CODEWITHSUNDEEP
python
user3150100
user3150100

Reputation: 45

Confusing simple python list behavior with +=

list4=[7,8]
def proc3(mylist):
    mylist+=[9]


print list4
proc3(list4)
print list4

Why does this code produce the answers [7,8] and [7,8,9]? I expected [7,8] and [7,8]. Doesn't mylist+=[9] concatenate the number 9 and create a new list as a local variable only? Why does the "print list4" after the running proc3(list4), but outside the procedure, not result in the original list? I must be missing something obvious here. Thanks in advance for any help.

Upvotes: 0

Views: 91

Answers (3)

user2555451
user2555451

Reputation:

+= does not return a new list object. Instead, it modifies the original list object in-place.

In other words, this line inside proc3:

mylist+=[9]

is modifying the list object referenced by mylist. As a result, it is no longer [7, 8] but [7, 8, 9].

Upvotes: 3

Emanuele Paolini
Emanuele Paolini

Reputation: 10162

the += operator modifies the object. Notice that 

x += y

is much different from

x = x + y

(check with your test code)

Upvotes: 0

Christian Tapia
Christian Tapia

Reputation: 34146

Quoting this answer in another post:

Everything is a reference in Python. If you wish to avoid that behavior you would have to create a new copy of the original with list(). If the list contains more references, you'd need to use deepcopy():

Let's:

list4 = [7, 8]
def proc3(mylist):
    print id(list4)
    mylist += [9]
    print id(list4)


print list4
print id(list4)
proc3(list4)
print id(list4)
print list4

This would print:

[7, 8]
36533384
36533384
36533384
36533384
[7, 8, 9]

So as you can see, it's the same list in every moment.

Upvotes: 1

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