Beginner Assembly Question

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ragnar_graybeard87

Joined: 28 Mar 2016
Posts: 4

ragnar_graybeard87 28 Mar 2016, 01:13

Hey guys,

I included the source for the program in question. Basically, I'm following the UDEMY asm tutorial by XorPD and I'm stuck on one of the questions. It says to get a 32-bit integer as input then decompose it into 4 bytes then add them. Obviously the AL and AH were easy to get at but it says to figure out the extended 4 bytes by using DiV.

My approach was to save AL and AH to another reg then zero out AX then keep dividing EAX by ECX (inc'ing by 1 till they matched) in hopes that it'd save the bytes into CX but it seems to place them into the extended portionl. I guess all I'm asking is how could I access those top bytes without using a fancy instruction which we haven't been introduced to yet? I'm sure its simpler than I'm making it out to be however I can't figure it out on my own so any suggestions and pointers would be highly appreciated!

Thanks in advance Smile

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28 Mar 2016, 01:13

AsmGuru62

Joined: 28 Jan 2004
Posts: 1661
Location: Toronto, Canada

AsmGuru62 28 Mar 2016, 01:50

Isn't it better to store 32-bit value into memory and then add the separate bytes from that memory location?

Code:

        ;
        ; EAX is the input
        ;
        push    eax             ; store it on stack
        mov     esi, esp        ; point ESI to the value (4 bytes of it) you just stored
        ;
        ; Add the bytes at ESI
        ;
        xor     eax, eax        ; EAX = 0
        xor     ecx, ecx        ; ECX = 0
        ;
        ;  Do it for all 4 bytes
        ;
        mov     cl, [esi]
        add     eax, ecx

        mov     cl, [esi + 1]
        add     eax, ecx

        mov     cl, [esi + 2]
        add     eax, ecx

        mov     cl, [esi + 3]
        add     eax, ecx
        ;
        ; EAX now has the sum of 4 bytes from initial input
        ;
        pop     ecx             ; restore stack pointer

Or you can also do it by shifting the 32-bit value by 8 bits:

Code:

        ;
        ; EAX is the input value
        ;
        xor     ecx, ecx        ; ECX = 0
        mov     cl, al          ; 1st byte into ECX
        xor     edx, edx        ; EDX = 0

        shr     eax, 8
        mov     dl, al
        add     ecx, edx        ; ECX is a sum of 2 bytes

        shr     eax, 8
        mov     dl, al
        add     ecx, edx        ; ECX is a sum of 3 bytes

        shr     eax, 8
        mov     dl, al
        add     ecx, edx        ; ECX is a final sum

EDITED:
Oh, I see, you need to use DIV.
If you divide EAX just once (no need for a loop) by 10000h - you will get the higher 16 bits into AX, so then use AH,AL to add them.

Code:

        ;
        ; EAX = input
        ;
        xor     ecx, ecx        ; ECX = 0
        xor     edx, edx        ; EDX = 0

        mov     cl, al
        mov     dl, ah
        add     ecx, edx        ; ECX = sum of AH + AL
        ;
        ; Divide EAX by 10000h
        ;
        mov     esi, 10000h
        xor     edx, edx
        div     esi
        ;
        ; Repeat the trick with AH + AL
        ;
        xor     edx, edx

        mov     dl, ah
        add     ecx, edx

        mov     dl, al
        add     ecx, edx
        ;
        ; ECX is the final sum
        ;

28 Mar 2016, 01:50

ragnar_graybeard87

Joined: 28 Mar 2016
Posts: 4

ragnar_graybeard87 28 Mar 2016, 02:41

That's amazing, I've been pulling my hair out trying to figure it out. I wouldn't have guessed the 10000h in a million years. In fact I realize I wasn't even doing the addition properly. I thought I'd be able to do this once I got the bytes in the right containers:

add ah,ch
add al,cl

then the sum of all 4 would be inside AH and AL which should be AX so then I'd:

movzx EAX,AX
call print_eax

but I never got the desired result. Thank you again for helping me with this its very very appreciated. Now I can figure out why this code does what it does and start learning bitwise operations Smile

28 Mar 2016, 02:41

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