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?
Python variables are scoped to the innermost function, class, or module in which they're assigned. Control blocks like if
and while
blocks don't count, so a variable assigned inside an if
is still scoped to a function, class, or module.
(Implicit functions defined by a generator expression or list/set/dict comprehension do count, as do lambda expressions. You can't stuff an assignment statement into any of those, but lambda parameters and for
clause targets are implicit assignment.)
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.
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.
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