mutable_default

Mutable Default Arguments

One of Python's most famous gotchas — default values are evaluated once at function definition time.

Files

File	Description
example.py	The trap, the fix, and related lambda bug

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Descriptive Example

Scenario

A function that accumulates items — why does the second call remember the first?

def append_bad(item, target=[]):
    target.append(item)
    return target

print(append_bad(1))   # [1]
print(append_bad(2))   # [1, 2]  ← surprise! expected [2]

The list [] is created once when Python defines the function. Every call shares the same list object.

The fix

def append_good(item, target=None):
    if target is None:
        target = []
    target.append(item)
    return target

print(append_good(1))   # [1]
print(append_good(2))   # [2]  ← correct

None is immutable and safe as a sentinel. A fresh list is created per call when no target is passed.

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Interview Q&A

Q1: Why does def f(a=[]) behave unexpectedly?
A: Default arguments are evaluated once at definition time. The same list object is reused across all calls that omit a.

Q2: How do you safely use mutable defaults?
A: Use None as default, then create the mutable inside the function: if x is None: x = [].

Q3: Does this affect immutable defaults like def f(a=0)?
A: No. Integers are immutable. Rebinding a inside the function doesn't affect other calls. The issue is mutating a shared object in place.

Q4: Does this apply to class attributes?
A: Yes! class Foo: items = [] is shared across all instances. Fix in __init__: self.items = [].

Q5: How do dataclasses handle this?
A: @dataclass with field(default_factory=list) — factory called per instance, not shared.

Q6: What is the related lambda-in-loop trap?
A: [lambda x: i*x for i in range(4)] — all lambdas bind to final i. Fix: lambda x, i=i: i*x.

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Run

python3 example.py