Could anybody tell me why is it returning None value at end other than the required numbers?

Question

def collatz(a):
    if a==1:
        return a
    elif a%2==0:
        print(a//2)
        collatz(a//2)
    else:
        print(a*3+1)
        collatz(a*3+1)

a = int(input("Enter no. "))
print(collatz(a))

your `elif` and `else` block aren't returning anything. you need to add return while calling `return collatz(...)` — Moinuddin Quadri, Mar 21 '21 at 15:07
Instead of `print(collatz(a))` try just `collatz(a)` at the end — dimitris_ps, Mar 21 '21 at 15:09
You may think about [accepting an answer](https://stackoverflow.com/help/someone-answers) to reward those how helped you, or at least comment to explain what's missing ;) — azro, Mar 27 '21 at 11:44

score 1 · Answer 1 · answered Mar 21 '21 at 15:09

First see Function returns None without return statement

Then your method has nothing to return, because the purpose is to see the path of values and the stop value is always 1

def collatz(a):
    print(a)
    if a == 1:
        return             # just the stop case
    elif a % 2 == 0:
        collatz(a // 2)
    else:
        collatz(a * 3 + 1)

So call it without a print, you don't expect anything to be returned

a = int(input("Enter value:"))
collatz(a)

You could even combine in

def collatz(a):
    print(a)
    if a % 2 == 0:
        collatz(a // 2)
    elif a > 1:
        collatz(a * 3 + 1)

score 0 · Answer 2 · answered Mar 21 '21 at 15:12

0

You can return the functions directly:

def collatz(a):
    if int(a)==1:
        return a
    elif a%2==0:
        print(a//2)
        return collatz(a//2)
    else:
        print(a*3+1)
        return collatz(a*3+1)

answered Mar 21 '21 at 15:12

berkayln

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Could anybody tell me why is it returning None value at end other than the required numbers?

2 Answers2