Python function directly altering list, not returning it

Question

[SOLVED]

So the code goes like:

>>> a = [1,3,2]
>>> my_func(a)
>>> a == []
True

Where my_func alters the list without returning it. I know how to do this in C with pointers, but really confused over a python solution.

Thanks in advance!!

EDIT: So I am doing a radix sort which has a helper function and the helper function returns the sorted list. I want the main function to alter the original list instead of returning it:

def radix(a):
    base = ...
    temp = radix_helper(a, index, base)
    a[:] = []
    a.extend(temp)

So it would run as:

>>> a = [1,3,4,2]
>>> radix(a)
>>> a
[1,2,3,4]

Are you asking why this is allowed to happen, or do you want to know how to write a function that does this? It's not clear. — SethMMorton
– SethMMorton, Commented Nov 1, 2013 at 22:34

Ignacio Vazquez-Abrams · Accepted Answer · 2013-11-01 22:34:46Z

2

Lists are mutable, so all you need to do is mutate the list within the function.

def my_func(l):
  del l[:]

answered Nov 1, 2013 at 22:34

Ignacio Vazquez-Abrams

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Peter DeGlopper · Accepted Answer · 2013-11-01 22:35:32Z

1

Python passes parameters by value, but the value of a is a reference to a list object. You can modify that list object in your function:

def my_func(a):
    a.append('foobar')

This can be the cause of unplanned side effects if you forget that you're working directly with the object in question.

answered Nov 1, 2013 at 22:35

Peter DeGlopper

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1 Comment

SethMMorton Over a year ago

No, if you do a = [] inside your function, a now refers to a different object than what was passed to the function. This will not give the effect the OP wants.

Simeon Visser · Accepted Answer · 2013-11-01 22:42:29Z

0

If the identifier is fixed then you can use the global keyword:

a = [1,2,3]

def my_func():
    global a
    a = []

Otherwise you can modify the argument directly:

a = [1,2,3]

def my_func(a):
    a.clear()

answered Nov 1, 2013 at 22:42

Simeon Visser

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