Continuing our series on “Fundamentals Exercises”, we have some more reverse engineering for you! Again, these exercises are 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, to avoid spoilers.
Examine the following code, registers, and stack values to determine the following:
1. What is the return value from DemoFunction2?
2. What is the purpose of DemoFunction2?
3. Bonus: Both the last exercise and this week’s exercise involved accessing data at ebp+8. Why ebp+8?
1. You probably don’t want to manually walk through every instruction that executes in the loop. Instead, walk through a few iterations to determine the intent of the code.
2. The bracket notation  in the assembly means to treat the value in brackets as a memory address, and access the value at that address.
3. 32-bit integer return values are stored in eax
0:000> uf 010024d0
010024d0 55 push ebp
010024d1 8bec mov ebp,esp
010024d3 8b5508 mov edx,dword ptr [ebp+8]
010024d6 33c0 xor eax,eax
010024d8 b920000000 mov ecx,20h
010024dd d1ea shr edx,1
010024df 7301 jnc asmdemo2!DemoFunction2+0x12 (010024e2)
010024e1 40 inc eax
010024e2 e2f9 loop asmdemo2!DemoFunction2+0xd (010024dd)
010024e4 5d pop ebp
010024e5 c3 ret
eax=80002418 ebx=7ffd7000 ecx=00682295 edx=00000000 esi=80002418 edi=00000002
eip=010024d0 esp=0006fe98 ebp=0006fea8 iopl=0 nv up ei pl zr na pe nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000246
0:000> dps esp
0006fe98 0100251c asmdemo2!main+0x20
0006feac 01002969 asmdemo2!_mainCRTStartup+0x12c
[Update: our answer. Posted 12/04/2008]
We had a great response to this exercise! It was good to see so many of you going through this. There were some readers that found this a good exercise for beginners, and others were looking for a return to Puzzlers. As an FYI, we may do more Puzzlers in the future, but for now we are going to continue on the “Fundamentals Exercise” track to help all our readers build up a solid foundation for debugging.
It was interesting to read how several of you not only gave the answers, but made suggestions for how the code could be optimized! I want to point out that the code we post for these exercises isn’t intended to be the optimal solution; it is written as a learning tool. That said, keep that in-depth feedback coming; I think everyone will benefit from a discussion of optimization.
Answers to exercise 2:
1. The return value is 1
2. The function calculates number of set bits in dword that is sent as a parameter.
3. Not sure but:
ebp+0 - return address
ebp+4 - this
ebp+8 - the parameter
1) Returned value is 5
2) It is counting the number of 1 bit in the passed in 80002418h parameter value.
The return value will be 0x5.
This function is returning number of 1's in the given input's binary representation.
ebp+8 is the first parameter passed to this function.
What is the return value from DemoFunction2
It seems to be 5
What is the purpose of DemoFunction2? : No idea ?
Why ebp+8? : It contains the first parameter to function...
Anways, it was fun. Keep posting.
#2) PopCount (Binary Counter)
#3) Paramater #1 after stack frame is setup
Finally, Same code written in assembly
popcount FRAME dwValue
DemoFunction2 takes paramter one, a 32 bit paramater, and then continually shifts it 32 times (20h), each time it is shifted the carry flag is checked to see if the shifted bit was set, if it wasn't the final result isn't incremented, if it was, the final result is incremented,, surmising that if each bit causes an increment to the return value the function simply counts the bits of the first paramater passed to our function.
PopCount does the same thing but as if the value needed folded upon itself, take every other bit and add them, then take every two bits and add them, then take every four bits and add them together, all the way until you add 16 bits to 16 bits. The reason for the subtraction instead of addition on first call is the reduction of overflow without the need for anding.
OK ... here is my try.
I think the code snippet is for counting how many 0 bits in the given 32bit unsigned number. The pesudo-C++ code would be like:
UINT DemoFunctions(UINT x)
UINT nRet = 0;
for (int i = 0; i < 32; ++i)
if (x&1 == 0)
x = x >> 1;
From the raw call stack the parameter is [0006fe9c] = 80002418, so the answer should be 27.
ret = count of non zero bits
1. CountBitsSetInDword(DWORD dwData = 80002418) = 5
3. ebp is the value of esp, ebp+4 is the return ip, ebp+8 is the first parameter of the function.
It appears that the function counts the number of bits that are set to 1 in the argument passed to the function.
In this example, the argument 80002418 is passed in and I expect the function would return a value of 5 because there are 5 bits set in the argumnet.
Oops - it is ebp+8 (not ebp+4).
The function counts the number of bits set in the ebp+8 location. ebp+8 contains a 00000002 which means EAX should have a 1 in it upon completion of the function.
I incorrectly looked at 80002418 as the routine argument (which is at ebp+4).
1. The return value from DemoFunction2 is 5.
2. The function's purpose is to count up the number of bits in the function's first parameter.
3. The data at ebp+8 is always where a function's first parameter is as long as the function was compiled to have a stack frame set up.
That was pretty fun! :-)
1. return 5.
2. return the number of 1 bit in arg.