Interesting results with the '+=' increment operator [duplicate]

Question

I had learned that n = n + v and n += v are the same. Until this;

def assign_value(n, v): n += v print(n) l1 = [1, 2, 3] l2 = [4, 5, 6] assign_value(l1, l2) print(l1)

The output will be:

[1, 2, 3, 4, 5, 6] [1, 2, 3, 4, 5, 6]

Now when I use the expanded version:

def assign_value(n, v): n = n + v print(n) l1 = [1, 2, 3] l2 = [4, 5, 6] assign_value(l1, l2) print(l1)

The output will be:

[1, 2, 3, 4, 5, 6] [1, 2, 3]

Using the += has a different result with the fully expanded operation. What is causing this?

In python this is because __add__ and __iadd__ can be implemented independently and differently. — luk2302
– luk2302, Commented Jan 4, 2023 at 11:46
I had learned that n = n + v and n += v are the same. They are not the same. — Martijn Pieters
– Martijn Pieters, Commented Jan 4, 2023 at 12:17

Dan · Accepted Answer · 2023-01-04 11:44:51Z

Thats because in the first implementation you are editing the list n itself (and therefore the changes still apply when leaving the function), while on the other implementation you are creating a new temporary list with the same name, so when you leave the function the new list disappears and the variable n is linked to the original list.

the += operator works similarly to x=x+y for immutable objects (since they always create new objects), but for mutable objects such as lists they work differently. x=x+y creats a new object x while x+=y edits the current object.

bereal · Accepted Answer · 2023-01-04 11:50:08Z

It may seem counter-intuitive, but they are not always the same. In fact,

a = a + b means a = a.__add__(b), creating a new object
a += b means a = a.__iadd__(b), mutating the object

__iadd__, if absent, defaults to the __add__, but it also can (and it does, in the case of lists) mutate the original object in-place.

fhunter · Accepted Answer · 2023-01-04 11:48:13Z

This works on how python treats objects and passes variables into functions. Basically - in first example (with += ) You are passing n and v into function by "pass-by-assignment" So n gets modified and it will be also modified out of function scope.

In second example - n is reassigned inside of the function to a new list. Which is not seen outside of the function.

Yash Mehta · Accepted Answer · 2023-01-04 11:51:11Z

In your 1st code. You changes list n itself see the below image..!

In your 2nd code. you just created a temporary list which is cleared when function call ends.. see the below images..!

In the next step when function ends the temporary list clear!!

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