From a book that programmers can learn ... Verification of rune checksum (variable length) Part 2

Getting started with programming in Python Continued from Last time

It seems that most of the parts I did last time will be diverted, Speaking of what was solved last time

・ Obtain 6-digit identification number from standard input ・ Add the acquired 6-digit identification numbers (hereinafter referred to as special matters) (Of the 6 digits, the odd digits are multiplied by 2, and if the calculation result of the multiplication is 2 digits, each digit is divided individually. Add up)

Here is the next problem, but in the problem from the beginning it is "arbitrary number of digits" It's not always a 6-digit number.

Code corresponding to an identification number of any length The answer in the book is judged by the number of the input / output result by ASCII character code in C ++. (End the loop when the ASCII character code "line feed" is acquired)

Book answer

char digit;
int checksum = 0;
int position = 1;
cout << "Enter a number with an even number of digits: ";
digit = cin.get();
while(digit != 10) {
    if (position % 2 == 1) checksum += digit - '0';###The odd numbers remain the same
    else checksum += doubleDigitValue(digit - '0');###Double the even number
    digit = cin.get();
    position++;
}

cout << "Checksum is " << checksum << ". \n";
if (checksum % 10 == 0){
    cout << "Checksum is divisible by 10. Valid. \n";
}else{
    cout << "Checksum is not divisible by 10. Invalid. \n";
}
・ ・ ・

Your answer I am changing it with reference to the part that received the comment. (Also, in the original problem, the odd number is doubled, but in the first task, the even number is doubled. Since the even number is doubled on the reference site of the rune checksum algorithm as well. That point has changed)

test02.py

#!/usr/bin/env python
#coding:utf-8

import sys
from ConsoleOut import cout
from test import doubleDigitValue


def number():
    checksum = 0
    position = 0
    cout << "Enter a number with an even number of digits:"
digits = input()
    while position < len(digits):
        digits = digits[position]
        if position % 2 == 1:
            checksum += int(digit)
        else:
            checksum += doubleDigitValue(int(digit))
        position += 1
            
    cout << "Checksum is " + str(checksum) + ". \n";
    if checksum % 10 == 0:
        cout << "Checksum is divisible by 10. Valid .\n"
    else:
        cout << "Checksum is not divisible by 10. Invalud. \n"

・ ・ ・(Terminal output result)
>>> from test02 import number
>>> number()
Enter a number with an even number of digits:49927398716
Checksum is 70. 
Checksum is divisible by 10. Valid .

I was thinking about how to find the odd number by summing up the number of characters entered. A very simple and easy-to-understand method of only even numbers and only odd numbers that you commented on last time I used it because I was taught.

Also, try using raw_input () that you taught me in standard input. I tried, but because I was using Python3.5, I got a NameError and got a NameError. Since it was on the reference site that sys.stdin.readline () can be used in Python3 I tried it, but it didn't work and the code is the same as last time.

In this article "Python Tips: I want to get a string from standard input" sys.stdin. There was a way to use readline ().