Unexpected behavior of Python's variable scope

Question

I'm a bit confused as to how Python's variable scope system works. Say I have situation like this:

a = 10
def test():
    print(a)

Then everything works just as I expect. Python first looks for a local variable a, fails to find it and then searches for a global variable.

However, in a situation like this:

a = 10
def test():
    print(a)
    a += 1
    print(a)

Python throws an UnboundLocalError exception, apparently originating from line 3 (print(a)). To me it seems that at least to this line nothing has changed, and I don't understand why there is an exception anyway.

@vaultah well sometimes do meet some stubborn people who need's to be answered :) — therealprashant, May 29 '15 at 06:53
@therealprashant they're answered in the dupe, which is *the whole point* of identifying it as such. If you have a better answer, *write it there* so all the dupes pointing to it benefit. If not, sit on your hands. — jonrsharpe, May 29 '15 at 06:56
@jonrsharpe got that. Thanks for the feedback. I will make sure that doesn't happens later — therealprashant, May 29 '15 at 06:57

score 0 · Answer 1 · answered May 29 '15 at 06:53

0

Since python does not have variable declarations, every variable assignment inside the scope of a function is considered local. So, you always have to specify that that variable is a global one:

a = 10
def test():
    global a
    print(a)
    a += 1
    print(a)
test()

answered May 29 '15 at 06:53

Ericson Willians

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Recommending the use of `global` is a bad idea. – Matthias May 29 '15 at 07:09

Unexpected behavior of Python's variable scope

1 Answers1