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Asked: 2026-05-18T19:57:53+00:00

for the code below: def printList(L): if L: print L[0] printList(L[1:]) I can have

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for the code below:

def printList(L):

   if L:

       print L[0]

       printList(L[1:])

I can have sequence diagram like this:

# NON PYTHON PSEUDO CODE

PrintList([1,2,3])

  prints [1,2,3][0] => 1

  runs printList([1,2,3][1:]) => printList([2,3])

  => we're now in printList([2,3])

        prints [2,3][0] => 2

        runs printList([2,3][1:]) => printList([3])

    => we are now in printList([3])

          prints [3][0] => 3

          runs printList([3][1:]) => printList([])

          => we are now in printList([])

                "if L" is false for an empty list, so we return None

    => we are back in printList([3])

          it reaches the end of the function and returns None

  => we are back in printList([2,3])

    it reaches the end of the function and returns None

=> we are back in printList([1,2,3])

  it reaches the end of the function and returns None

So my question is if I change the code to:

def printList(L):

   if L:
       print L[0]
       printList(L[1:])
       print L[0]

How would the sequence diagram change, I want to understand what exactly happens during the execution of this code.

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1 Answer

  1. Editorial Team

    The print statement called after the recursive calls will all get hit “on the way back up”. That is, each of your statements: “it reaches the end of the function and returns None” can be changed to “it prints the current value of L[0], reaches the end of the function, and returns None”, which will be 3, 2, and 1 respectively.

    Like so:

    PrintList([1,2,3])
prints [1,2,3][0] => 1
runs printList([1,2,3][1:]) => printList([2,3])
=> we're now in printList([2,3])
    prints [2,3][0] => 2
    runs printList([2,3][1:]) => printList([3])
    => we are now in printList([3])
        prints [3][0] => 3
        runs printList([3][1:]) => printList([])
        => we are now in printList([])
            "if L" is false for an empty list, so we return None
        => we are back in printList([3])
        prints [3][0] => 3
        it reaches the end of the function and returns None
    => we are back in printList([2,3])
   prints [2,3][0] => 2
   it reaches the end of the function and returns None
=> we are back in printList([1,2,3])
prints [1,2,3][0] => 1
it reaches the end of the function and returns None
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