Return-to-libc Attack Lab Solved

    • Lab Overview

The learning objective of this lab is for students to gain the first-hand experience on an interesting variant of buffer-overflow attack; this attack can bypass an existing protection scheme currently implemented in major Linux operating systems. A common way to exploit a buffer-overflow vulnerability is to overflow the buffer with a malicious shellcode, and then cause the vulnerable program to jump to the shellcode that is stored in the stack. To prevent these types of attacks, some operating systems allow programs to make their stacks non-executable; therefore, jumping to the shellcode will cause the program to fail.

Unfortunately, the above protection scheme is not fool-proof; there exists a variant of buffer-overflow attack called the Return-to-libc attack, which does not need an executable stack; it does not even use shell-code. Instead, it causes the vulnerable program to jump to some existing code, such as the system() function in the libc library, which is already loaded into a process’s memory space.

In this lab, students are given a program with a buffer-overflow vulnerability; their task is to develop a Return-to-libc attack to exploit the vulnerability and finally to gain the root privilege. In addition to the attacks, students will be guided to walk through some protection schemes that have been implemented in Ubuntu to counter against the buffer-overflow attacks. This lab covers the following topics:

    • Buffer overflow vulnerability and attack

    • Stack layout in a function invocation

    • Non-executable stack

    • Address randomization

    • The libc functions

Readings and related topics. Detailed coverage of the return-to-libc attack can be found in Chapter 5 of the SEED book, Computer Security: A Hands-on Approach, by Wenliang Du. A topic related to this lab is the general buffer-overflow attack, which is covered in a separate SEED lab, as well as in Chapter 4 of the SEED book.

Lab environment. This lab has been tested on our pre-built Ubuntu 12.04 VM and Ubuntu 16.04 VM, both of which can be downloaded from the SEED website.

    • Lab Tasks

2.1    Turning Off Countermeasures

You can execute the lab tasks using our pre-built Ubuntu virtual machines. Ubuntu and other Linux distributions have implemented several security mechanisms to make the buffer-overflow attack difficult. To

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simplify our attacks, we need to disable them first.

Address Space Randomization. Ubuntu and several other Linux-based systems uses address space ran-domization [?] to randomize the starting address of heap and stack. This makes guessing the exact addresses difficult; guessing addresses is one of the critical steps of buffer-overflow attacks. In this lab, we disable this feature using the following command:

$ sudo sysctl -w kernel.randomize_va_space=0


The StackGuard Protection Scheme. The GCC compiler implements a security mechanism called Stack-Guard to prevent buffer overflows. In the presence of this protection, buffer overflow attacks will not work. We can disable this protection during the compilation using the -fno-stack-protector option. For example, to compile a program example.c with StackGuard disabled, we can do the following:

$ gcc -fno-stack-protector example.c


Non-Executable Stack. Ubuntu used to allow executable stacks, but this has now changed: the binary images of programs (and shared libraries) must declare whether they require executable stacks or not, i.e., they need to mark a field in the program header. Kernel or dynamic linker uses this marking to decide whether to make the stack of this running program executable or non-executable. This marking is done automatically by the recent versions of gcc, and by default, stacks are set to be non-executable. To change that, use the following option when compiling programs:


For executable stack:

$ gcc -z execstack    -o test test.c

For non-executable stack:

$ gcc -z noexecstack    -o test test.c

Because the objective of this lab is to show that the non-executable stack protection does not work, you should always compile your program using the "-z noexecstack" option in this lab.

Configuring /bin/sh (Ubuntu 16.04 VM only). In both Ubuntu 12.04 and Ubuntu 16.04 VMs, the /bin/sh symbolic link points to the /bin/dash shell. However, the dash program in these two VMs have an important difference. The dash shell in Ubuntu 16.04 has a countermeasure that prevents itself from being executed in a Set-UID process. Basically, if dash detects that it is executed in a Set-UID process, it immediately changes the effective user ID to the process’s real user ID, essentially dropping the privilege. The dash program in Ubuntu 12.04 does not have this behavior.

Since our victim program is a Set-UID program, and our attack uses the system() function to run a command of our choice. This function does not run our command directly; it actually invokes /bin/sh to run our command. Therefore, the countermeasure in /bin/dash will immediately drop the Set-UID privilege before executing our command, making our attack more difficult. Therefore, we will link /bin/sh to another shell that does not have such a countermeasure. We have installed a shell program called zsh in our Ubuntu 16.04 VM. We use the following commands to link /bin/sh to zsh (there is no need to do these in Ubuntu 12.04):


$ sudo rm /bin/sh

$ sudo ln -s /bin/zsh /bin/sh
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It should be noted that the countermeasure implemented in dash can be easily circumvented with a little bit more effort. We use zsh just to make the task relatively easier to conduct.

2.2    The Vulnerable Program

/* retlib.c */


/* This program has a buffer overflow vulnerability. */ /* Our task is to exploit this vulnerability */ #include <stdlib.h>

#include <stdio.h>

#include <string.h>


int bof(FILE *badfile)
{

char buffer[12];

/* The following statement has a buffer overflow problem */ fread(buffer, sizeof(char), 40, badfile);


return 1;

}

int main(int argc, char **argv)

{
FILE *badfile;

badfile = fopen("badfile", "r");

bof(badfile);


printf("Returned Properly\n");

fclose(badfile);

return 1;

}

The above program has a buffer overflow vulnerability. It first reads an input of size 40 bytes from a file called badfile into a buffer of size 12. Since the function fread() does not check the buffer boundary, a buffer overflow will occur. This program is a root-owned Set-UID program, so if a normal user can exploit this buffer overflow vulnerability, the user might be able to get a root shell. It should be noted that the program gets its input from a file called badfile, which is provided by users. Therefore, we can construct the file in a way such that when the vulnerable program copies the file contents into its buffer, a root shell can be spawned.

Let us first compile the code and turn it into a root-owned Set-UID program. Do not forget to in-clude the -fno-stack-protector option (for turning off the StackGuard protection) and the "-z noexecstack" option (for turning on the non-executable stack protection). It should also be noted that changing ownership must be done before turning on the Set-UID bit, because ownership change will cause the Set-UID bit to be turned off.


    • gcc -fno-stack-protector -z noexecstack -o retlib retlib.c

    • sudo chown root retlib
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$ sudo chmod 4755 retlib

2.3    Task 1: Finding out the addresses of libc functions

In Return-to-libc attacks, we need to jump to some existing code that has already been loaded into the memory. We will use the system() and exit() functions in the libc library in our attack, so we need to know their addresses. There are many ways to do that, but using the GNU gdb debugger is probably the easiest method. Let us pick an arbitrary program a.out to debug:

$ gdb
a.out

(gdb) b main
À
(gdb) r
`
(gdb) p system
´
$1 =
{<text variable, no debug info>} 0xb7c56da0 <__libc_system>
(gdb) p exit
ˆ
$2 =
{<text variable, no debug info>} 0xb7c4a9d0 <__GI_exit>




In the above gdb commands, we first set a break point at the main function (Line À); we then run the debugged program using command r (Line `). The program will be stopped at the break point. We can now print out the address of the system() and exit() functions (Lines ´ and ˆ). From the outcome, we can see that the address for the system() function is 0xb7c56da0, and the address for the exit() function is 0xb7c4a9d0. The actual addresses in your VM might be different from these numbers.

2.4    Task 2: Putting the shell string in the memory

Our attack strategy is to jump to the system() function and get it to execute an arbitrary command. Since we would like to get a shell prompt, we want the system() function to execute the "/bin/sh" program. Therefore, the command string "/bin/sh" must be put in the memory first and we have to know its address (this address needs to be passed to the system() function). There are many ways to achieve these goals; we choose a method that use environment variables. Students are encouraged to use other approaches.

When we execute a program from a shell prompt, the shell actually spawns a child process to execute the program. All the exported shell variables will become the environment variable of the child process. This creates an easy way for us to put some arbitrary string in the child process’s memory. Let us define a new shell variable MYSHELL, and let it contain the string "/bin/sh". From the following commands, we can verify that the string gets into the child process, and it is printed out by the env command running inside the child process.


    • export MYSHELL=/bin/sh

    • env | grep MYSHELL MYSHELL=/bin/sh


We will use the address of this variable as an argument to system() call. The location of this variable in the memory can be found out easily using the following program:


void main(){
char* shell =    getenv("MYSHELL");
if (shell)

printf("%x\n", (unsigned int)shell);

}

If the address randomization is turned off, you will find out that the same address is printed out. However, when you run the vulnerable program retlib, the address of the environment variable might not be exactly

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the same as the one that you get by running the above program; such an address can even change when you change the name of your program (the number of characters in the file name makes difference). The good news is, the address of the shell will be quite close to what you print out using the above program. Therefore, you might need to try a few times to succeed.

2.5    Task 3: Exploiting the Buffer-Overflow Vulnerability

We are ready to create the content of badfile. Since the content involves some binary data (e.g., the address of the libc functions), we wrote a C program to do the construction. We provide you with a skeleton of the code, with the essential parts left for you to fill out.
/* exploit.c */


#include <stdlib.h>

#include <stdio.h>

#include <string.h>
int main(int argc, char **argv)
{

char buf[40];
FILE *badfile;

badfile = fopen("./badfile", "w");

/* You
need to decide the addresses and


the
values for
X, Y, Z. The order of the following
three statements
does not imply the order of X,
Y, Z.
Actually, we intentionally scrambled the order.
*/
*(long
*) &buf[X]
=
some address ;
//
"/bin/sh"
P
*(long
*) &buf[Y]
=
some address ;
//
system()
P
*(long
*) &buf[Z]
=
some address ;
//
exit()
P


fwrite(buf, sizeof(buf), 1, badfile);

fclose(badfile);

}

You need to figure out the addresses in lines marked by P, as well as to find out where to store those addresses (i.e., the values for X, Y, and Z). If your values are incorrect, your attack might not work. In your report, you need to describe how you decide the values for X, Y and Z. Either show us your reasoning, or if you use a trial-and-error approach, show your trials.
After you finish the above program, compile and run it; this will generate the contents for badfile. Run the vulnerable program retlib. If your exploit is implemented correctly, when the function bof() returns, it will return to the system() function, and execute system("/bin/sh"). If the vulnerable program is running with the root privilege, you can get the root shell at this point.

$
gcc -o exploit exploit.c
$
./exploit
// create the badfile
$
./retlib
// launch the attack by running the vulnerable program
# <----
You’ve got a root shell!






Attack variation 1: Is the exit() function really necessary? Please try your attack without including the address of this function in badfile. Run your attack again, report and explain your observations.

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Attack variation 2: After your attack is successful, change the file name of retlib to a different name, making sure that the length of the new file name is different. For example, you can change it to newretlib. Repeat the attack (without changing the content of badfile). Will your attack succeed or not? If it does not succeed, explain why.

2.6    Task 4: Turning on Address Randomization

In this task, let us turn on the Ubuntu’s address randomization protection and see whether this protection is effective against the Return-to-libc attack. First, let us turn on the address randomization:

$ sudo sysctl -w kernel.randomize_va_space=2

Please run the same attack used in the previous task. Can you succeed? Please describe your observation and come up with your hypothesis. In the exploit.c program used in constructing badfile, we need to provide three addresses and the values for X, Y, and Z. Which of these six values will be incorrect if the address randomization is turned on. Please provide evidences in your report.

If you plan to use gdb to conduct your investigation, you should be aware that gdb by default disables the address space randomization for the debugged process, regardless of whether the address randomiza-tion is turned on in the underlying operating system or not. Inside the gdb debugger, you can run "show disable-randomization" to see whether the randomization is turned off or not. You can use "set disable-randomization on" and "set disable-randomization off" to change the set-ting.

    • Guidelines: Understanding the function call mechanism

3.1    Understanding the Stack Layout

To know how to conduct the Return-to-libc attack, it is essential to understand how the stack works. We use a small C program to understand the effects of a function invocation on the stack. More detailed explanation can be found in the SEED book, Computer Security: A Hands-on Approach, by Wenliang Du.
/* foobar.c */

#include<stdio.h>

void foo(int x)

{

printf("Hello world: %d\n", x);

}


int main()

{

foo(1);

return 0;

}

We can use "gcc -S foobar.c" to compile this program to the assembly code. The resulting file foobar.s will look like the following:


......

8 foo:

    • pushl   %ebp

10    movl    %esp, %ebp
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11
subl
$8, %esp

12
movl
8(%ebp), %eax





13
movl
%eax, 4(%esp)

14
movl
$.LC0, (%esp)
: string "Hello world: %d\n"
15
call
printf

16
leave


17
ret



......






21
main:


22
leal
4(%esp), %ecx

23
andl
$-16, %esp

24
pushl
-4(%ecx)

25
pushl
%ebp

26
movl
%esp, %ebp





27
pushl
%ecx

28
subl
$4, %esp

29
movl
$1, (%esp)

30
call
foo

31
movl
$0, %eax

32
addl
$4, %esp





33
popl
%ecx

34
popl
%ebp

35
leal
-4(%ecx), %esp

36
ret








3.2    Calling and Entering foo()

Let us concentrate on the stack while calling foo(). We can ignore the stack before that. Please note that line numbers instead of instruction addresses are used in this explanation.

    • Line 28-29:: These two statements push the value 1, i.e. the argument to the foo(), into the stack. This operation increments %esp by four. The stack after these two statements is depicted in Fig-ure 1(a).

    • Line 30: call foo: The statement pushes the address of the next instruction that immediately follows the call statement into the stack (i.e the return address), and then jumps to the code of foo(). The current stack is depicted in Figure 1(b).

    • Line 9-10: The first line of the function foo() pushes %ebp into the stack, to save the previous frame pointer. The second line lets %ebp point to the current frame. The current stack is depicted in Figure 1(c).

    • Line 11: subl $8, %esp: The stack pointer is modified to allocate space (8 bytes) for local variables and the two arguments passed to printf. Since there is no local variable in function foo, the 8 bytes are for arguments only. See Figure 1(d).

3.3    Leaving foo()

Now the control has passed to the function foo(). Let us see what happens to the stack when the function returns.
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bfffe764
Parameters
00000001



bfffe760



esp






Parameters

Return addr

esp




00000001

080483dc



bfffe764
Parameters
bfffe760

Return addr
bfffe75c

Old ebp


esp




00000001

080483dc

bfffe768


bfffe764

bfffe760

bfffe75c

bfffe758

ebp

    (a) Line 28: subl $4, %esp

Line 29: movl $1, (%esp)





bfffe764
Parameters
00000001
bfffe760



Return addr
080483dc
bfffe75c



Old ebp
bfffe768
bfffe758



Local

ebp
variables

bfffe750



esp


(d) Line 11: subl $8, %esp


(b) Line 30: call foo
(c) Line 9: push %ebp








Line 10: movl %esp, %ebp


bfffe764



bfffe764
Parameters


Parameters
00000001








00000001



bfffe760


bfffe760











Return addr


esp




080483dc







bfffe75c











esp












(e) Line 16: leave    (f) Line 17: ret

Figure 1: Entering and Leaving foo()


    • Line 16: leave: This instruction implicitly performs two instructions (it was a macro in earlier x86 releases, but was made into an instruction later):

mov    %ebp, %esp

pop    %ebp

The first statement release the stack space allocated for the function; the second statement recover the previous frame pointer. The current stack is depicted in Figure 1(e).

    • Line 17: ret: This instruction simply pops the return address out of the stack, and then jump to the return address. The current stack is depicted in Figure 1(f).

    • Line 32: addl $4, %esp: Further resotre the stack by releasing more memories allocated for foo. As you can clearly see that the stack is now in exactly the same state as it was before entering the function foo (i.e., before line 28).

    • Submission

You need to submit a detailed lab report, with screenshots, to describe what you have done and what you have observed. You also need to provide explanation to the observations that are interesting or surprising. Please also list the important code snippets followed by explanation. Simply attaching code without any explanation will not receive credits.