Unfurl's debug mode cannot be disabled due to string config parsing (Werkzeug debugger exposure)

Summary

### Summary The Unfurl web app enables Flask debug mode even when configuration sets `debug = False`. The config value is read as a string and passed directly to `app.run(debug=...)`, so any non-empty string evaluates truthy. This leaves the Werkzeug debugger active by default. ### Details - `unfurl/app.py:web_app()` reads `debug` via `config['UNFURL_APP'].get('debug')`, which returns a string. - `UnfurlApp.__init__` passes that string directly to `app.run(debug=unfurl_debug, ...)`. - If `unfurl.ini` omits `debug`, the default argument is the string `"True"`. - As a result, debug mode is effectively always on and cannot be reliably disabled via config. ### PoC 1. Create a local `unfurl.ini` with `debug = False` under `[UNFURL_APP]`. 2. Run the server using `unfurl_app` (or `python -c 'from unfurl.app import web_app; web_app()'`). 3. Observe server logs showing `Debug mode: on` / `Debugger is active!`. 4. The included PoC script `security_poc/poc_debug_mode.py --spawn` automates this check. ### PoC Script (inline) ```python #!/usr/bin/env python3 """ Unfurl Debug Mode PoC (Corrected) ================================ This PoC demonstrates that Unfurl's Flask debug mode is effectively **always enabled by default** due to string parsing of the `debug` config value. Even `debug = False` in `unfurl.ini` evaluates truthy when passed to `app.run(debug=...)`. Two modes: 1) --spawn (default): launch a local Unfurl server with debug=False in a temp config and inspect logs for "Debug mode: on". 2) --target: attempt a remote indicator check (best-effort; may be silent if no exception is triggered). """ import argparse import os import subprocess import sys import tempfile import textwrap import time def run_spawn_check() -> None: repo_root = os.path.abspath(os.path.join(os.path.dirname(__file__), '..')) ini_contents = textwrap.dedent(""" [UNFURL_APP] host = 127.0.0.1 port = 5055 debug = False remote_lookups = false [API_KEYS] bitly = macaddress_io = """).strip() + "\n" with tempfile.TemporaryDirectory() as tmp: ini_path = os.path.join(tmp, 'unfurl.ini') with open(ini_path, 'w') as f: f.write(ini_contents) env = os.environ.copy() env['PYTHONPATH'] = repo_root cmd = [sys.executable, '-c', 'from unfurl.app import web_app; web_app()'] proc = subprocess.Popen( cmd, cwd=tmp, env=env, stdout=subprocess.PIPE, stderr=subprocess.PIPE, text=True ) # Allow server to start and emit logs time.sleep(2) proc.terminate() try: out, err = proc.communicate(timeout=2) except subprocess.TimeoutExpired: proc.kill() out, err = proc.communicate() output = (out or "") + (err or "") print("\n[+] Debug mode spawn check") print(" Config: debug = False") if "Debug mode: on" in output or "Debugger is active" in output: print(" ✅ Debug mode is ON despite debug=False (vulnerable)") else: print(" ⚠️ Debug mode not detected in logs (check output below)") if output.strip(): print("\n--- server output (truncated) ---") print("\n".join(output.splitlines()[:15])) print("--- end ---") def run_remote_probe(target: str) -> None: import requests print("\n[+] Remote debug indicator probe (best-effort)") print(f" Target: {target}") # This app does not easily throw exceptions from user input, so # absence of indicators does NOT prove debug is off. probe_urls = [ f"{target.rstrip('/')}/__nonexistent__", ] detected = False for url in probe_urls: try: resp = requests.get(url, timeout=10) if "Werkzeug Debugger" in resp.text or "Traceback" in resp.text: detected = True print(" ✅ Debug indicators found") break except Exception as e: print(f" ⚠️ Probe failed: {e}") if not detected: print(" ⚠️ No debug indicators found (this is not definitive)") def main(): parser = argparse.ArgumentParser(description='Unfurl debug mode PoC (corrected)') parser.add_argument('--spawn', action='store_true', help='Run local spawn check (default)') parser.add_argument('--target', help='Target Unfurl URL for remote probe') args = parser.parse_args() if args.target: run_remote_probe(args.target) else: run_spawn_check() if __name__ == '__main__': main() ``` ### Impact If the service is exposed beyond localhost (bound to 0.0.0.0 or reverse-proxied), an attacker can access the Werkzeug debugger. This can disclose sensitive information and may allow remote code execution if a debugger PIN is obtained. At minimum, stack traces and environment details are exposed on errors.