Counting number of digits of input using python

Your error mainly happened here:

countnumber=countnumber/10 

Note that you are intending to do integer division. Single-slash division in Python 3 is always "float" or "real" division, which yields a float value and a decimal part if necessary.

Replace it with double-slash division, which is integer division: countnumber = countnumber // 10. Each time integer division is performed in this case, the rightmost digit is cut.

You also have to watch out if your input is 0. The number 0 is considered to be one digit, not zero.

I would not convert that beautiful input to int to be honest.

print(len(input()) 

would be sufficient.

An easily understandable one liner that no one can complain about.

• But of course, if negative sign bothers you like wisty said,

  len(str(abs(int (v)))) 

  will be safer for sure.

• Again, if you are worried about the non numeric inputs like mulliganaceous said, you better cover that case.

  str = input() if str.isnumeric(): print(len(str(abs(v)))) else: print("bad input") 

The reason is that in python 3 the division of two integers yields a floating point number. It can be fixed using the // operator:

number = int(input()) digits_count = 0 while number > 0: digits_count += 1 number = number // 10 

You must be using Python3, logically your function is right. You just have to change

countnumber = countnumber // 10

because Python3, // is floor division, meanwhile / is true division.

>>>print(1 / 10) 0.1 >>>print(1 // 10) 0 

Btw, as @chrisz said above, you can just simply using the len() function to get the number of digits of the input

>>>print(len(input()) 

num = int(input()) count = 0 while num > 0: count += 1 num = num // 10 print(count) 

def digits(number): number = str(number) lenght = len(number) return lenght print(digits(25)) # Should print 2 print(digits(144)) # Should print 3 print(digits(1000)) # Should print 4 print(digits(0)) # Should print 1