Passing list as an argument to a function . Can someone explain this behavior?

Question

def test(lister):
    lister.append('why am i being shown in output?')

def pass_the_list():
    x = []
    test(x)
    print(x)

pass_the_list()

Output

['why am i being shown in output?']

Process finished with exit code 0

how to pass string as a reference without return?

...what? What does that have to do with what you've asked? Also Python passes everything the same way, but strings are *immutable*. — jonrsharpe, Nov 04 '19 at 08:12

score 2 · Answer 1 · answered Nov 04 '19 at 07:59

2

The list x is passed as a reference not value, therefore changes made in the function will preserve also when the function finishes.

answered Nov 04 '19 at 07:59

L. Letovanec

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how to pass a string as a reference? – GokusSG Nov 04 '19 at 08:11
strings are natively passed as value, so you would need to modify it in function then return it and assign to the variable again like ... x = test(x) – L. Letovanec Nov 04 '19 at 08:18
Thanks for the awnser that explains my question but why is the behavior not consistent with all python objects? – GokusSG Nov 04 '19 at 08:42
1

It is consistent, but some objects are mutable (E.g. lists) and some are not (E.g. strings) – L. Letovanec Nov 04 '19 at 09:04

Martin · Answer 2 · 2019-11-04T10:39:13.390

You can pass a string as a refence, but it doesnt matter, because string, unlike list, is immutable. So any change you do to the string will cause creation of a new object.

test_string = 'Test my immutability'
test_string[0]='B'
>> TypeError: 'str' object does not support item assignment

You can modify string like this though (by putting it into something mutable like list)

test_string = 'Test my immutability'

test_list = [test_string]


def reference_test(test_list):
    test_list[0]=test_list[0].replace('Test','Fest')

reference_test(test_list)

print(test_list)
>> ['Rest my immutability']

Passing list as an argument to a function . Can someone explain this behavior?

2 Answers2