How do I implicitly define a parameter to an empty list?

Question

In the code below, I’ve been wondering how we could tell if a parameter, b, is given.

The problem is that the third call of func does not retain [5] in a newly created list but rather some pointer in the start that b is pointing to. I’m guessing this is defined before in the program stack entering func call itself, so the calling and returning of func would not change b…?

Any insight is appreciated.

def func(a, b=[]):

    b.append([a])
    print(b)
    return b

func(3)
func(4, [])
func(5)

Asked By: meesinlid

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Source

Answer 1

The best way is to assign the default value of b to something arbitrary (usually None) then check if b is defined that way:

def func(a, b=None):
    if b is None:
        b = []

    b.append([a])
    print(b)
    return b

func(3)
func(4, [])
func(5)

Answered By: Alec

Answer 2

You can define b to a default value, e.g. b=None, and then pick either the value of b if given, or pick an empty list.

def func(a, b=None):

    lst = b or []
    lst.append([a])
    print(lst)
    return lst

func(3)
#[[3]]
func(4, [])
#[[4]]
func(5)
#[[5]]

Answered By: Devesh Kumar Singh

How do I implicitly define a parameter to an empty list?

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