Python recursive counter

Question

I need a recursive counter in python counting from 1 to n.

  def countup(n):
       a=1
       def hoch(a,n):
          if a<=n:
             print(a)
             a+=1
             hoch(a,n)
       hoch(a,n)

As someone from Germany you should know the wise words "Wieso einfach, wenn es auch kompliziert geht?" (*Why simple if it can be done complicated?*) — Klaus D., Mar 11 '18 at 15:23
What's wrong with `for x in range(1,n): print(x)` ? How is this 'recursive' ? — match, Mar 11 '18 at 15:27

score 1 · Accepted Answer · edited Mar 14 '18 at 02:20

1

Here is a solution.

def countup(n):
    if n >= 0:
        countup(n - 1)
        print(n)
countup(10)

Basically, if the number passed into countup is greater than 0, it recursively runs countup again, passing into it the next number below.

It only uses 1 function.

P.S. It already existed here:

edited Mar 14 '18 at 02:20

Pang

answered Mar 11 '18 at 15:28

kale

Olivier Melançon · Answer 2 · 2018-03-14T02:34:43.090

0

You can do very simple without recursion, which is not the ideal way to go.

def countup(n):
    print(*range(n + 1), sep='\n')

edited Mar 14 '18 at 02:34

answered Mar 14 '18 at 02:23

Olivier Melançon

score 0 · Answer 3 · answered Mar 14 '18 at 02:26

0

Assuming you have to do this recursively (which isn't the best way to do it), another option is to pass a to countup() as an optional argument.

def countup(n, a=1):
    if a <= n:
        print(a)
        countup(n, a+1)

countup(10)

answered Mar 14 '18 at 02:26

SCB

3 Answers3