Iterator protocol in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of custom iterator class

What is the output of this Python code using a custom iterator?

Python

class CountDown:
    def __init__(self, start):
        self.current = start
    def __iter__(self):
        return self
    def __next__(self):
        if self.current <= 0:
            raise StopIteration
        self.current -= 1
        return self.current + 1

for number in CountDown(3):
    print(number)

A 1\n2\n3

B 2\n1\n0

C 4\n3\n2

D 3\n2\n1

💡 Hint

Look at how __next__ decreases self.current and what it returns.

The iterator starts at 3. Each call to __next__ returns the current value before decreasing it. It stops when current is 0.

❓ Predict Output

intermediate

Output of iter and next with list

What is the output of this code snippet?

Python

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

A 10\n20\nStopIteration

B 10\n20\nTraceback (most recent call last): StopIteration

C 10\n20\nNone

D 10\n20\n0

💡 Hint

What happens when next() is called but no items remain?

Calling next() on an exhausted iterator raises a StopIteration exception, causing a traceback if not caught.

🧠 Conceptual

advanced

Understanding iterator protocol methods

Which method must a Python object implement to be considered an iterator?

A __iter__ only

B __next__ only

C __iter__ and __next__

D __call__

💡 Hint

Think about what Python expects when using a for-loop on an iterator.

An iterator must have both __iter__ returning itself and __next__ to get the next item.

❓ Predict Output

advanced

Output of generator and iterator interaction

What is the output of this code?

Python

def gen():
    yield 1
    yield 2

g = gen()
print(next(g))
print(iter(g) is g)
print(next(g))

A 1\nTrue\n2

B 1\nTrue\nStopIteration

C 1\nFalse\n2

D 1\nFalse\nStopIteration

💡 Hint

Generators are their own iterators. What does iter(g) return?

Calling iter() on a generator returns the generator itself, so iter(g) is g is True.

❓ Predict Output

expert

Output of nested iterator with StopIteration handling

What is the output of this code?

Python

class Repeater:
    def __init__(self, value, times):
        self.value = value
        self.times = times
        self.count = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.count >= self.times:
            raise StopIteration
        self.count += 1
        return self.value

r = Repeater('Hi', 2)
it = iter(r)
try:
    while True:
        print(next(it))
except StopIteration:
    print('Done')

A Hi\nHi\nDone

B Hi\nDone

C Hi\nHi

D Done

💡 Hint

How many times does the iterator yield before stopping?

The iterator yields 'Hi' twice, then raises StopIteration, caught by the except block which prints 'Done'.

Practice

1. What does the __iter__ method do in the iterator protocol?

easy

A. Returns the iterator object itself

B. Returns the next item in the sequence

C. Stops the iteration

D. Creates a list from the iterable

Solution

1. Step 1: Understand the role of __iter__

   The __iter__ method is called to get an iterator object from an iterable.

2. Step 2: Identify what __iter__ returns

   It returns the iterator object itself, which has the __next__ method to fetch items.

3. Final Answer:

   Returns the iterator object itself -> Option A

4. Quick Check:

   __iter__ returns iterator object [OK]

Hint: Remember: __iter__ returns the iterator itself [OK]

Common Mistakes:

• Confusing __iter__ with __next__
• Thinking __iter__ returns the next item
• Assuming __iter__ stops iteration

2. Which of the following is the correct way to define an iterator class in Python?

easy

A. class MyIter:
def __next__(self):
pass

B. class MyIter:
def next(self):
pass

C. class MyIter:
def __iter__(self):
return self
def __next__(self):
pass

D. class MyIter:
def iter(self):
return self

Solution

1. Step 1: Check required methods for iterator

   An iterator class must have __iter__ returning self and __next__ to get next item.

2. Step 2: Match methods with options

   class MyIter: def __iter__(self): return self def __next__(self): pass correctly defines both __iter__ and __next__ methods.

3. Final Answer:

   Defines both __iter__ and __next__ methods -> Option C

4. Quick Check:

   Iterator class needs __iter__ and __next__ [OK]

Hint: Iterator class must have __iter__ and __next__ methods [OK]

Common Mistakes:

• Using next() instead of __next__()
• Missing __iter__ method
• Defining iter() instead of __iter__()

3. What will be the output of this code?

class Count:
    def __init__(self, limit):
        self.limit = limit
        self.num = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.num < self.limit:
            self.num += 1
            return self.num
        else:
            raise StopIteration

for i in Count(3):
    print(i, end=' ')

medium

A. Error: StopIteration not handled

B. 0 1 2

C. 1 2 3 4

D. 1 2 3

Solution

1. Step 1: Understand the iterator behavior

   The Count class starts num at 0 and returns num+1 until it reaches limit 3.

2. Step 2: Trace the loop output

   Loop prints 1, 2, 3 then raises StopIteration to end loop.

3. Final Answer:

   1 2 3 -> Option D

4. Quick Check:

   Count(3) yields 1 to 3 [OK]

Hint: StopIteration ends loop; count from 1 to limit [OK]

Common Mistakes:

• Starting count from 0 instead of 1
• Expecting 4 as output
• Thinking StopIteration causes error

4. Identify the error in this iterator implementation:

class MyIter:
    def __init__(self):
        self.data = [1, 2, 3]
        self.index = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.index <= len(self.data):
            result = self.data[self.index]
            self.index += 1
            return result
        else:
            raise StopIteration

medium

A. The condition should be self.index < len(self.data)

B. Missing return self in __iter__

C. Should raise StopIteration before returning result

D. Index should start at 1, not 0

Solution

1. Step 1: Analyze the index condition

   Index goes from 0 to len(data)-1. Using <= allows index == len(data), causing IndexError.

2. Step 2: Correct the condition

   Change condition to self.index < len(self.data) to avoid out-of-range access.

3. Final Answer:

   The condition should be self.index < len(self.data) -> Option A

4. Quick Check:

   Index must be less than length to avoid error [OK]

Hint: Use < to avoid index out of range errors [OK]

Common Mistakes:

• Using <= instead of < in index check
• Forgetting to return self in __iter__
• Starting index at 1 instead of 0

5. You want to create a custom iterator that returns only even numbers from 0 up to a given limit (exclusive). Which implementation correctly follows the iterator protocol and filters evens?

class EvenIterator:
    def __init__(self, limit):
        self.limit = limit
        self.current = 0
    def __iter__(self):
        return self
    def __next__(self):
        while self.current < self.limit:
            val = self.current
            self.current += 1
            if val % 2 == 0:
                return val
        raise StopIteration

hard

A. Fails because __iter__ should return a new object each time

B. Correct implementation returning even numbers up to limit

C. Incorrect because it returns odd numbers instead

D. Raises StopIteration too early, missing some evens

Solution

1. Step 1: Check iterator protocol methods

   Class defines __iter__ returning self and __next__ with loop and StopIteration.

2. Step 2: Verify filtering logic

   Inside __next__, it loops until limit, returns only even values, skipping odds.

3. Final Answer:

   Correct implementation returning even numbers up to limit -> Option B

4. Quick Check:

   Iterator filters evens correctly [OK]

Hint: Use while loop inside __next__ to skip unwanted items [OK]

Common Mistakes:

• Returning odd numbers by mistake
• Not raising StopIteration when done
• Returning new iterator in __iter__ each time