Returning list in Python

Question

I'm studying Python and am hitting some Pythonic oddity. I can't figure out why in returning, this function throws away data.

def foo(mylist=None):
    print "Input: {}".format(mylist)
    if mylist is None:
        mylist = list()
    if len(mylist) == 3:
        return mylist
    else:
        mylist.append(len(mylist))
        foo(mylist)

print "Output: {}".format(foo())

This prints:

Input: None
Input: [0]
Input: [0, 1]
Input: [0, 1, 2]
Output: None

I would guess that it has to do with pointing to a list that no longer exists, but I don't get that in a simpler example:

def simple_foo():
    to_return = [1, 2, 3]
    return to_return

print "Simple output: {}".format(simple_foo())

I've even tried (in foo) deep-copying my list into a to_return variable, then returning that, but that doesn't seem to work either. Can anybody shed some light on this? I would appreciate it.

Dan D. · Accepted Answer · 2016-11-02 04:19:50Z

8

You have to keep returning:

Your recursive call:

foo(mylist)

needs its return value returned:

return foo(mylist)

answered Nov 2, 2016 at 4:19

Dan D.

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1 Comment

T.J. Gaffney Over a year ago

Thanks! That's not even a Python thing; I should have caught that.

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