Python Coding Challenge - Question with Answer (ID -260526)

 

Explanation:

🔹 Step 1: Create Empty List

x = []

An empty list is created:

[]

This list will store lambda functions.

🔹 Step 2: Start Loop

for i in range(3):

range(3) gives:

0,1,2

Loop runs 3 times.

🔹 Step 3: First Iteration

i = 0

Execute:

x.append(lambda: i)

Lambda created:

lambda: i

⚠️ Important:

Lambda does NOT store current value immediately 😈

It stores REFERENCE to variable i.

List now:

[lambda]

🔹 Step 4: Second Iteration

i = 1

Another lambda added:

lambda: i

List now:

[lambda, lambda]

BUT both lambdas still reference SAME variable:

i

🔹 Step 5: Third Iteration

i = 2

Third lambda added.

List becomes:

[lambda, lambda, lambda]

All lambdas reference SAME final variable:

i

🔹 Step 6: Loop Ends

After loop finishes:

i = 2

⚠️ Final value survives outside loop 😈

🔹 Step 7: Access x[1]

x[1]

This gives second lambda function.

Then:

x[1]()

calls lambda.

Lambda checks CURRENT value of:

i

Current value:

2

So:

x[1]() → 2

🔹 Step 8: Print Result

print(2)

👉 Final Output:

2

🔥 Final Output

2

Book: PYTHON LOOPS MASTERY