Understanding the decompilation of an object to source code - c

Question

First of all, I am a student, I do not yet have extensive knowledge about C, C ++ and assembler, so I am making a extreme effort to understand it.
I have this piece of assembly code from an Intel x86-32 bit processor.
My goal is to transform it to source code.
0x80483dc <main>: push ebp
0x80483dd <main+1>: mov ebp,esp
0x80483df <main+3>: sub esp,0x10
0x80483e2 <main+6>: mov DWORD PTR [ebp-0x8],0x80484d0
0x80483e9 <main+13>: lea eax,[ebp-0x8]
0x80483ec <main+16>: mov DWORD PTR [ebp-0x4],eax
0x80483ef <main+19>: mov eax,DWORD PTR [ebp-0x4]
0x80483f2 <main+22>: mov edx,DWORD PTR [eax+0xc]
0x80483f5 <main+25>: mov eax,DWORD PTR [ebp-0x4]
0x80483f8 <main+28>: movzx eax,WORD PTR [eax+0x10]
0x80483fc <main+32>: cwde
0x80483fd <main+33>: add edx, eax
0x80483ff <main+35>: mov eax,DWORD PTR [ebp-0x4]
0x8048402 <main+38>: mov DWORD PTR [eax+0xc],edx
0x8048405 <main+41>: mov eax,DWORD PTR [ebp-0x4]
0x8048408 <main+44>: movzx eax,BYTE PTR [eax]
0x804840b <main+47>: cmp al,0x4f
0x804840d <main+49>: jne 0x8048419 <main+61>
0x804840f <main+51>: mov eax,DWORD PTR [ebp-0x4]
0x8048412 <main+54>: movzx eax,BYTE PTR [eax]
0x8048415 <main+57>: cmp al,0x4b
0x8048417 <main+59>: je 0x804842d <main+81>
0x8048419 <main+61>: mov eax,DWORD PTR [ebp-0x4]
0x804841c <main+64>: mov eax,DWORD PTR [eax+0xc]
0x804841f <main+67>: mov edx, eax
0x8048421 <main+69>: and edx,0xf0f0f0f
0x8048427 <main+75>: mov eax,DWORD PTR [ebp-0x4]
0x804842a <main+78>: mov DWORD PTR [eax+0x4],edx
0x804842d <main+81>: mov eax,0x0
0x8048432 <main+86>: leave
0x8048433 <main+87>: ret
This is what I understand from the code:
There are 4 variables:
a = [ebp-0x8] ebp
b = [ebp-0x4] eax
c = [eax + 0xc] edx
d = [eax + 0x10] eax
Values:
0x4 = 4
0x8 = 8
0xc = 12
0x10 = 16
0x4b = 75
0x4f = 79
Types:
char (8 bits) = 1 BYTE
short (16 bits) = WORD
int (32 bit) = DWORD
long (32 bits) = DWORD
long long (32 bit) = DWORD
This is what I was able to create:
#include <stdio.h>
int main (void)
{
   int a = 0x80484d0;
   int b
   short c;
   int d;
   c + b?
if (79 <= al) {
instructions
} else {
instructions
}
   return 0
}
But I'm stuck. Nor can I understand what the sentence "cmp al .." compares to, what is "al"?
How do these instructions work?
EDIT1:
That said, as you comment the assembly seems to be wrong or as someone comments say, it is insane!
The code and the exercise are from the following book called: "Reversing, Reverse Engineering" on page 140 (3.8 Proposed Exercises). It would never have occurred to me that it was wrong, if so, this clearly makes it difficult for me to learn ...
So it is not possible to do a reversing to get the source code because it is not a good assembly? Maybe I am not oppressed? Is it possible to optimize it?
EDIT2:
Hi!
I did ask and finally she says this should be the c code:
inf foo(void){
char *string;//ebp-0x8
unsigned int *pointerstring//[ebp-0x4]
unsigned int *position;
*position = *(pointerstring+0xc);
unsigned char character;
character=(unsigned char) string[*position];
if ((character != 0x4)||(character != 0x4b))
{
*(position+0x4)=(unsigned int)(*position & 0x0f0f0f0f);
}
return(0);
}
Does it have any sense at all for you?, could someone please explain this to me?
Does anyone really program like this?
Thanks very much!

Answer 1

Your assembly is completely insane. This is roughly equivalent C:
int main() {
int i = 0x80484d0; // in ebp-8
int *p = &i; // in ebp-4
p[3] += (short)p[4]; // add argc to the return address(!)
if((char)*p != 0x4f || (char)*p != 0x4b) // always true because of || instead of &&
p[1] = p[3] & 0xf0f0f0f; // note that p[1] is p
return 0;
}
It should be immediately obvious that this is horrifically bad code that almost certainly won't do what the programmer intended.

Answer 2

The x86 assembly language follows a long legacy and has mostly kept compatibility. We need to go back to the 8086/8088 chip where that story starts. These were 16 bit processors, which means that their register had a word size of 16 bits. The general purpose registers were named AX, BX, CX and DX. The 8086 had instructions to manipulate the upper and lower 8-bit parts of these registers that were then named AH, AL, BH, BL, CH, CL, DH and DL. This Wikipedia page describes this, please take a look.
The 32 bit versions of these registers have an E in front: EAX, EBX, ECX, etc.
The particular instruction you mention, e.g, cmp al,0x4f is comparing the lower byte of the AX register with 0x4f. The comparison is effectively the same as a subtraction, but does not save the result, only sets the flags.
For the 8086 instruction set, there is a nice reference here. Your program is 32 bit code, so you will need at least a 80386 instruction reference.

Answer 3

You have analyzed variables, and that's a good place to start. You should try to add type annotations to them, size, as you started, and, when used as pointers (like b), pointers to what kind/size.
I might update your variable chart as follows, knowing that [ebp-4] is b:
c = [b + 0xc]
d = [b + 0x10]
e = [b + 0], size = byte
Another thing to analyze is the control flow. For most instructions control flow is sequential, but certain instructions purposefully alter it. Broadly speaking, when the pc is moved forward, it skips some code and when the pc is moved backward it repeats some code it already ran. Skipping code is used to construct if-then, if-then-else, and statements that break out of loops. Jumping back is used to continue looping.
Some instructions, called conditional branches, on some dynamic condition being true: skip forward (or backwards) and on being false do the simple sequential advancement to the next instruction (sometimes called conditional branch fall through).
The control sequences here:
...
0x8048405 <main+41>: mov eax,DWORD PTR [ebp-0x4] b
0x8048408 <main+44>: movzx eax,BYTE PTR [eax] b->e
0x804840b <main+47>: cmp al,0x4f b->e <=> 'O'
0x804840d <main+49>: jne 0x8048419 <main+61> b->e != 'O' skip to 61
** we know that the letter, a->e, must be 'O' here
0x804840f <main+51>: mov eax,DWORD PTR [ebp-0x4] b
0x8048412 <main+54>: movzx eax,BYTE PTR [eax] b->e
0x8048415 <main+57>: cmp al,0x4b b->e <=> 'K'
0x8048417 <main+59>: je 0x804842d <main+81> b->e == 'K' skip to 81
** we know that the letter, a->e must not be 'K' here if we fall thru the above je
** this line can be reached by taken branch jne or by fall thru je
0x8048419 <main+61>: mov eax,DWORD PTR [ebp-0x4] ******
...
The flow of control reaches this last line tagged we know that either the letter is either not 'O' or it is not 'K'.
The construct where the jne instruction is used to skip another test is a short-circuit || operator. Thus the control construct is:
if ( a->e != 'O' || a->e != 'K' ) {
then-part
}
As that these two conditional branches are the only flow control modifications in the function, there is no else part of the if, and there are no loops or other if's.
This code appears to have a slight problem.
If the value is not 'O', the then-part will fire from the first test. However, if we reach the 2nd test, we already know the letter is 'O', so testing it for 'K' is silly and will be true ('O' is not 'K').
Thus, this if-then will always fire.
It is either very inefficient, or, there is a bug that perhaps it is the next letter along in the (presumably) string should be tested for 'K' not the same exact letter.