[python] What's the scope of a variable initialized in an if statement?

I'm new to Python, so this is probably a simple scoping question. The following code in a Python file (module) is confusing me slightly:

if __name__ == '__main__':
    x = 1

print x

In other languages I've worked in, this code would throw an exception, as the x variable is local to the if statement and should not exist outside of it. But this code executes, and prints 1. Can anyone explain this behavior? Are all variables created in a module global/available to the entire module?

This question is related to python variables if-statement scope local-variables

The answer is


Scope in python follows this order:

  • Search the local scope

  • Search the scope of any enclosing functions

  • Search the global scope

  • Search the built-ins

(source)

Notice that if and other looping/branching constructs are not listed - only classes, functions, and modules provide scope in Python, so anything declared in an if block has the same scope as anything decleared outside the block. Variables aren't checked at compile time, which is why other languages throw an exception. In python, so long as the variable exists at the time you require it, no exception will be thrown.


As Eli said, Python doesn't require variable declaration. In C you would say:

int x;
if(something)
    x = 1;
else
    x = 2;

but in Python declaration is implicit, so when you assign to x it is automatically declared. It's because Python is dynamically typed - it wouldn't work in a statically typed language, because depending on the path used, a variable might be used without being declared. This would be caught at compile time in a statically typed language, but with a dynamically typed language it's allowed.

The only reason that a statically typed language is limited to having to declare variables outside of if statements in because of this problem. Embrace the dynamic!


Yes. It is also true for for scope. But not functions of course.

In your example: if the condition in the if statement is false, x will not be defined though.


Yes, they're in the same "local scope", and actually code like this is common in Python:

if condition:
  x = 'something'
else:
  x = 'something else'

use(x)

Note that x isn't declared or initialized before the condition, like it would be in C or Java, for example.

In other words, Python does not have block-level scopes. Be careful, though, with examples such as

if False:
    x = 3
print(x)

which would clearly raise a NameError exception.


you're executing this code from command line therefore if conditions is true and x is set. Compare:

>>> if False:
    y = 42


>>> y
Traceback (most recent call last):
  File "<pyshell#6>", line 1, in <module>
    y
NameError: name 'y' is not defined

Unlike languages such as C, a Python variable is in scope for the whole of the function (or class, or module) where it appears, not just in the innermost "block". It is as though you declared int x at the top of the function (or class, or module), except that in Python you don't have to declare variables.

Note that the existence of the variable x is checked only at runtime -- that is, when you get to the print x statement. If __name__ didn't equal "__main__" then you would get an exception: NameError: name 'x' is not defined.


And note that since Python types are only checked at runtime you can have code like:

if True:
    x = 2
    y = 4
else:
    x = "One"
    y = "Two"
print(x + y)

But I'm having trouble thinking of other ways in which the code would operate without an error because of type issues.


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