Variables declared outside function [duplicate]

python

Unlike languages that employ 'true' lexical scoping, Python opts to have specific 'namespaces' for variables, whether it be global, nonlocal, or local. It could be argued that making developers consciously code with such namespaces in mind is more explicit, thus more understandable. I would argue that such complexities make the language more unwieldy, but I guess it's all down to personal preference.

Here are some examples regarding global:-

>>> global_var = 5
>>> def fn():
...     print(global_var)
... 
>>> fn()
5
>>> def fn_2():
...     global_var += 2
...     print(global_var)
... 
>>> fn_2()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in fn_2
UnboundLocalError: local variable 'global_var' referenced before assignment
>>> def fn_3():
...     global global_var
...     global_var += 2
...     print(global_var)
... 
>>> fn_3()
7

The same patterns can be applied to nonlocal variables too, but this keyword is only available to the latter Python versions.

In case you're wondering, nonlocal is used where a variable isn't global, but isn't within the function definition it's being used. For example, a def within a def, which is a common occurrence partially due to a lack of multi-statement lambdas. There's a hack to bypass the lack of this feature in the earlier Pythons though, I vaguely remember it involving the use of a single-element list...

Note that writing to variables is where these keywords are needed. Just reading from them isn't ambiguous, thus not needed. Unless you have inner defs using the same variable names as the outer ones, which just should just be avoided to be honest.

When Python parses a function, it notes when a variable assignment is made. When there is an assignment, it assumes by default that that variable is a local variable. To declare that the assignment refers to a global variable, you must use the global declaration.

When you access a variable in a function, its value is looked up using the LEGB scoping rules.

So, the first example

x = 1
def inc():
    x += 5
inc()

produces an UnboundLocalError because Python determined x inside inc to be a local variable,

while accessing x works in your second example

def inc():
    print x

because here, in accordance with the LEGB rule, Python looks for x in the local scope, does not find it, then looks for it in the extended scope, still does not find it, and finally looks for it in the global scope successfully.

The local names for a function are decided when the function is defined:

>>> x = 1
>>> def inc():
...     x += 5
...     
>>> inc.__code__.co_varnames
('x',)

In this case, x exists in the local namespace. Execution of x += 5 requires a pre-existing value for x (for integers, it's like x = x + 5), and this fails at function call time because the local name is unbound - which is precisely why the exception UnboundLocalError is named as such.

Compare the other version, where x is not a local variable, so it can be resolved at the global scope instead:

>>> def incg():
...    print(x)
...    
>>> incg.__code__.co_varnames
()

Similar question in faq: http://docs.python.org/faq/programming.html#why-am-i-getting-an-unboundlocalerror-when-the-variable-has-a-value

Variables declared outside function [duplicate]

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