CODEWITHSUNDEEP
pythonscopeinterpreted-language
amdson
amdson

Reputation: 1

How does python determine whether to categorize a variable as local or non-local in a function?

Running this code prints 2.

x = 2
def foo():
    print(x)
foo()

However, running this code returns an UnboundLocalVariable exception when I attempt to print x. (Line 3).

x = 2
def foo():
    print(x)
    x = 10
foo()

Python is an interpreted language, so I don't understand how it can 'know' that I will assign x a value as a local variable on line 5, when it gives an error on line 4. Is there some sort of pseudo-compiling when I define the function?

Upvotes: 0

Views: 48

Answers (1)

VintiumDust
VintiumDust

Reputation: 105

In python, variable definition is attached to the scope that it's in. In this example, x is attached to the scope of foo at the time foo is compiled. Python gives you an UnboundLocalError on line 3 because x is attached to foo's scope but has not been bound yet. The same error can be seen if no global x is present:

>>> def foo():
...     print(x)
...     x = 10
...
>>> foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in foo
UnboundLocalError: local variable 'x' referenced before assignment

Exploring further, we can see that when compiling line-by-line (outside of a function, in the interactive REPL), python can't look ahead to see if x will be defined:

>>> print(x)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'x' is not defined

And we get a NameError instead. UnboundLocalError is a subclass of NameError because it indicates the same problem, just with a more helpful name and message.

Upvotes: 0

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