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⛓️ Tricky Python Interview Question

> What will this code output and why?

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

for f in funcs:
    f()

❓ Question: Why does this code print 2, 2, 2 instead of 0, 1, 2?

🔍 Explanation:

This is a classic example of late binding in Python closures. The lambda functions don't capture the value of i at each step of the loop. Instead, they all hold a reference to the same variable i.

🧠 What happens:

• The for loop completes. By the end of the loop, the variable i holds the value 2.
• The funcs list contains three functions, but each one is defined to print the value of whatever i is at the time of execution.
• When you call each function f(), it looks up the current value of i in its enclosing scope, which is 2.

👉 Result:

2
2
2

✅ How to fix:

You can force the lambda to capture the value of i at the time of definition by using a default argument.

funcs = []
for i in range(3):
    # The default argument captures i's value at definition time
    funcs.append(lambda num=i: print(num))

for f in funcs:
    f()

This trick works because default arguments are evaluated when the function is defined, not when it's called. This will correctly print:

0
1
2

This is a common interview question that tests your understanding of closures, scope, and late binding in Python.

━━━━━━━━━━━━━━━
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Interview question

What will this code output and why?

a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)

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

a += (4, 5)

print(a)
print(b)

Try to answer yourself first, then check the explanation.

Part 1. Lists
a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)

Step-by-step:

a = [1, 2, 3]

Variable a refers to the list [1, 2, 3].

b = a
Now b refers to the same list as a.
Not a copy, but the very same object in memory.

a += [4, 5]
Important: for lists, += works as an in-place modification:
this is roughly the same as:

a.extend([4, 5])


That is, elements 4 and 5 are added to the existing list that both a and b refer to.

Therefore:

print(a)  # [1, 2, 3, 4, 5]
print(b)  # [1, 2, 3, 4, 5]

Both point to the same modified list.

Part 2. Tuples
a = (1, 2, 3)
b = a

a += (4, 5)

print(a)
print(b)


Tuples are immutable, and here the magic begins.

a = (1, 2, 3)
a refers to the tuple (1, 2, 3).

b = a
b refers to the same tuple (1, 2, 3).

a += (4, 5)
For tuples, += cannot modify the object in place (they are immutable).
So Python does this:

a = a + (4, 5)


That is, a new tuple (1, 2, 3, 4, 5) is created and variable a is reassigned to it.

b still points to the old tuple (1, 2, 3).

Therefore:

print(a)  # (1, 2, 3, 4, 5)
print(b)  # (1, 2, 3)

The trick

For the list, a += [4, 5] mutates the object in place, visible through all variables referencing it.

For the tuple, a += (4, 5) creates a new object and reassigns only a.

In summary:

# List part:
[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5]

# Tuple part:
(1, 2, 3, 4, 5)
(1, 2, 3)

If you want, I can explain another tricky question — about mutable default arguments in functions or about loops and closures.

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