CODEWITHSUNDEEP
pythonfunctionooplinked-list
THE_GO_ODFATHER
THE_GO_ODFATHER

Reputation: 53

How are the objects and variables are being used inside the function when they are declared outside the function

I made an object and a variable outside a function but I am able to use it inside a function I was just messing around the code and I noticed this ps I was trying to make a linked list

enter image description here

this is the rest of the code I cant add another image here

ll=Linkedlist()
first=Node(1)
Second=Node(2)
Third=Node(3)
ll.head=first
first.next=Second
Second.next=Third
ll.tail=Third
x=4
def display():
    ll.trav()
    print(x)

display()

my question now is how am I able to use ll and x=4 in the function display when both are defined outside the function

Upvotes: 1

Views: 127

Answers (3)

mujjiga
mujjiga

Reputation: 16906

Official docs here

Sample (comments inline)

x = 123

def fun_1():
  # gloabl x, can read it but cannot overwrite it
  print ("fun_1\t", id(x), x)

def fun_2():
  # local x, can read/write
  x = 0
  print ("fun_2\t", id(x), x)

def fun_3():
  # Global x, can read/write into it
  global x
  x = 0
  print ("fun_3\t", id(x), x)


print ("main\t", id(x), x)
fun_1()
print ("main\t", id(x), x)
fun_2()
print ("main\t", id(x), x)
fun_3()
print ("main\t", id(x), x)

Output:

main     94861900917056 123
fun_1    94861900917056 123
main     94861900917056 123
fun_2    94861900913120 0
main     94861900917056 123
fun_3    94861900913120 0
main     94861900913120 0

Not that method calls on object are not overwriting the object but rather changing the some internal state. i.e obj.do_something(...) only changes the internal state of obj but the object pointed bu obj is still same which, so it is not an overwrite operation.

Upvotes: 1

pu239
pu239

Reputation: 748

The other answer tells you how to do it in another way (the way you have done it is also not wrong, as is proved by its working). However it doesn't answer the question

how am I able to use ll and x=4 in the function display when both are defined outside the function

The answer to this is from another post: link

In Python, variables that are only referenced inside a function are implicitly global. If a variable is assigned a new value anywhere within the function’s body, it’s assumed to be a local. If a variable is ever assigned a new value inside the function, the variable is implicitly local, and you need to explicitly declare it as ‘global’.
Though a bit surprising at first, a moment’s consideration explains this. On one hand, requiring global for assigned variables provides a bar against unintended side-effects. On the other hand, if global was required for all global references, you’d be using global all the time. You’d have to declare as global every reference to a built-in function or to a component of an imported module. This clutter would defeat the usefulness of the global declaration for identifying side-effects.

Upvotes: 2

Vroomy
Vroomy

Reputation: 21

If you just want to use ll and x in display(), you just pass them as parameters :

def display(xParam, llParam):
   llParam.trav()
   print(xParam)
display(x, ll)

Upvotes: 0

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