[Solved] What will be the output of this code?class MyIter: def __init__(self): self.n = 0 def __iter__(self): return self def __next__(self): if self.n | Python

Python - Magic Methods and Operator Overloading

What will be the output of this code?

class MyIter:
    def __init__(self):
        self.n = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.n < 3:
            self.n += 1
            return self.n
        else:
            raise StopIteration

it = MyIter()
print(list(it))
print(list(it))

A[1, 2, 3] followed by []

B[1, 2, 3] followed by [1, 2, 3]

C[] followed by [1, 2, 3]

D[] followed by []

Step-by-Step Solution

Solution:

Step 1: Understand iterator state
The iterator counts from 1 to 3, then stops. After first list(), the iterator is exhausted.
Step 2: Analyze second list() call
Since the iterator is exhausted, second list() returns an empty list.
Final Answer:
[1, 2, 3] followed by [] -> Option A
Quick Check:
Iterator exhausted after first use, second is empty [OK]

Quick Trick: Iterators remember state; exhausted means empty next time [OK]

Common Mistakes:

MISTAKES

Assuming iterator resets automatically
Expecting second list() to repeat values
Confusing iterable with iterator behavior

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What will be the output of this code?

Step 1: Understand iterator state

Step 2: Analyze second list() call

Final Answer:

Quick Check:

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