[Solved] Given this async Flask route, what will be the output when accessed?from flask import Flask import asyncio app = Flask(__name__) @app.route('/') async def home(): await asyncio.sleep(0.... | Flask

Flask - Ecosystem and Patterns

Given this async Flask route, what will be the output when accessed?

from flask import Flask
import asyncio
app = Flask(__name__)

@app.route('/')
async def home():
    await asyncio.sleep(0.1)
    return 'Done'

AThe route crashes because Flask does not support async.

BThe route returns immediately with 'Done'.

CThe route returns 'Done' after a short delay.

DThe route causes a syntax error due to await usage.

Step-by-Step Solution

Solution:

Step 1: Understand async route behavior
The route uses await asyncio.sleep(0.1) which pauses the function asynchronously for 0.1 seconds.
Step 2: Predict output after await
After the short async pause, the route returns the string 'Done'. So the client sees 'Done' after a slight delay.
Final Answer:
The route returns 'Done' after a short delay. -> Option C
Quick Check:
Await sleep delays then returns [OK]

Quick Trick: Await pauses async route before returning result [OK]

Common Mistakes:

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