Python 中递归结束时不需要的无

Question

when I try to print the all the natural numbers from 1 to n using recursion I'm getting None at the end of the result My code

def num_1_to_n(n):
   if n==1:
        return n
   print(n)
   print(num_1_to_n(n-1))
n=int(input(enter the n value))
num_1_to_n(n)

I wanted to get

4
4
3
2
1

but I get

4
4
3
2
1
None
None
None

Can anyone please tell me the reason behind the printing of None after 1

Answer 1

Recursion often occurs problems like this, so use for loops:

n = int(input("Enter the n value"))
for i in range(n, 1, -1):
    print(i)
print(1) # for the loop to not stop at 2

Or, shorter:

n = int(input("Enter the n value"))
for i in range(n, 0, -1):
    print(i)

Answer 2

try this code

def num_1_to_n(n):
   if n==1:
       print(1)
       return n
   print(n)
   num_1_to_n(n-1)
n=int(input("enter the n value"))
num_1_to_n(n)

Output:

enter the n value4
4
3
2
1

you gate those none because of print(num_1_to_n(n-1)) this print

Answer 3

In your code

def num_1_to_n(n):
   if n==1:
        return n
   print(n)
   print(num_1_to_n(n-1))

the statement print(num_1_to_n(n-1)) prints the return value of the function call num_1_to_n(n-1) . Your function returns 1 in case of n == 1 and None otherwise, since Python implicitly returns None when it reaches the end of a function body during execution (emphasis mine):

The return statement returns with a value from a function. return without an expression argument returns None. Falling off the end of a function also returns None .

So you will see three None for the three calls where n != 1 , when you start at 4 .

Answer 4

Also you can try it this way:

def num_1_to_n(n):
  if n > 1:
      print(n)
      num_1_to_n(n-1)
  else:
      print(1)

n=int(input('enter the n value: '))
num_1_to_n(n)

Answer 5

It is printing None because you of print(num_1_to_n(n-1)) when you only need to call the function like num_1_to_n(n-1) .

Final Code:

def num_1_to_n(n):
   if n==1:
        print(1)
        return n

   print(n)
   num_1_to_n(n-1)

n=int(input("enter the n value"))
num_1_to_n(n)