Python interview traces: explain aliasing, defaults and identity

Operation contract

The trace copies an outer list, checks the shared inner object, and mutates that object. It then uses a None-default function twice to prove that the two returned tag lists have distinct identity. A useful answer states when objects are allocated and which references still point to them before giving the output.

Failure and ownership boundary

Identity is not equality. Two different lists can compare equal, while one shared list can change beneath several names. Correcting a mutable default alone also does not copy a list supplied by a caller. Read Python function arguments: avoid shared mutable defaults, Python lists: slicing copies the outer sequence, not nested objects and Python closures: capture loop values at the intended time before translating these traces to concurrent work.

Working program

nested = [[125]]
outer_copy = nested.copy()
print(nested is outer_copy)
print(nested[0] is outer_copy[0])
outer_copy[0].append(75)
print(nested)

def new_tags(tags=None):
    return [] if tags is None else tags

print(new_tags() is new_tags())

Output

False
True
[[125, 75]]
False

Costs and limits

An outer copy is O(n) reference work; identity tests compare object identity without traversing all elements. Equality of containers may traverse their contents.

Common Mistakes

• Explain ownership before naming the output.
• Do not use is as general value equality.

Connected lessons

Python lists: slicing copies the outer sequence, not nested objects, Python function arguments: avoid shared mutable defaults, Python closures: capture loop values at the intended time.