CODEWITHSUNDEEP
python
my_question
my_question

Reputation: 3235

Can I make function to return value of a list instead its reference?

Here is the code:

class Foo:
  def __init__(self):
    self.l = [1, 2, 3]

  def get_list(self):
    return self.l

f = Foo()
l2 = f.get_list()
print(f.get_list())
l2.remove(2)
print(l2)
print(f.get_list())

Here is the run:

>python /tmp/1.py
[1, 2, 3]
[1, 3]
[1, 3]

So the function get_list() returns reference.

Is there a way to have have it return value?

Upvotes: 1

Views: 54

Answers (2)

Raghav Maheshwari
Raghav Maheshwari

Reputation: 19

You can return the copy of the list like this

def get_list(self):
    # It will return shallow copy of the list.
    return self.l.copy()

Or, You can also use deepcopy module for deepcopy like this

from copy import deepcopy 

def get_list(self):
    return deepcopy(self.l)

For eg.

a = [[1,2,3],[2,3]]
b = a.copy()

# If I append something to a, b will be unchanged.
a.append(1)
# a = [[1,2,3],[2,3], 1]
# b = [[1,2,3],[2,3]]

# If I modify any object inside a, it will also reflect in b. i.e. shallow copy.
a[0].append(5)
# a = [[1,2,3,5],[2,3], 1]
# b = [[1,2,3,5],[2,3]]

In case of deepcopy, b still remain unchanged i.e deepcopy also make copies of inner objects.

Upvotes: 1

nmpauls
nmpauls

Reputation: 174

Seems like a good place to use list copying.

class Foo:
  def __init__(self):
    self.l = [1, 2, 3]

  def get_list(self):
    return self.l.copy()

or

class Foo:
  def __init__(self):
    self.l = [1, 2, 3]

  def get_list(self):
    return self.l[:]

Upvotes: 1

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