Display contents of a list recursively

Question

We’re studying recursion in our programming course, and we have an exercise where we have to show in Idle the contents of a list using recursiveness. I did so:

def print_list_r(ls=[], n=0):
    if len(ls) == 0 or n < 0:
        return False
    if n < len(ls):
        print(ls[n])
    else:
        return
    print_list_r(ls, n+1)

Is this a good algorithm? It is possible to do this algorithm without using the parameter n?

Answer 1

Yes, it is a good algorithm. All relevant tests have been done (stop condition ok, limit of the indexes ok), incidentally the test by len(ls) == 0 is kind of redundant: if n >= 0 - necessary condition to pass through the first if - is true, so the condition n < len(ls) will only be true if len(ls) > 0. There is no unnecessary copy of objects (the same list is always used) and is still present at tail recursion - if the implementation takes advantage of this, your code can be as efficient as an iterative (unfortunately does not seem to be the case with Cpython).

What I don’t understand is why you return False if the list is empty and None otherwise. I even understand that you may want to take a different exit if n is invalid (negative), but an empty list, in my conception, should be a valid entry - treated indistinctly from other entries.

How to do without using the parameter n, is possible yes, although inefficiently:

def print_list_r(ls=[]):
    if len(ls) > 0:
        print(ls[0])
        print_list_r(ls[1:])

That is, if the list is not empty, print the first element and call the function recursively passing as argument to sublist with the indexes 1onwards (ending with an empty sublist).