Setting the value of a global variable in the context of a function raises UnboundLocalError?

Question

Maybe my coffee is not strong enough this morning, but this behavior is confusing me right now:

>>> a = 'foo'
>>> def func1():
...   print a
... 
>>> def func2():
...   print a
...   a = 'bar'
... 
>>> func1()
foo
>>> func2()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in func2
UnboundLocalError: local variable 'a' referenced before assignment

(Note that it's the print a statement that's raising the error in func2(), not a = 'bar'.)

Can somebody explain to me what's going on here?

possible duplicate of [Python scope](http://stackoverflow.com/questions/146359/python-scope) — Fred Foo, Mar 18 '11 at 16:34

score 2 · Accepted Answer · edited May 23 '17 at 10:33

2

Because a is being set inside func2, Python assumes it's a local variable. Put a global a declaration before the print statement:

def func2():
    global a
    print a
    a = 'bar'

See also this question about Python scoping rules.

edited May 23 '17 at 10:33

Community

1
1

answered Mar 18 '11 at 16:26

Fred Foo

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Thanks, that's exactly it. I so very rarely use global variables in Python that I have yet to stumble across this. – jsdalton Mar 18 '11 at 16:37

Setting the value of a global variable in the context of a function raises UnboundLocalError?

1 Answers1