Python Coding Challange - Question with Answer (01010925)

 

Let’s carefully break this down.

Code:

g = (i*i for i in range(3))
print(next(g))
print(next(g))

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Step 1: Generator Expression

g = (i*i for i in range(3))

• This creates a generator object.

• It will not calculate squares immediately, but will produce values one at a time when asked (lazy evaluation).

• range(3) → [0, 1, 2].

• So generator will yield:

  • First call → 0*0 = 0

  • Second call → 1*1 = 1

  • Third call → 2*2 = 4

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Step 2: First next(g)

• Asks the generator for its first value.

i = 0 → 0*0 = 0.

• Output: 0.

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Step 3: Second next(g)

• Generator resumes where it left off.

i = 1 → 1*1 = 1.

• Output: 1.

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Final Output:

0
1

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⚡ If you call next(g) one more time → you’ll get 4.
⚠️ If you call again after that → StopIteration error, since generator is exhausted.

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