Passing arguments down recursive functions

Question

I want to pass an argument from the first call of a recursive function down to the later ones:

Example:

def function(x):
    if base_case:
        return 4
    else:
        return function(x_from_the_first_call + x_from_this_call)

Is there any better way of doing this than a closure?

E.g.

def function(outer_x):
    def _inner(x)
        if base_case:
            return 4
        else:
            return function(outer_x + x)
    return _inner(outer_x)

Can you just make your function accept two arguments? Then just pass the two values separately (with a default value for the initial call) and add them inside the function. — BrenBarn
– BrenBarn, Commented Jan 17, 2014 at 20:55
@BrenBarn Would you consider that cleaner than a closure? My problem with doing that is it causes there to be an extra parameter which could confuse anyone calling the function who is unaware of the implementation, which strikes me as a bad thing. — rlms
– rlms, Commented Jan 17, 2014 at 20:58
I usually do it with a separate named argument that defaults to None. And "since we're all consenting adults" in python, presumably any caller who changes that arg has a good reason for doing so. — roippi
– roippi, Commented Jan 17, 2014 at 21:14

ndpu · Accepted Answer · 2014-01-17 21:16:06Z

1

If you will change x somehow in function, then this should work i think:

def function(x, *args):
    if base_case:
        return 4
    else:
        new_x = x+1 # some change to x

        if args:
            # in args all previous x values

        # remove if in case if you need all previous values
        if not args:
            args.append(x)

        return function(new_x, *args)

edited Jan 17, 2014 at 21:16

answered Jan 17, 2014 at 20:59

ndpu

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