inner function accessing outer function's variable

Question

Let's suppose to have this simple code:

def my_outer_function():
    outer_var = 123
    def my_inner_function():
        return outer_var + 1
    return my_inner_function

get_inner = my_outer_function()
get_inner()

I wonder there isn't any kind of runtime error. outer_var - the variable of the outer function - is available only when that function is running on, i.e. it vanishes when my_outer_function ends. But when I call get_inner(), my_outer_function() is already ended so I would have bet on a runtime error since my_inner_function couldn't find outer_var.

How do you explain all this?

Its called a [closure](http://stackoverflow.com/questions/13857/can-you-explain-closures-as-they-relate-to-python). — Paul Rooney, Jan 04 '17 at 22:44
This is called a "closure". When you return the inner function, it captures the binding of the outer variable. — Barmar, Jan 04 '17 at 22:44

score 3 · Accepted Answer · edited Sep 26 '17 at 22:08

the variable of the outer function is available only when that function is running on, i.e. it vanishes when my_outer_function ends.

That's not entirely true. The variable is available in my_outer_function's scope. my_inner_function has the scope of its own declarations and its parent's scope.

my_inner_function references a variable outside of its own scope, and so those references are bound to my_inner_function as a closure when its parent's scope is no longer available. To learn more about closures, see Can you explain closures (as they relate to Python)? (taken from Paul Rooney's comment)

inner function accessing outer function's variable

1 Answers1