Lambda syntax and behavior in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple lambda function

What is the output of this Python code?

Python

func = lambda x: x * 3
result = func(4)
print(result)

A 12

B 7

C Error

D None

💡 Hint

Remember that lambda functions return the expression result.

The lambda multiplies the input by 3. Input is 4, so 4 * 3 = 12.

❓ Predict Output

intermediate

Lambda with multiple arguments

What will this code print?

Python

adder = lambda a, b: a + b
print(adder(5, 10))

A 15

B 510

C TypeError

D None

💡 Hint

The lambda adds two numbers.

The lambda adds 5 and 10, resulting in 15.

❓ Predict Output

advanced

Lambda capturing variable in loop

What is the output of this code?

Python

funcs = []
for i in range(3):
    funcs.append(lambda: i)
results = [f() for f in funcs]
print(results)

A [0, 1, 2]

B [2, 2, 2]

C [0, 0, 0]

D TypeError

💡 Hint

Lambdas capture variables by reference, not by value.

All lambdas capture the same variable i, which ends at 2 after the loop.

❓ Predict Output

advanced

Correct lambda capturing loop variable

What does this code print?

Python

funcs = []
for i in range(3):
    funcs.append(lambda i=i: i)
results = [f() for f in funcs]
print(results)

A [0, 0, 0]

B [2, 2, 2]

C [0, 1, 2]

D SyntaxError

💡 Hint

Default arguments capture the current value at each loop iteration.

Using default argument i=i captures the current value of i for each lambda.

🧠 Conceptual

expert

Behavior of lambda with mutable default argument

Consider this code snippet. What will be printed?

Python

def make_funcs():
    funcs = []
    lst = []
    for i in range(3):
        lst.append(i)
        funcs.append(lambda x=lst: x)
    return funcs

funcs = make_funcs()
for f in funcs:
    print(f())

A TypeError

B

[0]
[0, 1]
[0, 1, 2]

C

[]
[]
[]

D

[0, 1, 2]
[0, 1, 2]
[0, 1, 2]

💡 Hint

Default arguments are evaluated once when the function is defined.

The default argument x=lst captures the same list object, which has all elements at the end.