[Solved] What will be the output of this code?class Manager: def __enter__(self): print('Start') return 'Resource' def __exit__(self, exc_type, exc_val, exc_tb): print('End') with Manager() as m: print(m) | Python

Python - Context Managers

What will be the output of this code?

class Manager:
    def __enter__(self):
        print('Start')
        return 'Resource'
    def __exit__(self, exc_type, exc_val, exc_tb):
        print('End')

with Manager() as m:
    print(m)

AEnd Start Resource

BResource Start End

CStart Resource End

DStart End Resource

Step-by-Step Solution

Solution:

Step 1: Trace __enter__ method call
When entering the with block, __enter__ prints 'Start' and returns 'Resource', assigned to m.
Step 2: Trace inside with block and __exit__ call
Inside the block, print(m) outputs 'Resource'. After block ends, __exit__ prints 'End'.
Final Answer:
Start Resource End -> Option C
Quick Check:
Execution order = __enter__, block, __exit__ [OK]

Quick Trick: Prints from __enter__, block, then __exit__ [OK]

Common Mistakes:

Assuming __exit__ runs before block
Confusing return value of __enter__

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What will be the output of this code?

Step 1: Trace enter method call

Step 2: Trace inside with block and exit call

Final Answer:

Quick Check:

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What will be the output of this code?

Step 1: Trace __enter__ method call

Step 2: Trace inside with block and __exit__ call

Final Answer:

Quick Check:

Want More Practice?

Step 1: Trace enter method call

Step 2: Trace inside with block and exit call