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Home/ Questions/Q 7854247
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T19:59:40+00:00 2026-06-02T19:59:40+00:00

Using FASM I’m trying to shift dword with a SHR instruction which should shift

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Using FASM I’m trying to shift dword with a SHR instruction which should shift bits to right appending zeroes. Here is the code:


format ELF executable 3

entry start

segment readable executable

start:
    cmp byte [counter],0
    jz fin

    mov eax,4
    mov ebx,1
    mov ecx,some
    mov edx,4
    int 0x80 ; write 4 bytes

    mov eax,4
    mov ebx,1
    mov ecx,splitter
    mov edx,1
    int 0x80 ; write a split byte (0x42)

    shr dword [some],4

    dec byte [counter]
    jmp start
fin:
    mov eax,1
    xor ebx,ebx
    int 0x80

segment readable writeable

some db 0x89,0xAB,0xCD,0xEF
splitter db 0x42
counter db 10

I expected to get something like that: 0x89ABCDEF -> 0x089ABCDE -> 0x0089ABCD -> 0x00089ABC -> 0x000089AB -> 0x0000089A -> 0x00000089 -> 0x00000008 -> 0x00000000
But I get this instead: 0x89ABCDEF -> 0xB8DAFC0E -> 0xABCDEF00 -> 0xDAFC0E00 -> 0xCDEF0000 -> 0xFC0E0000 -> 0xEF000000 -> 0x0E000000 -> 0x00000000

What am I doing wrong?

P. S. This values were gotten by opening the file in which I redirect the output in hex editor. (It’s not the output itself)

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  1. Editorial Team
    Editorial Team
    2026-06-02T19:59:41+00:00Added an answer on June 2, 2026 at 7:59 pm

    Your numbers are stored little endian. http://en.wikipedia.org/wiki/Endianness

    To clarify,
    0x89ABCDEF is stored as 0xEFCDAB89, so right shift produces 0x0EFCDAB8, which is the bytes you see, in reverse order

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