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Home/ Questions/Q 9053365
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T13:24:18+00:00 2026-06-16T13:24:18+00:00

what does this command do in assambly ? mov ebx, [eax+ecx*4+12] does it do

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what does this command do in assambly ?

mov ebx, [eax+ecx*4+12] 

does it do the same of the following ?

MOV EBX,ECX
ADD EBX,EBX ; *2
ADD EBX,EBX ; *4
ADD EBX,EAX
ADD EBX,12
MOV EBX,[EBX]

if yes why do they don’t give the same result in Jasmin when eax = 100h, ebx = 0, ecx = 100h

end.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-16T13:24:19+00:00Added an answer on June 16, 2026 at 1:24 pm

    lets say we have these commands

    mov eax, 0x1234 ; 1234h
    mov ebx, 0
    mov ecx, 0x1234 ; 1234h
    MOV EBX,ECX ; ebx = 0x1234
    ADD EBX,EBX ; *2
    ADD EBX,EBX ; *4
    ADD EBX,EAX ; ebx = 0x5B04
    ADD EBX,12  ; ebx = 0x5B10
    MOV EBX, [EBX] ; Error out of range
    
    ; Or
    mov ebx, [eax+ecx*4+12] ; => mov ebx, [0x1234+0x1234*4+12]
    

    => its actually mov ebx, [0x5B10] where 0x5B10 is the address number in the memory, but in my case the maximal address is 0xFFC < 0x5B10, so its out of range so the appeared result is the result of the before last command

    so by using smaller values like 0x14 we get the same result

    mov eax, 0x14 ; eax = 0x14
    mov ebx, 0    ; ebx = 0
    mov ecx, 0x14 ; ecx = 0x14
    MOV EBX,ECX   ; ebx = 0x14
    ADD EBX,EBX   ; *2
    ADD EBX,EBX   ; *4
    ADD EBX,EAX   ; ebx = 0x64
    ADD EBX,12    ; ebx = 0x70
    MOV EBX, [EBX] ; => mov ebx, [0x70]
    
    ; Or
    mov ebx, [eax+ecx*4+12] ; => mov ebx, [0x14+0x14*4+12] => mov ebx, [0x70]
    

    we get in both cases or methode:

    eax = 0x14
    ebx = 0   => its 0 because the address 0x70 in the memory is empty = 0x0000
    ecx = 0x14
    

    BUT if we add before the last line something like

    mov [0x70], 0x111
    

    we will get in ebx 0x111

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