $ - msg1 - 1
msg2 db 10,'Number 1: ',0
lmsg2 equ $ - msg2 - 1
msg3 db 'Number 2: ',0
lmsg3 equ $ - msg3 - 1
msg4 db 10,'1. Add',10,0
lmsg4 equ $ - msg4 - 1
msg5 db '2. Subtract',10,0
lmsg5 equ $ - msg5 - 1
msg6 db '3. Multiply',10,0
lmsg6 equ $ - msg6 - 1
msg7 db '4. Divide',10,0
lmsg7 equ $ - msg7 - 1
msg8 db 'Operation: ',0
lmsg8 equ $ - msg8 - 1
msg9 db 10,'Result: ',0
lmsg9 equ $ - msg9 - 1
msg10 db 10,'Invalid Option',10,0
lmsg10 equ $ - msg10 - 1
nlinea db 10,10,0
lnlinea equ $ - nlinea - 1
section .bss
; Spaces reserved for storing the values ??provided by the user.
opc: resb 2
num1: resb 2
num2: resb 2
result: resb 2
section .text
global _start
_start:
; Print on screen the message 1
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, lmsg1
int 80h
; Print on screen the message 2
mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, lmsg2
int 80h
; We get num1 value.
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 80h
; Print on screen the message 3
mov eax, 4
mov ebx, 1
mov ecx, msg3
mov edx, lmsg3
int 80h
; We get num2 value.
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 80h
; Print on screen the message 4
mov eax, 4
mov ebx, 1
mov ecx, msg4
mov edx, lmsg4
int 80h
; Print on screen the message 5
mov eax, 4
mov ebx, 1
mov ecx, msg5
mov edx, lmsg5
int 80h
; Print on screen the message 6
mov eax, 4
mov ebx, 1
mov ecx, msg6
mov edx, lmsg6
int 80h
; Print on screen the message 7
mov eax, 4
mov ebx, 1
mov ecx, msg7
mov edx, lmsg7
int 80h
; Print on screen the message 8
mov eax, 4
mov ebx, 1
mov ecx, msg8
mov edx, lmsg8
int 80h
; We get the option selected.
mov ebx,0
mov ecx,opc
mov edx,2
mov eax,3
int 80h
mov ah, [opc] ; Move the selected option to the registry ah
sub ah, '0' ; Convert from ascii to decimal
; We compare the value entered by the user to know what operation to perform.
cmp ah, 1
je add
cmp ah, 2
je subtract
cmp ah, 3
je multiply
cmp ah, 4
je divide
; If the value entered by the user does not meet any of the above
; conditions then we show an error message and we close the program.
mov eax, 4
mov ebx, 1
mov ecx, msg10
mov edx, lmsg10
int 80h
jmp exit
add:
; We keep the numbers in the registers eax and ebx
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub al, '0'
sub bl, '0'
; Add
add al, bl
; Conversion from decimal to ascii
add al, '0'
; We move the result
mov [result], al
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 2
int 80h
; We end the program
jmp exit
subtract:
; We keep the numbers in the registers eax and ebx
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub eax, '0'
sub ebx, '0'
; Subtract
sub al, bl
; Conversion from decimal to ascii
add al, '0'
; We move the result
mov [result], al
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
multiply:
; We store the numbers in registers ax and bx
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub al, '0'
sub bl, '0'
; Multiply. AL = AX x BX
mul bl
; Conversion from decimal to ascii
add ax, '0'
; We move the result
mov [result], ax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
divide:
; We store the numbers in registers ax and bx
mov al, [num1]
mov bl, [num2]
mov dx, 0
mov ah, 0
; Convert from ascii to decimall
sub al, '0'
sub bl, '0'
; Division. AL = AX / BX
div bl
; Conversion from decimal to ascii
add ax, '0'
; We move the result
mov [result], ax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
exit:
; Print on screen two new lines
mov eax, 4
mov ebx, 1
mov ecx, nlinea
mov edx, lnlinea
int 80h
; End the program
mov eax, 1
mov ebx, 0
int 80h
