See function stop from locating defined class in Python

Question

I'd like to define a class inside a function (for testing purpose) and put a value into a function variable:

def foo():
    myvar = None
    class myclass:
        def run(self):
            myvar = 5

    mm = myclass()
    mm.run()
    print(myvar)

The above prints None

Is there any way other than global to make the myvar variable accessible from the class? The correct answer would print 5

Nope, and that's isn't a good practive to do so. Don't try tricky manipulation — azro
– azro, Commented Nov 25, 2020 at 18:50
Why would you want to do this? This defeats the entire purpose of a class, which is to encapsulate state. As stated above though, you can use nonlocal myvar, which makes it assign to the closest enclosing scope, in this case, the local scope of the function. More reasonably, you should just have your run function return the value of myvar, and then simply capture the resulting value in the caller — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Nov 25, 2020 at 19:06
@juanpa.arrivillaga this is not a helpful answer. Please answer the question first and then provide your opinion. — Ray Salemi
– Ray Salemi, Commented Nov 26, 2020 at 14:52

ufe · Accepted Answer · 2020-11-25 19:00:26Z

1

It's not possible to assign a value to a variable outside the current scope without global. If you need to persist the value within the class you can define class variables instead. Example:

def foo():
    class Class:
        var_to_change = None
        def run (self):
            self.var_to_change = 5
    print (Class.var_to_change)
    instance = Class()
    instance.run()
    print (Class.var_to_change)

I haven't tested the above code but it should work in theory.

answered Nov 25, 2020 at 19:00

ufe

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2 Comments

juanpa.arrivillaga Over a year ago

this will also print None. Also, in this case, it is possible with nonlocal.

ufe Over a year ago

I had no idea there was something like nonlocal. Wow!

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