Exploit Reverse Engineering with LD_PRELOAD


18 Дек 2022
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	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)
	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 ;)


	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);
			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!

	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!
	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.


	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) {
				return 0;

                        memset(buf, 0x0, sizeof(buf));

                        plen = read(fd, buf, strlen(argv[1]));

                        if (plen != strlen(argv[1])) {

                                if (plen < 0) {

                                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]) {

                                        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

	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.


	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) {
				return 0;

			printf("On a long enough timeline, the survival rate for everyone drops to zero\n");

	-- 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.
	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
	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.


	Izik <[email protected]> [or] http://www.tty64.org

# milw0rm.com [2006-03-10]

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