从汇编中逆向工程C源代码

Question

I would like to know if anyone can help me out with a problem I am having when studying one of the lecture slides from an introductory assembly class that I am taking in school. The problem I am having is not understanding the assembly, it is how exactly the C source code is ordered based on the assembly. I will post the snippet I am talking about and maybe it will be clearer what I am talking about.

C Source given:

int arith(int x, int y, int z)
{ 
   int t1 = x+y;
   int t2 = z+t1;
   int t3 = x+4;
   int t4 = y * 48; 
   int t5 = t3 + t4;
   int rval = t2 * t5;
   return rval;
}

Assembly given:

arith:
pushl %ebp
movl %esp,%ebp

movl 8(%ebp),%eax
movl 12(%ebp),%edx
leal (%edx,%eax),%ecx
leal (%edx,%edx,2),%edx
sall $4,%edx
addl 16(%ebp),%ecx
leal 4(%edx,%eax),%eax
imull %ecx,%eax

movl %ebp,%esp
popl %ebp
ret

I am just confused as to how I am supposed to be able to discern for example that the adding of z + t1 ( z + x + y ) is listed on the second line(in the source) when in the assembly it comes after the y * 48 in the assembly code or for example that x + 4 is the 3rd line when in the assembly it is not even in a line by itself, its sort of mixed in with the last leal statement. It makes sense to me when I have the source but I am supposed to be able to reproduce the source for a test and I do understand that the compiler optimizes things but if anyone has a way of thinking about the reverse engineering that could help me out I would greatly appreciate it if they could walk me through their thought process.

Thanks.

Answer 1

I've broken down the disassembly for you to show how the assembly was produced from the C source.

8(%ebp) = x , 12(%ebp) = y , 16(%ebp) = z

arith:

Create the stack frame:

pushl %ebp
movl %esp,%ebp

---

Move x into eax , y into edx :

 movl 8(%ebp),%eax movl 12(%ebp),%edx 

---

t1 = x + y . leal (Load effective address) will add edx and eax , and t1 will be in ecx :

 leal (%edx,%eax),%ecx 

---

int t4 = y * 48; in two steps below, multiply by 3, then by 16. t4 will eventually be in edx :

Multiply edx by 2, and add edx to the result, ie. edx = edx * 3 :

 leal (%edx,%edx,2),%edx 

Shift left 4 bits, ie. multiply by 16:

 sall $4,%edx 

---

int t2 = z+t1; . ecx initially holds t1 , z is at 16(%ebp) , at the end of the instruction ecx will be holding t2 :

 addl 16(%ebp),%ecx 

---

int t5 = t3 + t4; . t3 was simply x + 4 , and rather than calculating and storing t3 , the expression of t3 is placed inline. This instruction essential does (x+4) + t4 , which is the same as t3 + t4 . It adds edx ( t4 ) and eax ( x ), and adds 4 as an offset to achieve that result.

 leal 4(%edx,%eax),%eax 

---

int rval = t2 * t5; Fairly straight-forward this one; ecx represents t2 and eax represents t5 . The return value is passed back to the caller through eax .

 imull %ecx,%eax 

---

Destroy the stack frame and restore esp and ebp :

 movl %ebp,%esp popl %ebp 

---

Return from the routine:

 ret 

---

From this example you can see that the result is the same, but the structure is a bit different. Most likely this code was compiled with some sort of optimization or someone wrote it themself like that to demonstrate a point.

As others have said, you can't go exactly back to the source from the disassembly. It's up to the interpretation of the person reading the assembly to come up with equivalent C code.

---

To help with learning assembly and understanding the disassembly of your C programs, you can do the following on Linux:

Compile with debug information ( -g ), which will embed the source:

 gcc -c -g arith.c 

If you're on a 64-bit machine, you can tell the compiler to create a 32-bit binary with the -m32 flag (I did so for the example below).

Use objdump to dump the object file with it's source interleaved:

gcc -c -g arith.c

-d = disassembly, -S = display source. You can add -M intel-mnemonic to use the Intel ASM syntax if you prefer that over the AT&T syntax that your example uses.

Output:

objdump -d -S arith.o

As you can see, without optimizations the compiler produces a larger binary than the example you have. You can play around with that and add a compiler optimization flag when compiling (ie. -O1 , -O2 , -O3 ). The higher the optimization level, the more abstract the disassembly's going to seem.

For example, with just level 1 optimization ( gcc -c -g -O1 -m32 arith.c1 ), the assembly code produced is a lot shorter:

arith.o:     file format elf32-i386


Disassembly of section .text:

00000000 <arith>:
int arith(int x, int y, int z)
{ 
   0:   55                      push   %ebp
   1:   89 e5                   mov    %esp,%ebp
   3:   83 ec 20                sub    $0x20,%esp
   int t1 = x+y;
   6:   8b 45 0c                mov    0xc(%ebp),%eax
   9:   8b 55 08                mov    0x8(%ebp),%edx
   c:   01 d0                   add    %edx,%eax
   e:   89 45 fc                mov    %eax,-0x4(%ebp)
   int t2 = z+t1;
  11:   8b 45 fc                mov    -0x4(%ebp),%eax
  14:   8b 55 10                mov    0x10(%ebp),%edx
  17:   01 d0                   add    %edx,%eax
  19:   89 45 f8                mov    %eax,-0x8(%ebp)
   int t3 = x+4;
  1c:   8b 45 08                mov    0x8(%ebp),%eax
  1f:   83 c0 04                add    $0x4,%eax
  22:   89 45 f4                mov    %eax,-0xc(%ebp)
   int t4 = y * 48; 
  25:   8b 55 0c                mov    0xc(%ebp),%edx
  28:   89 d0                   mov    %edx,%eax
  2a:   01 c0                   add    %eax,%eax
  2c:   01 d0                   add    %edx,%eax
  2e:   c1 e0 04                shl    $0x4,%eax
  31:   89 45 f0                mov    %eax,-0x10(%ebp)
   int t5 = t3 + t4;
  34:   8b 45 f0                mov    -0x10(%ebp),%eax
  37:   8b 55 f4                mov    -0xc(%ebp),%edx
  3a:   01 d0                   add    %edx,%eax
  3c:   89 45 ec                mov    %eax,-0x14(%ebp)
   int rval = t2 * t5;
  3f:   8b 45 f8                mov    -0x8(%ebp),%eax
  42:   0f af 45 ec             imul   -0x14(%ebp),%eax
  46:   89 45 e8                mov    %eax,-0x18(%ebp)
   return rval;
  49:   8b 45 e8                mov    -0x18(%ebp),%eax
}
  4c:   c9                      leave  
  4d:   c3                      ret

Answer 2

You can't reproduce the original source, you can only reproduce an equivalent source.

In your case the calculation for t2 can appear anywhere after t1 and before retval .

The source might even have been a single expression:

return (x+y+z) * ((x+4) + (y * 48));

Answer 3

When reverse engineering, you don't care about the original source code line by line, you care about what it does. A side effect is that you see what the code does, not what the programmer intended the code to do.

Answer 4

反编译并不是完全可以实现的：当从源代码（其中注释和名称给出了原始程序员的意图的线索）到二进制机器代码（其中指令将由处理器执行）时，存在一些知识损失。