Question about closures in python

Question

We don't inner functions have access to s,f
https://code.sololearn.com/cMq02ZJWjaLl/?ref=app

Answer 1

Frogged
Instead of changing the values of the variables 's' and 'f', new variables local to b() and c() named 's' & 'f' are being created.
I'm not sure about this part, but I think that Python thinks in the right-hand-side of
`f, s = s, f`
we are referencing the local variables 's' and 'f'.
This is because changing 'f' and 's' to any other name works
`a, b = f, s`
I don't think there is any way to mutate the 'f' and 's' variables of the a() function without making them global. But maybe someone else can suggest a way?
My suggestion is to make a list instead of the variables, like so
lst = [True, False]
And then replace every 's' with lst[0] and 'f' with lst[1]

Answer 2

Frogged yep , similar to how you can set attributes of classes and objects, you can also set attributes to functions.
Useful when you want the option to retrieve some variable, but not explicity return it with the return statement. Also attributes can be set from inside the function definition or from outside the function definition.

Answer 3

XXX yes..
I think, Python is not clean here....
Finally :
in a python closure all variables that are shared and changed by inner functions should be put into a mutable container.
or..Vitaly Sokol thanks..
alternatively each one is to be marked as non-local

Answer 4

https://code.sololearn.com/cnC9p1Sj395w/?ref=app

Answer 5

It is to understand closures

Answer 6

? its still a function. works the same no matter the scope.
you added a function inside of another functions scope, the variables still need to be passed just like if you created to variables and made a function.
you need to pass them or make them global

Answer 7

that goes with every variable

Answer 8

why wouldn't you just use a chain of elifs?

Answer 9

also you can do it with function attributes:
https://code.sololearn.com/chMKa1VykhF0/?ref=app

Answer 10

Vitaly Sokol that's great

Answer 11

Calvin Thomas they are declared in the outer function. not globally.
that's a point in closures:
Although the outer function is no more in the memory, its variables are accessible and can be shared by the functions created and returned from outer function.

Answer 12

because of namespaces.
you cant reference a variable inside a function without either passing the variable to the function or making the variable global

Answer 13

@slick but it is a closure
no prob for accessing i

Answer 14

How about i

Answer 15

But I have access

Answer 16

Why doesn't "global f, s" work here? Am I not making f and s global variables?

Answer 17

Frogged I see, thanks for the explanation.