- 34,644
- 0
- 18 Дек 2022
- EDB-ID
- 13233
- Проверка EDB
-
- Пройдено
- Автор
- IZIK
- Тип уязвимости
- PAPERS
- Платформа
- MULTIPLE
- CVE
- N/A
- Дата публикации
- 2006-03-10
Код:
---------------------------------------------------
Reverse Engineering with LD_PRELOAD
Izik <[email protected]>
---------------------------------------------------
This paper is about the LD_PRELOAD feature, and how it can be useful for reverse engineering dynamically linked executables.
This technique allows you to hijack functions/inject code and manipulate the application flow.
Compiling Methods
-----------------
Generally there are two ways of producing an executable. The first method is the static compilation. During the compilation process the
executable that will be produced, will be independent and will include all it needs to function properly. The advantage of this method
is mainly portability as the user is not required to install anything in order to use the application. The disadvantage is a relatively oversized
executable, also bugs that originate from an outside component will not be fixed, since the executable is not linked. The other method is
the dynamically produced executable. It is dependent on shared libraries to function, and corresponding to changes in the shared libraries.
For better or worse, this can be an advantage and a disadvantage. Dynamically linked executables by their nature are more lightweight than
the static produced executables.
The gcc compiler will by default produce a dynamically linked executable, unless you specify the '-static' parameter to gcc.
You can browse through the dynamically linked executable dependencies list, using the ldd utility. Following this:
root@magicbox:~# ldd /bin/ls
linux-gate.so.1 => (0xffffe000)
librt.so.1 => /lib/tls/librt.so.1 (0xb7fd7000)
libc.so.6 => /lib/tls/libc.so.6 (0xb7eba000)
libpthread.so.0 => /lib/tls/libpthread.so.0 (0xb7ea8000)
/lib/ld-linux.so.2 (0xb7feb000)
root@magicbox:~#
Each dependency presenting the library it depends on, while the adress we see, is the one it would be mapped to during runtime.
This article revolves around dynamically linked executables as they can be manipulated by the LD_PRELOAD.
Runtime Linker
--------------
Runtime linker's purpose is to bind the dynamically linked executables to its dependencies by all means. Taking care of resolving the
symbols and invoking the initialization functions of shared libraries as well as the finalization functions, if any. It also provides a
set of functions to allow an application to load and unload libraries at runtime. This is often refereed to as Application Plug-in feature.
LD_PRELOAD is an environment variable that affects the runtime linker. It allows you to put a dynamic object, that will create some sort
of a buffer layer, between the application references and the dependencies. It also grants you the possibility of linking with the
application and relocating symbols/references.
To simplify the situation. This is a man-in-the middle attack between the program and the needed libraries ;)
Limitations
-----------
The LD_PRELOAD option does not effect applications that have been chmod'ed with +s (setgid/setuid) flag. Unless the user that runs
the program already has a root privileges. Also on some platforms, to be able to use the LD_PRELOAD option the user should already
have root privileges. All the examples brought up in this paper assume that you will use your own machine as a part of your
reverse engineering framework. But it would work regardless, if you have root privileges elsewhere.
Hack in a box
-------------
The simplest thing that LD_PRELOAD allows is to hijack a function. In order to do it, we will create a shared library that will include
the implementation of the function that we wish to hijack. But first let's take a look at our challenge:
-- snip snip --
/*
* strcmp-target.c, A simple challenge that compares two strings
*/
#include <stdio.h>
#include <string.h>
int main(int argc, char **argv) {
char passwd[] = "foobar";
if (argc < 2) {
printf("usage: %s <given-password>\n", argv[0]);
return 0;
}
if (!strcmp(passwd, argv[1])) {
printf("Green light!\n");
return 1;
}
printf("Red light!\n");
return 0;
}
-- snip snip --
In this example we got a simple password-matching challenge. It uses the strcmp() function for comparing the two strings.
Lets attack it by hijacking the strcmp() function to see who's running against who and to ensure it always returns equal strings!
-- snip snip --
#include <stdio.h>
#include <string.h>
/*
* strcmp-hijack.c, Hijack strcmp() function
*/
/*
* strcmp, Fixed strcmp function -- Always equal!
*/
int strcmp(const char *s1, const char *s2) {
printf("S1 eq %s\n", s1);
printf("S2 eq %s\n", s2);
// ALWAYS RETURN EQUAL STRINGS!
return 0;
}
-- snip snip --
Now let's compile our strcmp-hijack.c snippet as a shared library, by doing this:
root@magicbox:/tmp# gcc -fPIC -c strcmp-hijack.c -o strcmp-hijack.o
root@magicbox:/tmp# gcc -shared -o strcmp-hijack.so strcmp-hijack.o
Before attacking, let's see if our challenge does what we expect it to do, by doing this:
root@magicbox:/tmp# ./strcmp-target redbull
Red light!
root@magicbox:/tmp#
Now let's attack it using LD_PRELOAD, by doing this:
root@magicbox:/tmp# LD_PRELOAD="./strcmp-hijack.so" ./strcmp-target redbull
S1 eq foobar
S2 eq redbull
Green light!
root@magicbox:/tmp#
Our evil shared library has done its job. We hijacked the strcmp() function and made it return a fixed value. Also we put a
debug print that shows what the two arguments are, that have been sent to the function. Now we know what the real password is as well.
Notice I am using bash shell. And LD_PRELOAD is an environment variable, which means it is up to your shell to set this variable. If you have
troubles to set it, you should refer to your shell man page, in order to find out how setting environment variable can be done.
Gatekeeper
----------
Hijacking functions is fun. But becoming a wrapper to a function is more powerful. As a wrapper function we will accept the arguments
that have been sent to the original function, pass 'em along to the original function and examine the result. Then we could decide if
we want to return the original value, or fix the return value. Calling the original function would be done by using a pointer to the original
function address. We will then use the runtime linker api, thus the dl*() function family.
This is our new challenge, which is a little bit more complex:
-- snip snip --
/*
* crypt-mix.c, Don't let crypt() get cought up in your mix!
*/
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <string.h>
int main(int argc, char **argv) {
char buf[256], alpha[34], beta[34];
int j, plen, fd;
if (argc < 2) {
printf("usage: %s <keyfile>\n", argv[0]);
return 1;
}
if (strlen(argv[1]) > 256) {
fprintf(stderr, "keyfile length is > 256, go fish!\n");
return 0;
}
fd = open(argv[1], O_RDONLY);
if (fd < 0) {
perror(argv[1]);
return 0;
}
memset(buf, 0x0, sizeof(buf));
plen = read(fd, buf, strlen(argv[1]));
if (plen != strlen(argv[1])) {
if (plen < 0) {
perror(argv[1]);
}
printf("Sorry!\n");
return 0;
}
strncpy(alpha, (char *)crypt(argv[1], "$1$"), sizeof(alpha));
strncpy(beta, (char *) crypt(buf, "$1$"), sizeof(beta));
for (j = 0; j < strlen(alpha); j++) {
if (alpha[j] != beta[j]) {
printf("Sorry!\n");
return 0;
}
}
printf("All your base are belong to us!\n");
return 1;
}
-- snip snip --
This challenge principle is quite simple. It performs a MD5 hash on the given filename then buffers up the same amount of bytes
as the filename length itself, and performs the same MD5 function on the payload. Then it compares the two, without explicitly
using the strcmp() function. This forces us to find a weak spot. This challenge's weak spot is the crypt() function which can
be hijacked. The plan is to buffer up the 1st hash returned by the 1st crypt() call, then fix up the 2nd crypt() return value so
it will match the 1st hash, and pass the comparing part smoothly.
This is our new evil shared library:
-- snip snip --
#define _GNU_SOURCE
#include <stdio.h>
#include <dlfcn.h>
/*
* crypt-mixup.c, Buffer up crypt() result to return later on.
*/
// Pointer to the original crypt() call
static char *(*_crypt)(const char *key, const char *salt) = NULL;
// Pointer to crypt() previous result
static char *crypt_res = NULL;
/*
* crypt, Crooked crypt function
*/
char *crypt(const char *key, const char *salt) {
// Initialize of _crypt(), if needed.
if (_crypt == NULL) {
_crypt = (char *(*)(const char *key, const char *salt)) dlsym(RTLD_NEXT, "crypt");
crypt_res = NULL;
}
// No previous result, continue as normal crypt()
if (crypt_res == NULL) {
crypt_res = _crypt(key, salt);
return crypt_res;
}
// We already got result buffered up!
_crypt = NULL;
return crypt_res;
}
-- snip snip --
We will start by simply testing the challenge, doing this:
root@magicbox:/tmp# gcc -o crypt-mix crypt-mix.c -lcrypt
root@magicbox:/tmp# echo "foobar" > mykey
root@magicbox:/tmp# ./crypt-mix mykey
Sorry!
Now let's try it again with our evil shared library, by doing this:
root@magicbox:/tmp# gcc -fPIC -c -o crypt-mixup.o crypt-mixup.c
root@magicbox:/tmp# gcc -shared -o crypt-mixup.so crypt-mixup.o -ldl
root@magicbox:/tmp# LD_PRELOAD="./crypt-mixup.so" ./crypt-mix mykey
All your base are belong to us!
Again using LD_PRELOAD, we just bypassed the mechanism and went straight on.
Cerberus
--------
After we did a few high level tricks. It is time to get our hands a little dirty. And this means involving Assembly in the mix.
The next challenge might look a little bit odd, check it out:
-- snip snip --
/*
* cerberus.c, Impossible statement
*/
#include <stdio.h>
int main(int argc, char **argv) {
int a = 13, b = 17;
if (a != b) {
printf("Sorry!\n");
return 0;
}
printf("On a long enough timeline, the survival rate for everyone drops to zero\n");
exit(1);
}
-- snip snip --
As you can see in this challenge our input doesn't count. The statement will always be incorrect. Regardless of any parameters that
will be passed to main(). But there is a way out!
To understand this trick, let's first disassemble the main function, following this:
(gdb) disassemble main
Dump of assembler code for function main:
0x080483c4 <main+0>: push %ebp
0x080483c5 <main+1>: mov %esp,%ebp
0x080483c7 <main+3>: sub $0x18,%esp
0x080483ca <main+6>: and $0xfffffff0,%esp
0x080483cd <main+9>: mov $0x0,%eax
0x080483d2 <main+14>: sub %eax,%esp
0x080483d4 <main+16>: movl $0xd,0xfffffffc(%ebp)
0x080483db <main+23>: movl $0x11,0xfffffff8(%ebp)
0x080483e2 <main+30>: mov 0xfffffffc(%ebp),%eax
0x080483e5 <main+33>: cmp 0xfffffff8(%ebp),%eax
0x080483e8 <main+36>: je 0x8048403 <main+63>
0x080483ea <main+38>: sub $0xc,%esp
0x080483ed <main+41>: push $0x8048560
0x080483f2 <main+46>: call 0x80482d4 <_init+56>
0x080483f7 <main+51>: add $0x10,%esp
0x080483fa <main+54>: movl $0x0,0xfffffff4(%ebp)
0x08048401 <main+61>: jmp 0x804841d <main+89>
0x08048403 <main+63>: sub $0xc,%esp
0x08048406 <main+66>: push $0x8048580
0x0804840b <main+71>: call 0x80482d4 <_init+56>
0x08048410 <main+76>: add $0x10,%esp
0x08048413 <main+79>: sub $0xc,%esp
0x08048416 <main+82>: push $0x0
0x08048418 <main+84>: call 0x80482e4 <_init+72>
0x0804841d <main+89>: mov 0xfffffff4(%ebp),%eax
0x08048420 <main+92>: leave
0x08048421 <main+93>: ret
0x08048422 <main+94>: nop
0x08048423 <main+95>: nop
0x08048424 <main+96>: nop
0x08048425 <main+97>: nop
0x08048426 <main+98>: nop
0x08048427 <main+99>: nop
0x08048428 <main+100>: nop
0x08048429 <main+101>: nop
0x0804842a <main+102>: nop
0x0804842b <main+103>: nop
0x0804842c <main+104>: nop
0x0804842d <main+105>: nop
0x0804842e <main+106>: nop
0x0804842f <main+107>: nop
End of assembler dump.
(gdb)
The IF statement takes place in <main+36>, the 'je' is not evaluated. Therefore we do not jump to <main+63> but rather continue
to <main+38>. There it pushes the string into the stack and calls printf() -- Hold on! Are you thinking what I'm thinking? *RET HIJACK* ;-)
-- snip snip --
#define _GNU_SOURCE
#include <stdio.h>
#include <dlfcn.h>
#include <stdarg.h>
/*
* megatron.c, Making the impossible possible!
*/
// Pointer to the original printf() call
static int (*_printf)(const char *format, ...) = NULL;
/*
* printf, One nasty function!
*/
int printf(const char *format, ...) {
if (_printf == NULL) {
_printf = (int (*)(const char *format, ...)) dlsym(RTLD_NEXT, "printf");
// Hijack the RET address and modify it to <main+66>
__asm__ __volatile__ (
"movl 0x4(%ebp), %eax \n"
"addl $15, %eax \n"
"movl %eax, 0x4(%ebp)"
);
return 1;
}
// Rewind ESP and JMP into _PRINTF()
__asm__ __volatile__ (
"addl $12, %%esp \n"
"jmp *%0 \n"
: /* no output registers */
: "g" (_printf)
: "%esp"
);
/* NEVER REACH */
return -1;
}
-- snip snip --
As always before attacking, we test the challenge, by doing this:
root@magicbox:/tmp# ./cerberus
Sorry!
It is a pretty straight forward challenge. Now let's attack this challenge using our evil shared library, doing this:
root@magicbox:/tmp# gcc -fPIC -o megatron.o megatron.c -c
root@magicbox:/tmp# gcc -shared -o megatron.so megatron.o -ldl
root@magicbox:/tmp# LD_PRELOAD="./megatron.so" ./cerberus
On a long enough timeline, the survival rate for everyone drops to zero
Our attack succeeds. Manipulating the printf() function return address has made the impossible possible. I will not get into details
about this attack. As it alone deserved a paper about it. I would say that I've cheated a little bit, as this code has created a stack corruption,
and I have used the exit() function to do the some dirty work for me. Notice that the printf() family functions are unusual in the way they
meant to accept unlimited number of parameters. The code above is meant to be a proof of concept to how it is possible to manipulate the
program flow dynamically.
To conclude this paper I would say that LD_PRELOAD is a powerful feature, and can be used on many purposes and Reverse Engineering is only one.
Contact
-------
Izik <[email protected]> [or] http://www.tty64.org
# milw0rm.com [2006-03-10]- Источник
- www.exploit-db.com
