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assemblyx86endiannessmov

ASM mov byte array

I need to know what the last two lines of code do,everything compiles and works as expected. My section declarations are:

SECTION .data
prompt: db      "Enter 10 digits: "
plen:   equ     $-prompt
SECTION .bss
digits: equ     10
inbuf:  resb    digits + 2
SECTION .text

The code snippet that I don't understand is: 

mov     al, ah              ; move AH to AL
add     al, '0'             ;add the ascii value of 0 (48) to al, store in al
mov     [inbuf+10], al      ;????
mov byte    [inbuf+11], 10  ;?????

Thank you

Solution

  • Since everything is a byte, there's no endianness involved.

    These lines just do:

    mov     [inbuf+10], al      ; store the digit
mov byte    [inbuf+11], 10  ; 10 (line feed) on the end

    By the way, if there are digits in inbuf + 0 through inbuf + 10, that means you have 11 of them.