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Mediumpythonfunctionsmutability

~90s

Mutating against rebinding

What does this Python program print?

def f(lst):
    lst.append(4)
    lst = [9, 9]
    lst.append(5)

a = [1, 2, 3]
f(a)
print(a)
Show the answer and the worked solution

Correct answer: B.

[1, 2, 3, 4]

Worked solution

  1. lst starts out referring to the caller's list, so lst.append(4) makes that list [1, 2, 3, 4].
  2. lst = [9, 9] rebinds the local name to a brand new list, breaking the link to the caller's object.
  3. The final append therefore lands on the new list, which is discarded when f returns, so a prints as [1, 2, 3, 4].

Answer: [1, 2, 3, 4].

The idea

Python passes references by value: mutating the object is visible to the caller, rebinding the name is not. The dividing line is the assignment statement in the middle.

Shortcut. Only what happens before lst = [9, 9] reaches the caller, and that is the append(4); everything after it touches a new list.

Common trap. "[1, 2, 3, 4, 5]" misses that the rebinding cuts the link, and so lets the append(5) land on the caller's list too.

Programming Fundamentals

37 of 40

Level

Medium

Target time

~90s

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Mutating against rebinding — Aptitude Question

Mutating against rebinding: What does this Python program print? Answer and worked solution on CodeKairo.

Question

What does this Python program print?

def f(lst):
    lst.append(4)
    lst = [9, 9]
    lst.append(5)

a = [1, 2, 3]
f(a)
print(a)

Options

  1. [1, 2, 3]
  2. [1, 2, 3, 4]
  3. [1, 2, 3, 4, 5]
  4. [9, 9, 5]
Show the answer and the worked solution

Correct answer: B. [1, 2, 3, 4]

Worked solution

  1. lst starts out referring to the caller's list, so lst.append(4) makes that list [1, 2, 3, 4].
  2. lst = [9, 9] rebinds the local name to a brand new list, breaking the link to the caller's object.
  3. The final append therefore lands on the new list, which is discarded when f returns, so a prints as [1, 2, 3, 4].

Answer: [1, 2, 3, 4].

The idea

Python passes references by value: mutating the object is visible to the caller, rebinding the name is not. The dividing line is the assignment statement in the middle.

Shortcut. Only what happens before lst = [9, 9] reaches the caller, and that is the append(4); everything after it touches a new list.

Common trap. "[1, 2, 3, 4, 5]" misses that the rebinding cuts the link, and so lets the append(5) land on the caller's list too.

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