Python Function returning null

Question

def counter(number,count):
    if (number!=1 and number%2==0):
        a=number/2
        count=count+1
        counter(a,count)
    elif (number!=1 and number%2!=0):
        a=3*(number)+1
        count=count+1
        counter(a,count)        
    else:
        print count
        return count

z=counter(13,0)
print z

count is evaluated to 9 and it does print it, but won't return it? Says None when printing 'z'

Martijn Pieters · Accepted Answer · 2015-04-26 16:52:55Z

3

You are ignoring the recursive calls; add return statements where you call counter() in counter itself:

def counter(number,count):
    if (number!=1 and number%2==0):
        a=number/2
        count=count+1
        return counter(a,count)
    elif (number!=1 and number%2!=0):
        a=3*(number)+1
        count=count+1
        return counter(a,count)        
    else:
        print count
        return count

Recursive calls are just like any other function call, if you don't do anything with the return value it is just discarded. Recursive calls don't magically pass on their result to the outermost call.

answered Apr 26, 2015 at 16:52

Martijn Pieters

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Ignacio Vazquez-Abrams · Accepted Answer · 2015-04-26 16:53:02Z

2

You forgot to return the result of recursion.

return counter(a,count)

answered Apr 26, 2015 at 16:53

Ignacio Vazquez-Abrams

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