July, 06 2005г. | Izik |
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.
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.
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
-- 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.oBefore 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.
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.
-- 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 zeroOur 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.