Python Coding Challenge - Question with Answer (ID -270126)

 

1️⃣ async def foo()

async def foo(): 
     return 5

• This defines an asynchronous function.

• Calling foo() does NOT execute it immediately.

• It returns a coroutine object.

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2️⃣ asyncio.run(foo())

asyncio.run(foo())

• asyncio.run():

  • Creates an event loop

  • Executes the coroutine foo()

  • Waits until it finishes

  • Returns the final result

✅ Since foo() returns 5, this line evaluates to:

5

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3️⃣ + 1

asyncio.run(foo()) + 1

• Now it becomes:

5 + 1

• Result:

6

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4️⃣ print(...)

print(6)

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✅ Final Output

6

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⚠️ Important Trick

If you wrote this instead:

print(foo() + 1)

❌ It would raise an error because:

• foo() returns a coroutine, not an integer

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💡 Key Takeaway

• async def → returns a coroutine

• asyncio.run() → executes it and returns the result

• You can only use the returned value after awaiting/running

900 Days Python Coding Challenges with Explanation