Why reassignment to an argument inside function body does not alter mutable data types in python?

Question

I decided to get know the Python data model and mutability concept, and after that "gotta" moment one test crashed my happiness.

I am completely ok with this example:

x = ['1']
def func(a):
    b = 'new value'
    a.append(b)   

func(x)
print(x) # expected  ['1', 'new value']

BUT ! What is hapenning here ?

x = ['1']
def func(a):
    b = 'new value'
    a = b
    print(a)
    # expected  'new value'
    # factual  'new value'


func(x)
print(x)
# expected  'new value'
# factual ['1'] !!!

Why x does not equal to 'new value' ?

**Python does not support call by reference semantics at all** — juanpa.arrivillaga, Jun 08 '20 at 17:58
`a = b` assigns to a *local variable `a`*. Read the following: https://nedbatchelder.com/text/names.html — juanpa.arrivillaga, Jun 08 '20 at 17:59
Re-assignment to a local variable **never affects another variable**. Note, the mutability of the object you passed in is totally irrelevant. Python evaluation strategy and assignment semantics is exactly the same *for all types of objects* — juanpa.arrivillaga, Jun 08 '20 at 18:01

score 1 · Accepted Answer · answered Jun 08 '20 at 18:03

In the first example, both x and a are pointing to the same list. So when you append a value to that list using the a reference, you see that side effect after the function returns.

In the second example, you reassign the local variable a to point to a new value, but x still points to the original list. After the function returns, a goes away, but x still points to the list.

Why reassignment to an argument inside function body does not alter mutable data types in python?

1 Answers1