Function don't update the list realtime python

Question

def func(a,b):
a=a.append(b)
if a is not None:
    for i in a:
        print(i)
li=[1,2,3]
def testcall():
    c=10
    func(li,c)
if __name__=='__main__':
        test()

Why it is not printing the updated list
even though is I update the list in test function and then sent it, it’ll not print anything. Does anyone have any clue why the function is having this strange behavior.
If I let the "func" to only accept the one argument and sent update the list in "test" It’ll still don’t show anything.

Asked By: Faizan Ali

||

Source

Answer 1

To answer your main concern of "why doesn’t it print anything", when you perform a = a.append(b), the inner a.append(b) returns nothing, so a is nothing. Also, a major issue exists with your code because it lacks correct indentation and you’ve misnamed your call to testcall() as test(). This is probably what you want.

def func(a,b):
  print(a) # [1,2,3]
  a.append(b) 
  print(a) # [1,2,3,10]
  if a is not None:
    for i in a:
      print(i)

li=[1,2,3]

def testcall():
  c=10
  func(li,c)

if __name__=='__main__':
  testcall()

Some experiments you can do from within IDLE:

>>> a = [1,2,3]
>>> b = [4,5,6]
>>> c = a.append(b)
>>> c
>>> a
[1, 2, 3, [4, 5, 6]]
>>>

Answered By: Porkfu

Function don't update the list realtime python

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