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How to use scanf in NASM?

Tags:

assembly

nasm

I'm trying to figure out how to use scanf to get user input. I know to use printf: all I have to do is push the data I want to write on the screen into the stack like this:

global _main
extern _printf
extern _scanf

section .data
msg db "Hi", 0

section .text
_main:
  push ebp
  mov ebp, esp  

  push msg
  call _printf

  mov esp, ebp
  pop ebp
ret

But I can't figure out how to use scanf. Can someone please just give me the simplest possible source code you can for scanf? I really just want to put what the user types in.

I'm not used to 32bit Assembly. I've only ever used 16bit, and I know in 16bit (DOS) you can just do this:

mov ah, 3fh
mov dx, input
int 21h

input rb 100d

And whatever you type in will the placed at the address of "input."

like image 575
user1432532 Avatar asked Jun 11 '12 01:06

user1432532


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How does Scanf work in assembly?

sscanf takes as arguments a string to scan, a format string, and pointers to variables where to store the results of the scan. So, your 0 and 1 numbers would be stored on the corresponding variables, assuming the scan was successful. EAX contains the return value.

How do you take user input in NASM?

You're only reading the characters without storing them. Instead of storing into that 'input', you should store AL either directly into StudentID/InBoxID/Assignment/Version. You could take advantage of their relative positions in memory and write a single loop to fill all of them, as in a contiguous space.


2 Answers

I found this 'Programming in NASM.PDF'

; add1.asm
SECTION .data
    message1: db "Enter the first number: ", 0
    message2: db "Enter the second number: ", 0
    formatin: db "%d", 0
    formatout: db "%d", 10, 0 ; newline, nul terminator
    integer1: times 4 db 0 ; 32-bits integer = 4 bytes
    integer2: times 4 db 0 ;
SECTION .text
   global _main 
   extern _scanf 
   extern _printf     

_main:

   push ebx ; save registers
   push ecx
   push message1
   call printf

   add esp, 4 ; remove parameters
   push integer1 ; address of integer1 (second parameter)
   push formatin ; arguments are right to left (first parameter)
   call scanf

   add esp, 8 ; remove parameters
   push message2
   call printf

   add esp, 4 ; remove parameters
   push integer2 ; address of integer2
   push formatin ; arguments are right to left
   call scanf

   add esp, 8 ; remove parameters

   mov ebx, dword [integer1]
   mov ecx, dword [integer2]
   add ebx, ecx ; add the values          ; the addition
   push ebx
   push formatout
   call printf                            ; call printf to display the sum
   add esp, 8                             ; remove parameters
   pop ecx
   pop ebx ; restore registers in reverse order
   mov eax, 0 ; no error
   ret

Which is the asm version of this C function:

#include <stdio.h>
int main(int argc, char *argv[])
{
    int integer1, integer2;
    printf("Enter the first number: ");
    scanf("%d", &integer1);
    printf("Enter the second number: ");
    scanf("%d", &integer2);
    printf("%d\n", integer1+integer2);
    return 0;
}
like image 122
Preet Sangha Avatar answered Sep 22 '22 17:09

Preet Sangha


Thanks Preet. I made a simple example based on your code to illustrate the use of scanf.

Program that requests an integer and prints it out to the screen:

section .text
  global main
  extern printf
  extern scanf

section .data
  message: db "The result is = %d", 10, 0
  request: db "Enter the number: ", 0
  integer1: times 4 db 0 ; 32-bits integer = 4 bytes
  formatin: db "%d", 0

main:
  ;  Ask for an integer
  push request
  call printf
  add esp, 4    ; remove the parameter

  push integer1 ; address of integer1, where the input is going to be stored (second parameter)
  push formatin ; arguments are right to left (first  parameter)
  call scanf
  add esp, 8    ; remove the parameters

  ; Move the value under the address integer1 to EAX
  mov eax, [integer1]

  ; Print out the content of eax register
  push eax
  push message
  call printf
  add esp, 8

  ;  Linux terminate the app
  MOV AL, 1
  MOV EBX, 0 
  INT 80h 
like image 37
mateuszb Avatar answered Sep 22 '22 17:09

mateuszb