CODEWITHSUNDEEP
assemblyx86stdoutnasm
kubas500
kubas500

Reputation: 33

NASM - adding two numbers (one from stdin, another hardcoded)

I am newbie in NASM. I want to add two numbers one from stdin and second hardcoded and after print result on the screen. But the result i got is question mark (�). Here is code:

section .bss
    buf:   resb    1
    res:   resb    1
section .text
    global  _start

_read:
    mov     eax,    3       ; sys_read
    mov     ebx,    0       ; stdin
    mov     ecx,    buf    ; buffer
    mov     edx,    1       ; read byte count
    int     80h

_adding:   
    add ecx, 10
    sub ecx, '0'
    mov [res], ecx 

_write:    
    mov     eax,    4       ; sys_write
    mov     ebx,    1       ; stdout
    mov     ecx,    res    ; buffer
    mov     edx,    1       ; write byte count
    int     80h

_exit:
    mov     eax,    1       ; exit
    mov     ebx,    0       ; exit status
    int     80h

I tryed like samples shows, adding like this:

_adding:
    sub ecx, '0'
    mov eax, '1'            ; put 1
    sub eax, '0'
    add eax, ecx             ; adding 1+'user input'
    add eax, '0'
    mov [res], eax

but with that i get nothing in output. For any help thank you.

Upvotes: 0

Views: 3062

Answers (3)

Welcor
Welcor

Reputation: 3083

Here is a working example of your code. It works only, when the result is <10 aka when your result is only 1 digit long. 

section .bss
    buf:   resb    1
    res:   resb    1
section .text
    global  _start
_start:             ; you made the label _start global, 
                    ; but you forgot to add the label.
_read:
    mov     eax, 3      ; sys_read
    mov     ebx, 0      ; stdin
    mov     ecx, buf    ; buffer (memory address, where read should save 1 byte)
    mov     edx, 1      ; read byte count
    int     80h

_adding:   
    mov     cl, [buf]   ; copy 1 byte (the ASCII-char) from address buf into cl
    sub     cl, '0'     ; same as "sub cl, 30h"; changes ASCII number into binary number. (This is optional)
    add     cl, 1       ; it will not work, when the result is >9! 
                        ; Because then you get 2 digits
    add     cl, '0'     ; convert binary number back to ascii-char. 
                        ; (This is optional)
    mov     [res], cl   ; you could use buf instead of res, too.

_write:    
    mov     eax,    4           ; sys_write
    mov     ebx,    1           ; stdout
    mov     ecx,    res         ; buffer
    mov     edx,    1           ; write byte count
    int     80h

_exit:
    mov     eax,    1           ; exit
    mov     ebx,    0           ; exit status
    int     80h

Upvotes: 1

Jester
Jester

Reputation: 58762

You forgot to even use the value you read. It's in [buf] of course. Also, it's a byte not a dword (although the latter accidentally happens to work in this case). Furthermore, the simple way you are trying only works for single digit numbers, so adding 10 is already doomed. Also, you adjust the input by subtracting '0' but don't add that back for output (like it's done in your example at the bottom).

Upvotes: 0

Mark Lakata
Mark Lakata

Reputation: 20818

It seems you are reading ASCII from the terminal (1 character) then adding 10 to that one character and subtracting '0' (0x30). What character are you entering? If you are entering a digit ('0' to '9' which is 0x30 to 0x39), the result will be the ASCII code of 0x0A to 0x13, which are unprintable control characters.

Try changing the add acx,10 to add acx,1 and remove the line that has sub ecx,'0'. If you enter '0', you should get '1'.

If that doesn't work, then you have additional bugs in your IO. I don't know enough about int 80h to help with that.

Upvotes: 0

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