I have made this function that is meant to print each digit of a number one-by-one without using loops (comprehensions are not included).I have done a good job thus far the only thing im missing is that my return statement completely ommits the number 0 (ex.print_digits(2019) = 2 1 9)

def print_digits(x):
    ver = [u for u in str(x)] 
    if x < 10:
        print(x)
    else:
        print(ver[0])
        ver.pop(0)
        a_string = "".join(ver)
        inter = int(a_string)
        return print_digits(inter)

CodePudding user response：

After

ver.pop(0)

a_string = '019', so

inter = int(a_string)

evaluates to 19.

You can still print 0 by adding

if a_string[0] == '0':
    print(0)

between

a_string = "".join(ver)

and

inter = int(a_string)

CodePudding user response：

Modified version

def print_digits(x):
    ver = [u for u in str(x)]

    # change 'x' to 'int' before comparison
    if int(x) < 10: 
        print(x)
    else:
        ver.pop(0)
        a_string = "".join(ver)
       
        # comment this line. Now the function argument can be string.
        # inter = int(a_string)
        
        # give 'a_string' (instead of 'inter') to the function
        return print_digits(a_string)
print_digits(2019)

CodePudding user response：

It was happening because when the number "09" gets converted to integer it changes to 9 and when the condition x<10 gets checked it turns out to be true. I have fixed your code.

def print_digits(x):
  ver = [u for u in str(x)]
  if int(x) < 10 and x[0]!="0":
  
      print(x)
  else:
      print(ver[0])
      ver.pop(0)

  a_string = "".join(ver)
  return print_digits(a_string)

CodePudding user response：

You could avoid recursion altogether using something like:

def print_digits(x):
    print("\n".join(list(str(x))))