Variable-length keyword arguments (**kwargs) in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of function using **kwargs with default values

What is the output of this Python code?

Python

def greet(**kwargs):
    name = kwargs.get('name', 'Guest')
    greeting = kwargs.get('greeting', 'Hello')
    return f"{greeting}, {name}!"

print(greet(name='Alice'))

A Hello, Alice!

B Alice, Hello!

C Hello, Guest!

D TypeError

💡 Hint

Look at how kwargs.get() provides default values if the key is missing.

The function uses kwargs.get('name', 'Guest') to get the 'name' key or default to 'Guest'. Since 'name' is provided as 'Alice', it returns 'Hello, Alice!'.

❓ Predict Output

intermediate

Counting keys in **kwargs

What will be printed by this code?

Python

def count_keys(**kwargs):
    return len(kwargs)

print(count_keys(a=1, b=2, c=3))

A TypeError

B 0

C 3

D 1

💡 Hint

Count how many keyword arguments are passed to the function.

The function returns the number of keys in kwargs. Since three keyword arguments are passed, it returns 3.

🔧 Debug

advanced

Identify the error with **kwargs unpacking

What error does this code raise when run?

Python

def show_info(**kwargs):
    print(kwargs['age'])

info = {'name': 'Bob', 'age': 25}
show_info(info)

A TypeError

B KeyError

C SyntaxError

D No error, prints 25

💡 Hint

Check how the dictionary is passed to the function expecting keyword arguments.

The function expects keyword arguments, but info is passed as a single positional argument, causing a TypeError.

🚀 Application

advanced

Which call prints all keyword arguments correctly?

Given this function, which call will print all keyword arguments as expected?

Python

def print_all(**kwargs):
    for key, value in kwargs.items():
        print(f"{key}: {value}")

A print_all({'x': 1, 'y': 2})

B print_all([('x', 1), ('y', 2)])

C print_all(*{'x': 1, 'y': 2})

D print_all(x=1, y=2)

💡 Hint

Remember how to pass keyword arguments to a function expecting **kwargs.

Only option D passes keyword arguments correctly. Option D passes a dict as a positional argument, B passes a list as a positional argument, and C tries to unpack a dict as positional args.

🧠 Conceptual

expert

Behavior of **kwargs with overlapping keys

Consider this code snippet. What will be the output?

Python

def func(a, **kwargs):
    print(a)
    print(kwargs)

func(1, a=2, b=3)

A

2
{'b': 3}

B TypeError

C

1
{'a': 2, 'b': 3}

D

1
{'b': 3}

💡 Hint

Think about how Python handles duplicate argument names in function calls.

Python raises a TypeError because the argument 'a' is given both positionally and as a keyword argument.