Reason For Unintuitive Unboundlocalerror Behaviour 2

Following up on Reason for unintuitive UnboundLocalError behaviour (I will assume you've read it). Consider the following Python script: def f(): # a+=1 # 1 aa=a

Solution 1:

Unlike strings and integers, lists in Python are mutable objects. This means they are designed to be changed. The line

c[0] += 'c'

is identical to saying

c.__setitem__(0, c.__getitem__(0) + 'c')

which doesn't make any change to what the name c is bound to. Before and after this call, c is the same list – it's just the contents of this list that have changed.

Had you said

c += ['c']
c = [42]

in the function f(), the same UnboundLocalError would have occured, because the second line makes c a local name, and the first line translates to

c = c + ['c']

requiring the name c to be already bound to something, which (in this local scope) it isn't yet.

Solution 2:

The important thing to think about is this: what object does a (or b or c) refer to? The line a += 1 is changing which integer a refers to. Integers are immutable, so when a changes from 1 to 2, it's really the same as a = a + 1, which is giving a an entirely new integer to refer to.

On the other hand, c[0] += 'c' doesn't change which list c refers to, it merely changes which string its first element refers to. Lists are mutable, so the same list can be modified without changing its identity.