I am new to assembly language programming and I wrote a small program to print the integer using sys_write system call. Here's my code :
section .data
N: dw 216
chr: dw ,0,0,0,0x0a
section .bss
section .text
global _start
_start:
xor ax, ax
mov ax, word [N]
mov cx, 10
mov ebx,4
shift_while: div cx
add dx, 0x0030
mov word [chr+ebx],dx
sub ebx, 2
xor dx, dx
cmp ax, 0
jne shift_while
call printchar
exit: mov eax, 1
mov ebx, 0
int 80h
printchar: pushad
mov eax, 4
mov ebx, 1
mov ecx, chr
mov edx, 8
int 80h
popad
ret
I have hard coded 216, the number to be printed and I am getting the correct output. However what I am bemused by is the "mov word [chr+ebx],dx" instruction. dx contains 0x0032 in the first iteration so at the address [chr+ebx] this value should be stored as 32 00 (hex). But when I examined chr memory using gdb, it showed:
(gdb) x /5hx 0x80490d2
0x80490d2 <chr>: 0x0032 0x0031 0x0036 0x000a
what I expected was 0x3200 0x3100 0x3600 x0a00 and thought I'd have to do further memory manipulation to get the right result. Am I going wrong somewhere with this. Are there things I can't seem to see. I'd really appreciate a little help here. This is my first first post on stackoverflow.
It's just a representation thing - what you have in memory from a byte-wise perspective is
32 00 31 00 26 00 0a 00
but when you view this as 16 bit values it's
0032 0031 0026 000a
Similarly, if you viewed it as 32 bit values it would be:
00310032 000a0026
Such is the weirdness of little endianness. ;-)
gdb
is helping you out here.
You asked for the h
(halfword
) format, on a little-endian platform, so it is decoding the memory as 16-bit little endian-values for you.
If you use the b
format instead, you'll see something more like you expected.
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