I'm super new to assembly, and I've spent literally hours trying to figure this out. The program I'm making is supposed to take two numbers from zero to nine, then multiply them and show the correct answer (ex: 9 times 3 is 27). I need to have the 2 numbers be input from the user with the notation of "space number space number" (an example could be " 5 3"). I tried making the code without input (hardcode the numbers) and that works, but when I try to make the numbers come from input, the result is wrong. The program is supposed to take two numbers from zero to nine, then multiply them. I really can't figure out what's wrong...Any help at all would be very much appreciated.
When I enter: " 5 3" it outputs "? 4" (it should be 15)
When I enter: " 9 3" it still outputs "? 4" (it should be 27)
When I enter: "5 3" it outputs "?6"
When I enter: "9 3" it outputs "??4"
When I check var1 and var2 (by making it print them after them being read), the numbers are correct. As in, if " 5 3" is entered var1 does print 5 and var2 does print 3. Even if the spaces are changed, var1 and var2 print the correct numbers (so if I enter "5 3" var1 is still 5, and var2 is still 3).
My professor just explained what to do to get input, but I'm not 100% sure how it works. I find it odd that the output is different depending on where the spaces are, as var1 and var2 should both be accepting 2 bytes.
My code that's incorrect:
;this program multiplies two numbers input by the user, being able to solve with two digits
;the notation for the input should be space number space number (ex:" 9 3")
section .data
section .bss
var1 resb 2
var2 resb 2
result resb 2
tensdigit resb 2
onesdigit resb 2
section .text
global _start
_start:
;reads & saves user input for var1
mov eax, 3
mov ebx, 0
mov ecx, var1
mov edx, 2 ;2 bytes of info
int 80h
;reads & saves user input for var2
mov eax, 3
mov ebx, 0
mov ecx, var2
mov edx, 2 ;2 bytes of info
int 80h
mov al, [var1]
; sub al, '0' ;not sure if this is needed or not
mov bl, [var2]
; sub bl, '0'
;multiply bl by al
mul bl
; sub ax, '0'
mov [result], ax
;divide by ten
mov ax, 0 ;set the register to 0! Very important to avoid errors
mov ax, [result]
mov bx, 10
div bx ;divide by ten
mov [tensdigit], al ;this should have 1 if using 12 (2 if 24, etc)
mov ax, [result]
sub ax, 10
mov [result], ax ;move new number into result
mov al, [result]
mov bl, 10
div bl
;find the remainder
mov [onesdigit], ah
add [tensdigit], word '0' ;add 0 to ensure it's considered a number
add [result], word '0'
add [onesdigit], word '0'
;print tensdigit
mov eax, 4
mov ebx, 1
mov ecx, tensdigit
mov edx, 2
int 80h
;print onesdigit
mov eax, 4
mov ebx, 1
mov ecx, onesdigit
mov edx, 2
int 80h
;system exit
mov eax,1
mov ebx,0
int 80h;
This is my code when the numbers are hardcoded (this one multiplies the numbers correctly):
;this program adds two numbers, being able to solve with two digits
section .data
section .bss
var1 resb 2
var2 resb 2
result resb 2
tensdigit resb 2
onesdigit resb 2
section .text
global _start
_start:
;setup the variables
mov [var1], word 9
mov [var2], word 3
mov al, [var1]
mov bl, [var2]
mul bl
mov [result], ax
;divide by ten
mov ax, 0 ;set the register to 0! Very important to avoid errors
mov ax, [result]
mov bx, 10
div bx ;divide by ten
mov [tensdigit], al ;this should have 1 if using 12 (2 if 24, etc)
mov ax, [result]
sub ax, 10
mov [result], ax ;move new number into result
mov al, [result]
mov bl, 10
div bl
;find the remainder
mov [onesdigit], ah
add [tensdigit], word '0' ;add 0 to ensure it's considered a number
add [result], word '0'
add [onesdigit], word '0'
;print tensdigit
mov eax, 4
mov ebx, 1
mov ecx, tensdigit
mov edx, 2
int 80h
;print onesdigit
mov eax, 4
mov ebx, 1
mov ecx, onesdigit
mov edx, 2
int 80h
mov eax,1
mov ebx,0
int 80h;
