Debug Fundamentals Exercise 1: Reverse engineer a function

Debug Fundamentals Exercise 1: Reverse engineer a function

  • Comments 38


Hello ntdebuggers!  We’ve seen a lot of interest in our Puzzlers, and we’ve also seen requests and interest in topics covering debugging fundamentals.  So we’ve decided to combine the two topics and post a series of “Fundamentals Exercises”.  These exercises will be designed more as learning experiences rather than simply puzzlers.  We hope you find them interesting and educational!


Feel free to post your responses here, but we won’t put them on the site until after we post the “official” responses, so as to avoid spoilers.



Examine the following code, registers, and stack values to determine the following:


1.       When the function “DoTheWork” returns, what is the return value from that function?

2.       Bonus: what is the mathematical operation that “DoTheWork” performs?



1.       The bracket notation [] in the assembly means to treat the value in brackets as a memory address, and access the value at that address.

2.       32-bit integer return values are stored in eax



// Code

0:000> uf eip


0040101c 55              push    ebp

0040101d 8bec            mov     ebp,esp

0040101f 8b4d08          mov     ecx,dword ptr [ebp+8]

00401022 8bc1            mov     eax,ecx

00401024 49              dec     ecx

00401025 0fafc1          imul    eax,ecx

00401028 83f902          cmp     ecx,2

0040102b 7ff7            jg      demo2!DoTheWork+0x8 (00401024)

0040102d 5d              pop     ebp

0040102e c3              ret


// Current register state

0:000> r

eax=00000007 ebx=7ffd9000 ecx=ffffffff edx=00000007 esi=00001771 edi=00000000

eip=0040101c esp=0012fe9c ebp=0012feac iopl=0         nv up ei pl nz na po nc

cs=001b  ss=0023  ds=0023  es=0023  fs=003b  gs=0000             efl=00000202


0040101c 55              push    ebp


// Current stack values for this thread

0:000> dps esp

0012fe9c  00406717 demo2!main+0x27

0012fea0  00000007

0012fea4  82059a87

0012fea8  00000007

0012feac  0012ff88

0012feb0  004012b2 demo2!mainCRTStartup+0x170

0012feb4  00000002

0012feb8  00980e48

0012febc  00980e80

0012fec0  00000094

0012fec4  00000006

0012fec8  00000000

0012fecc  00001771

0012fed0  00000002

0012fed4  76726553

0012fed8  20656369

0012fedc  6b636150

0012fee0  00003120

0012fee4  00000000

0012fee8  00000000

0012feec  00000000

0012fef0  00000000

0012fef4  00000000

0012fef8  00000000

0012fefc  00000000

0012ff00  00000000

0012ff04  00000000

0012ff08  00000000

0012ff0c  00000000

0012ff10  00000000

0012ff14  00000000

0012ff18  00000000


[Update: our answer. Posted 11/19/2008]

Wow - what a great response from our readers on this exercise!  It is great to see the various approaches to reverse engineering this code.  As for the answer, the numerical result (stored in eax) is 5040, and the corresponding mathematical operation is a factorial.  So 7! is the result calculated, given that 7 was passed to the function.  Congratulations to all of you that got it right! 


Many of you posted some C code to give an idea of what the original source of DoTheWork() might have looked like.  The original source was actually written in assembly!  However, it was written to be called from C, and it uses ebp in the same way that compiled C code might.  This function wasn’t written with optimal performance in mind, but rather for learning about reverse engineering.



Leave a Comment
  • Please add 2 and 8 and type the answer here:
  • Post
  • I'm betting on the factorial function .

  • DoTheWork performs the factorial operation

    the stack says we're in DoTheWork(7) so the return value is 7! = 5040

    nice puzzlers, keep it up !

  • DoTheWork is the factorial function; as such it'll return 7!.

  • [ebp] -> Old Ebp == 0012ff88

    [ebp+4] -> return address == 004012b2

    [ebp+8] -> Param 1 == 2

    In this case:

    2 * (2 - 1) = 2

    int DoTheWork(int n) {

    int i = n;

    do {


     i = i * n;

    } while (n > 2);

    return (i);


    BTW, why not make this a fastcall :)?

  • (1) At function exit, EAX = 0x13b0

    (2) The function computes the factorial of its argument.

  • Looks like factorial, assuming the input is positive (doesn't handle 0!)... result will be 7! = 5040.


    int DoTheWork(int x)


    int r = x;

    do {

    r *= --x;

    } while (x > 2);

    return r;



  • 1) Return value should be 5040 (in decimal)

    2) Looks like the !n function called with n=7

  • Doh! didn't look at esp

    1.) 7!

    2.) Factorial

  • I'm going to refrain from posting my answer since I don't want to spoil the challenge for anyone.  But this was a really fun post (to me, at least), so thanks for it!

  • This might be completly wrong but off the bat I would translate it to something like this:

    void* DoTheWork(void* e, int p)




       p = ((--p) * p;


     while (p != 2);

    return e;


    How close did i got it? :=)

  • I guess this is this:

    int DoTheWork(int p1)


    int eax;

    int ecx;

       ecx = p1;

       eax = p1;



           ecx -= 1;

           eax *= ecx;

       } while (ecx > 2);

       return eax;


    0012feb4  00000002

    Here is the argument you used.

    The function returns 2.

  • It calculates the number 7 factorial. The return value will be 7! = 5040.


  • Great idea with an exercise!

    The function calculates the factorial of the argument. The result in the example is 7!.

    int fact(int arg1) {

       int retval = arg1;

       do {

           retval *= --arg1;

       } while (arg1 > 2);

       return retval;


  • I assume the result is 5040. About the function I am not quiet sure.

  • I assume the result is 5040. About the math function I am not sure.

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