please explain the lost output | python

Question

def chval(x):
x*=3
print(x)
x=3
chval(x)
print(x)
x=[1,2]
chval(x)
print(x)

Code output:
9
3
[1,2,1,2,1,2]
[1,2,1,2,1,2]

the lost out-put was little confusing some one please explain that please

Answer 1

list is a mutable type. Mutable types behave as if they are passed by reference while immutable types behave as if they are passed value.

More info:
https://realpython.com/JUMP_LINK__&&__python__&&__JUMP_LINK-pass-by-reference/

Check the id's of the objects. Also, note I changed the local variables name the y in the function.

def chval(y):
y*=3
print("id: y", id(y))
print(y)
x=3
chval(x)
print("id: x", id(x))
print(x)
x=[1,2]
chval(x)
print("id: x", id(x))
print(x)

Answer 2

Take a look at function body

x *= 3
print(x)

What happens to <x> in the `chval` function is contextual. Meaning, what happens to <x> when <x> was an `int` will be different to that when <x> was a `list`.

When <x> was an `int`, its value will be multiplied by 3. Now remember, `int` is an immutable type (class), meaning, the modification to `int` <x> in function `chval` will not reflect to the actual object outside the `chval` function.

When <x> was a `list`, the function will replicate the `list` contents three times. This is why [1, 2] becomes [1, 2, 1, 2, 1, 2]. And also remember that `list` is a mutable type (class), meaning, the changes written to the `list` will reflect to the actual object outside the `chval` function.