Is Python call by value or call by reference?

How are parameters passed to functions in Python?

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#python #call-by-value #call-by-reference

[varunnadipudi]

I was exploring for this the whole Internet, for the simple and precise explanation. luckily found you and you made it all clear. Thank you dude.

[shivamshinde]

very nice explanation...couldn't find out better explanation than this on YouTube..

[ChinmayRGharat]

Still i am confused, i am writing a simulator where i am passing objects of various class to various other classes, and then use those objects to assign certain values to variables in that class, it's working but according to you it should not!

[akhils]

BASIC - In Python, variables inside functions are considered local if and only if they appear in the left side of an assignment statement. so immutable objects are acts like it its were passed by values

# TODO: acts as a pass by value (IMMUTABLE OBJECTS) but it's passed by reference only
def increment(a):
print('inside function init id(a)', id(a)) # id(a) is same as id(num) so everything is passed by reference only
a += 1 # assignment causes assigning the new value to a ( so a become local to function scope )
print('inside function val of a', a)
print('inside function id(a)', id(a)) # id(a) is not same as id(num) cause of assignment involved
return a


num = 1
print('outside function id(num)', id(num))
increment(num)
print('outside function val of num', num)


But incase of Mutable objects

# TODO: acts as a pass by reference (MUTABLE OBJECTS)
def extend(a):
print('inside function init id(a)', id(a)) # id(a) is same as id(li)
a.append(5) # since no assignment involved any changes to the a( reference of li) also affects the value of the variable outside the function.
print('inside function val of a', a)
print('inside function id(a)', id(a)) # id(a) is same as id(li)
return a


li = [1, 2, 3, 4]
print('outside function id(li)', id(li))
extend(li)
print('outside function val of li', li)

[aashitabansal]

Thank you so much I was struggling in this concept and you helped to clear it...🙌

[krishnasoni]

Unique information is provided by you, thanks

[swadhikarc]

and if in case you seriously intended to empty the list you should:

def clean_it(mylist):
mylist.clear()

[malharjajoo]

So basically it's the way the assignment operator works as opposed to member functions (like append) that can mutate the object.

[raviteja]

Thank you. A very nice explanation, better than many videos.

[arpitjain]

another important thing += and = are different, list_ += [1] will modify the same object but list_ = list_ + 1 will create new object . This is only true for mutables, for strings/tuples/int which are immutable both will create new objects

[sujitsutar]

God level explanation very well explained thank you it was helpful.

[codecompass]

I have a few things to ask.

1. you created a function added a code block that execute a certain operation but where is the output going?
2. Since you haven't returned the output how can you say that your var i.e. num will be incremented just by passing it into a function without getting a return?

[mustafabohra]

Thanks buddy!

Never have thought above it even though using the function calls passing the lists on a daily basis.

[tejaskasare]

Crystal Clear bro....you got one more subscriber

[parshvasoni]

due seriously perfect explanation....liked it ....thanks

[niladrighosh]

I got ur point but can you describe me the 1st example where you use the num variable ....it was not incremented??

[SapnaSaini]

really liked the way you explained it .Thanks !

[sadiqsatwilkar]

Thanks, that was a very good explanation.
Is it true to say then that if a mutable object is passed as a parameter then it is call by reference whereas if an immutable object is passed as a parameter then it is call by value?

[ChinmayRGharat]

So objects are passed by reference, no multiple instances of obj created?

[subeesha]

Keep rocking bro..fantastic explanation...