The function, when it undoes recursion, returns None , and I'm not sure why.
def fat(n):
if n == 0 or n == 1:
return 1
else:
return n * fat(n - 1)
def superfat (n, x = 1):
if n > 0:
x*=fat(n)
n-=1
return superfat(n, x)
número = int(input("Digite um número para descobrir o seu superfatorial: "))
print("O fatorial de {} é {}.".format(número, superfat(número)))
See your code.
def superfat (n, x = 1):
if n > 0:
x*=fat(n)
n-=1
return superfat(n, x)
When n > 0 , the following excerpt will be executed:
x*=fat(n)
n-=1
return superfat(n, x)
But when n <= 0 , what will be executed?
Nothing!
You should write what should be returning when that if is false.
For example ...
def superfat (n, x = 1):
if n > 0:
x*=fat(n)
n-=1
return superfat(n, x)
return 1
In this case, I would rewrite to:
def superfat (n, x = 1):
if n <= 0:
return 1
x*=fat(n)
n-=1
return superfat(n, x)
Well, the moment the recursion is over is clearer (but this is a matter of preference).