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

Do you mean?

def gs(a,b):
    def ry():
        if a==b:
            return a
    return ry()

Answer 2

you return ry() explicitly instead of just calling it.

Answer 3

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.

Check for that then:

def gs(a,b):
    def ry():
        if a==b:
            return a
    ret = ry()
    if ret: return ret
    # do other stuff

Answer 4

As you mention "steps" in a function, it almost seems like you want a generator:

def gs(a,b):
  def ry():
    if a==b:
      yield a
  # If a != b, ry does not "generate" any output
  for i in ry():
    yield i
  # Continue doing stuff...
  yield 'some other value'
  # Do more stuff.
  yield 'yet another value'

(Generators can now also act as coroutines, since Python 2.5, using the new yield syntax.)

Answer 5

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]

Answer 6

I had a similar problem, but solved it by simply changing the order of the call.

def ry () if a==b gs()

in some languages like javascript you can even pass a function as a variable in a function:

function gs(a,b, callback) { if (a==b) callback(); }

gs (a,b, ry);