In Python, variables inside if conditions hide the global scope even if they are not executed?

Question

def do_something():
    print 'doing something...'

def maybe_do_it(hesitant=False):
    if hesitant:
        do_something = lambda: 'did nothing'
    result = do_something()
    print result

maybe_do_it()

The result of this code is:

  File "scope_test.py", line 10, in <module>
    maybe_do_it()
  File "scope_test.py", line 7, in maybe_do_it
    result = do_something()
UnboundLocalError: local variable 'do_something' referenced before assignment

But this code prints "did something..." as expected:

def do_something():
    print 'doing something...'

def maybe_do_it(hesitant=False):
    result = do_something()
    print result

maybe_do_it()

How did the function get overridden even though the condition inside the if statement never executed? This happens in Python 2.7 -- is it the same in Python 3?

Answer 1

How did the function get overridden even though the condition inside the if statement never executed?

The decision whether the variable is local or global is made at compile time. If there is an assignment to a variable anywhere in the function, it's a local variable, no matter if the assignment is ever executed.

This happens in Python 2.7 -- is it the same in python 3?

Yes.

By the way, in Python 2, you can override this behavior by using exec (not recommended):

def do_something():
    print 'doing something...'

def maybe_do_it(hesitant=False):
    if hesitant:
        exec "do_something = lambda: 'did nothing'"
    result = do_something()
    print result

maybe_do_it(False)    # doing something...
maybe_do_it(True)    # did nothing

An exec inside a function will, loosely speaking, postpone the decision whether to look up the variable globally or locally to execution time.