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GO Classes DPP | GATE DA | Python Programming | List Iterators

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Consider the following Python code:

nums = [1, 2, 3]
it = iter(nums)

first = next(it)
nums.append(4)
second = list(it)

nums.append(5)
third = list(it)

print(first, second, third)

What is the output of the code above?

  1. 1 [2, 3] [4, 5]
  2. 1 [2, 3, 4] []
  3. 1 [2, 3, 4] [5]
  4. 1 [2, 3] []

1 Answer

0 0 votes

The statement $\texttt{it = iter(nums)}$ creates an iterator over the list.

The statement $\texttt{first = next(it)}$ takes the first value from the iterator.

first = 1

Now the iterator is positioned after $\texttt{1}$.

Then $\texttt{nums.append(4)}$ adds $\texttt{4}$ before the remaining iterator values are consumed.

So $\texttt{second = list(it)}$ collects the remaining values:

[2, 3, 4]

After this, the iterator is exhausted.

Even though $\texttt{nums.append(5)}$ adds a new value later, the exhausted iterator cannot be reused.

Therefore, $\texttt{third = list(it)}$ gives an empty list.

The final output is:

1 [2, 3, 4] []

Correct Option: B

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