Mutable default arguments

Python is so duck-typing. However, mutable default arguments of function call isn’t one of them. Here is an example:

def compute(inputs: list=[]):
    inputs.append("something")
    patterns = ["a list based on"] + inputs
    return patterns

MISUNDERSTANDING: Function call are always idempotent. inputs will be initialized to an empty list when lacking parameters. Illusion of this function call:

>>> compute()
["a list based on", "something"]

>>> compute()
["a list based on", "something"] #Illusion, Not the reality

Reality:

>>> compute()
["a list based on", "something"]

>>> compute()
["a list based on", "something", "something"]

There 2 things under the hood.

Default variable is an attribute of the function object. Yes, your function is just an object and your default variable is an attribute of that object. It’s an attribute of the object for all function calls.

Assignment always creates a new local variable in the innermost scope. Consult this for more.

There are two types of local variables: function default attribute and function call’s new local by assignment. Both variables(function attribute and variable of function call are both local). It doesn’t involve closure.

def func(foo=[]):
  if foo == [] or foo == [1]:
    print(id(foo))
    foo.append(1)
    foo = []
    print(id(foo))
  print(func.__defaults__)
  print(locals())
  print(foo)
  return foo

func()
print()
func()
print()
func()

Output

140674899168960
140674899568704
([1],)
{'foo': []}
[]

140674899168960
140674899568704
([1, 1],)
{'foo': []}
[]

([1, 1],)
{'foo': [1, 1]}
[1, 1]

Important

foo = [] assignment will create a new variable instead of changing default foo. if foo == [] or foo == [1]: is checking default argument foo’s value. But foo = [] created a new local foo variable.