variable scope in python nested functions

Question

As I am studying decorators, I noticed something strange :

def f():
... msg='aa'
... def a():
... print msg
... msg='bb'
... def b():
... print msg
... return a,b
... 
>>> a,b = f()
>>> a()
bb
>>> b()
bb
>>> 

Why a() returns 'bb' and not 'aa' ??

Answer 1

Because a and b have the same outer scope, in which msg is bound to 'bb'. Put them in separate functions if you want them to have separate scopes.

• Eric: yes but when a() is defined, msg = 'aa' then.Does that means functions are really defined at the end of the parent function ?
• Ignacio Vazquez-Abrams: The scope is bound to the function when the function is defined. But the name within the scope is rebound to a different string.
• AndiDog: @Eric: The variable msg is not defined as a constant "aa". The variable survives in the local scope of f, and the inner functions a/b have access to that scope at any later time. As you overwrite the value of msg immediately, i.e. before calling a or b, they will read the new value.
• Eric: Yes, I believed it was like ADA language template mechanism. But, yes, that sound finally very logical. Thank you all

Answer 2

Both a and b have read access to the outer scope (the local scope of f). As you overwrite the value of msg, the later call to a/b will read the new value.

• Eric: OK, I believed that It was like ADA language template mechanism, in fact a/b are using the still living msg variable in its scope, so having the value 'bb' at the time a/b are executed.