Practical Reverse Engineering Page 22 Chapter 1 Exercise 1

1. This function uses a combination SCAS and STOS to do its work. First, explain what is the type of the [EBP+8] and [EBP+C] in line 1 and 8, respectively. Next, explain what this snippet does.
01: 8B 7D 08         mov   edi, [ebp+8]
02: 8B D7            mov   edx, edi
03: 33 C0            xor   eax, eax
04: 83 C9 FF         or    ecx, 0FFFFFFFFh
05: F2 AE            repne scasb
06: 83 C1 02         add   ecx, 2
07: F7 D9            neg   ecx
08: 8A 45 0C         mov   al, [ebp+0Ch]
09: 8B FA            mov   edi, edx
10: F3 AA            rep stosb
11: 8B C2            mov   eax, edx”

[EBP+8] is pointing to a string.
[ebp+0Ch] is pointing to a character.
The above pointers are likely to be arguments passed into the function.

Line 1-5 is check length of the string before it is NULL terminated. Count will be in ECX.
Line 6-7 Prepare the count in ECX for use in the 2nd part of the function.
Line 8-10 over write the string pointed by [ebp+8] with character pointed by [ebp+0Ch].
Line 11 the pointer to the overwritten string is returned by Eax.

Feel free to leave comments on my answers. :)