Difference between assignment and compound operators in Python

Question

Please look into the code below

def double(arg):
    print("Before: ", arg)
    arg = arg * 2
    print("After: ", arg)

I was studying Head first Python, and I came to this section where they were discussing about pass by value and pass by reference. If we invoke above function with a list as argument such as:

num = [1,2,3]
double(num)
print(num)

The output is :-

Before:  [1, 2, 3]
After:  [1, 2, 3, 1, 2, 3]
[1, 2, 3]

Which seems fine, considering the fact that arg in function double is a new object reference. So the value of num did not change.

But if I use compound operators instead of assignment operators, things work differently as shown:

def double(arg):
    print("Before: ", arg)
    arg *= 2
    print("After: ", arg)

num = [1,2,3]
double(num)
print(num)

The output that I get for this is:

Before:  [1, 2, 3]
After:  [1, 2, 3, 1, 2, 3]
[1, 2, 3, 1, 2, 3]

Why does this happen? I used to think a*=2 and a = a*2 are same. But what's going on in here?

Thanks

Also related: [When is “i += x” different from “i = i + x” in Python?](https://stackoverflow.com/q/15376509/846892) — Ashwini Chaudhary, Jul 19 '17 at 19:06

score 2 · Accepted Answer · answered Jul 19 '17 at 19:01

This is a difference between mutable and immutable objects. A mutable object can implement obj *= something by actually modifying the object in place; an immutable object can only return a new object with the updated value (in which case the result is identical to obj = obj * something). The compound assignment statements can handle either case, it's entirely up to the object's implementation.

score 2 · Answer 2 · answered Jul 19 '17 at 19:01

2

a *= 2 modifies the structure itself (a) whereas a = a*2 reassigns a as a new variable.

(see this question)

answered Jul 19 '17 at 19:01

Cary Shindell

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Difference between assignment and compound operators in Python

2 Answers2