make a parent function return - super return?

Question

there is a check I need to perform after each subsequent step in a function, so I wanted to define that step as a function within a function.

>>> def gs(a,b):
...   def ry():
...     if a==b:
...       return a
...
...   ry()
...
...   a += 1
...   ry()
...
...   b*=2
...   ry()
... 
>>> gs(1,2) # should return 2
>>> gs(1,1) # should return 1
>>> gs(5,3) # should return 6
>>> gs(2,3) # should return 3

so how do I get gs to return 'a' from within ry? I thought of using super but think that's only for classes.

Thanks

There's been a little confusion... I only want to return a if a==b. if a!=b, then I don't want gs to return anything yet.

edit: I now think decorators might be the best solution.

Answer 1

This should allow you to keep checking the state and return from the outer function if a and b ever end up the same:

def gs(a,b):
    class SameEvent(Exception):
        pass
    def ry():
        if a==b:
            raise SameEvent(a)
    try:
        # Do stuff here, and call ry whenever you want to return if they are the same.
        ry()

        # It will now return 3.
        a = b = 3
        ry()

    except SameEvent as e:
        return e.args[0]