CODEWITHSUNDEEP
assemblyx86nasm
user2417339
user2417339

Reputation:

Subtraction in x86 assembly

I was looking through the following code from tutorial point: 

section .text
   global _start        ;must be declared for using GCC

_start:                 ;tell linker entry point
   sub     ah, ah
   mov     al, '9'
   sub     al, '3'
   aas
   or      al, 30h
   mov     [res], ax

   mov  edx,len         ;message length
   mov  ecx,msg         ;message to write
   mov  ebx,1           ;file descriptor (stdout)
   mov  eax,4           ;system call number (sys_write)
   int  0x80            ;call kernel

   mov  edx,1           ;message length
   mov  ecx,res         ;message to write
   mov  ebx,1           ;file descriptor (stdout)
   mov  eax,4           ;system call number (sys_write)
   int  0x80            ;call kernel

   mov  eax,1           ;system call number (sys_exit)
   int  0x80            ;call kernel

section .data
msg db 'The Result is:',0xa
len equ $ - msg
section .bss
res resb 1

I feel like I understand this code except for the line:

or al, 30h

I understand that the or is a bitwise or and 30h is 0011 0000 in binary. I don't understand why this is needed for the code to work though! Can someone explain this to me?

Upvotes: 1

Views: 8275

Answers (1)

Gene
Gene

Reputation: 46960

This isn't a great example.

After the subtraction, al contains 6.

The aas instruction does nothing because the high nibble of al is zero.

The ASCII code for 0 is 30h. "Or"ing this with the 6 produces 36h, which is ASCII "6". In general, this converts the binary value for a decimal digit to its ASCII code.

Sounds like you'll benefit from studying the difference between binary values and their ASCII representations. See for example an ASCII table.

But also note, this code is (AFAICS) erroneous because it stores a 16-bit word into a single reserved byte.

Upvotes: 1

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