Python Coding Challenge - Question with Answer (ID -010326)

 

🔍 Step 1: What does the loop do?

for i in range(3):

range(3) → 0, 1, 2

So the loop runs 3 times.

Each time we append:

lambda x: x + i

So it looks like we are creating:

x + 0
x + 1
x + 2

But… ❗ that’s NOT what actually happens.

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🔥 The Important Concept: Late Binding

Python does not store the value of i at the time the lambda is created.

Instead, the lambda remembers the variable i itself, not its value.

This is called:

🧠 Late Binding

The value of i is looked up when the function is executed, not when it is created.

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 After the Loop Finishes

After the loop ends:

i = 2

The loop stops at 2, so the final value of i is 2.

Now your list funcs contains 3 functions, but all of them refer to the SAME i.

Effectively, they behave like:

lambda x: x + 2
lambda x: x + 2
lambda x: x + 2

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🚀 Now Execution Happens

[f(10) for f in funcs]

Each function is called with 10.

Since i = 2:

10 + 2 = 12

For all three functions.

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✅ Final Output

[12, 12, 12]

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🎯 How To Fix It (Capture Current Value)

If you want expected behavior (10, 11, 12), do this:

funcs = []

for i in range(3):
    funcs.append(lambda x, i=i: x + i)

print([f(10) for f in funcs])

Why this works?

i=i creates a default argument, which stores the current value of i immediately.

Now output will be:

[10, 11, 12]

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💡 Interview Tip

If someone asks:

Why does Python give [12,12,12]?

Say confidently:

Because of late binding in closures — lambdas capture variables, not values.

Book: 🐍 50 Python Mistakes Everyone Makes ❌