Development server with debug mode in Flask - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What happens when Flask debug mode is enabled?

Consider a Flask app started with debug mode enabled. What is the main behavior you will observe during development?

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    return 'Hello, Flask!'

if __name__ == '__main__':
    app.run(debug=True)

A The server disables the debugger and requires manual restarts on code changes.

B The server automatically reloads on code changes and shows detailed error messages in the browser.

C The server runs in production mode with caching enabled.

D The server runs slower but disables all error messages for security.

💡 Hint

Think about what debug mode helps developers do easily.

Debug mode in Flask enables the interactive debugger and automatic reloads when code changes, making development faster and easier.

📝 Syntax

intermediate

Which code correctly enables debug mode in Flask?

Select the code snippet that correctly starts a Flask app with debug mode enabled.

A app.run(debug='True')

B app.run(debug=1)

C app.run(debug='yes')

D app.run(debug=True)

💡 Hint

Debug expects a boolean value, not a string.

The debug parameter expects a boolean True or False. Passing a string or integer may not work as intended.

🔧 Debug

advanced

What error occurs if you run Flask with debug=True but forget to set __name__ == '__main__'?

Given this code snippet, what error or behavior will occur? from flask import Flask app = Flask(__name__) @app.route('/') def home(): return 'Hi' app.run(debug=True)

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    return 'Hi'

app.run(debug=True)

A RuntimeError: The server is running but debug mode causes multiple reloads and crashes.

B The server starts but immediately crashes with RuntimeError about working outside the main thread.

C The server starts but runs the app twice due to the reloader causing duplicate output.

D The server starts normally but debug mode is ignored and no auto-reload happens.

💡 Hint

Think about how Flask's reloader works when not protected by the main guard.

Without the if __name__ == '__main__' guard, the app.run(debug=True) causes the reloader to spawn a child process, running the code twice.

❓ state_output

advanced

What is the output when Flask debug mode triggers an error in a route?

If a route function raises an exception while Flask is running with debug=True, what will the user see in the browser?

Flask

from flask import Flask
app = Flask(__name__)

@app.route('/')
def home():
    raise ValueError('Oops!')

if __name__ == '__main__':
    app.run(debug=True)

A A detailed interactive debugger page showing the error and stack trace.

B A plain 500 Internal Server Error page with no details.

C The browser shows a blank page with no response.

D The server crashes and stops running immediately.

💡 Hint

Debug mode helps developers see what went wrong directly in the browser.

With debug=True, Flask shows an interactive debugger with error details and stack trace in the browser when exceptions occur.

🧠 Conceptual

expert

Why should Flask debug mode never be enabled in production?

Select the best reason why running Flask with debug=True is unsafe in a production environment.

A Debug mode enables an interactive debugger that can execute arbitrary code, exposing the server to attacks.

B Debug mode disables all security features, allowing anyone to access the server files.

C Debug mode slows down the server drastically, causing denial of service.

D Debug mode automatically deletes logs and user data after each request.

💡 Hint

Think about what the interactive debugger allows a user to do.

The interactive debugger allows code execution from the browser, which is a major security risk if exposed publicly.