[Solved] What will this code print?lst = [10, 20, 30] it = iter(lst) print(next(it)) print(next(it)) print(next(it)) print(next(it)) | Python

Python - Magic Methods and Operator Overloading

What will this code print?

lst = [10, 20, 30]
it = iter(lst)
print(next(it))
print(next(it))
print(next(it))
print(next(it))

A10 20 30 None

B10 20 30 30

C10 20 30 StopIteration error

D10 20 30 40

Step-by-Step Solution

Solution:

Step 1: Understand iter() and next()
iter(lst) creates an iterator over the list. next(it) returns next item or raises StopIteration if none left.
Step 2: Trace the calls
First three next() calls print 10, 20, 30. Fourth next() raises StopIteration error.
Final Answer:
10 20 30 StopIteration error -> Option C
Quick Check:
next() raises StopIteration when exhausted [OK]

Quick Trick: next() raises StopIteration when no items left [OK]

Common Mistakes:

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