Function is returning NoneType in Python instead of integer

Question

def function(n):
    if n%4==1:
      return(n**2)
    else:
      n += 1
      function(n)

# Here for example n=6:

if function(6)%2==0:
    print(function(6))
else:
    print("Hey!")

This is showing the error unsupported operand type(s) for %: ‘NoneType’ and ‘int’. I have tried to convert NoneType with int() but that is telling me to give "str" or other data types as an argument.

When I am telling function(n) to return in both ‘if’ and ‘else’ conditions, only then it does not show the error.

Asked By: NightSky

||

Source

Answer 1

def function(n):
    if n%4==1:
      return(n**2)
    else:
      n += 1
      return function(n)

if function(6)%2==0:
    print(function(6))
else:
    print("Hey!")

in your else statement you return nothing. Put the return statement in there for your recursive function.

Answered By: CodeoftheWarrior

Answer 2

To elaborate on comments from Nin17 and molbdnilo:

Your function function(n) is supposed to return an integer value.
Like you did in if branch with return(n**2), do return in the else branch:

def function(n):
    if n % 4 == 1:
      return n**2
    else:      
      return function(n + 1)  # inlined the increased n and do return

I would recommend to give the function a meaningful name.

A function like yours that calls itself is a recursive function. One of it’s key-mechanisms is that it returns a value.

Answered By: hc_dev

Function is returning NoneType in Python instead of integer

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