Java: Adding Digits Of A String

// (1) WHY DOES THIS NOT WORK

because Integer.parseInt(...); is expecting a string as parameter not a char

// (2) MORE IMPORTANTLY WHY DOES THIS WORK

char c = n.charAt(i);

any char is nothing else as an integer mapped to a table of symbols...(ASCII table for example) so this (c - '0') is just another valid mathematical operation

1. charAt is not a valid method of the primitive type int.

2. '0' is the character 0, and the character encoding set that Java uses has 0 to 9 in a consecutive block. Therefore c - '0' yields the position of c in that consecutive block which is therefore the value of the digit. (Actually this sort of thing is idiomatic C - goes right back to the 1960s).

You should first convert String to Int.. Please check the below code:

 class MainClass {
        public static void main(String[] args) {
            System.out.println(D5PSum(11));
        }

        private static int D5PSum(int number) {
            String n = Integer.toString(number);
            System.out.println(n);
            int sum = 0;
            for (int i = 0; i < n.length(); i++) {
                // (1) WHY DOES THIS NOT WORK
                String str = String.valueOf(n.charAt(i));
                sum += Integer.parseInt(str);

                // (2) MORE IMPORTANTLY WHY DOES THIS WORK
    //            char c = n.charAt(i);
    //            sum += (c-'0');
                // (3) WHAT IN THE WORLD IS c-'0'
            }
            return sum;
        }
    }

1 It doesnt work because Integer.parseInt takes a String and String.charAt returns a char(actar). Integer.parseInt (Character.toString(n.charAt(i))) would Work.

2/3 A char represents a number between 0 and 65535. EACH digit-characters (0-9) has a number in that range which depends on the charset. All digits are typically in a row, for example the character 0 has the value 48, 1 49 and 9 57. So ich you want to know the digit value of a char you simply subtract the character value of 0 from your character. That is the Line c-'0'

        // (1) WHY DOES THIS NOT WORK
        //sum += Integer.parseInt(number.charAt(i));

number is a variable of primitive data type "int" so number.charAt(i) won't work.

        // (2) MORE IMPORTANTLY WHY DOES THIS WORK
        char c = n.charAt(i);

n is an instance of String and we are getting the character at i th position in the n string

        sum += (c-'0');
        // (3) WHAT IN THE WORLD IS c-'0'

for every character there is an ascii code assigned. '0' = 48, 'c' = 99. That's the reason why it works here. when 'c'-'0' is executed, it's equivalent to 99-48

Why convert to a string in the first place? The simplest and fastest way to solve this is without deviation to strings:

private static int D5PSum(int number) {
    int v = number, sum = 0;
    while (v != 0) {
        sum += v % 10;
        v /= 10;
    }
    return sum;
}

If you want your code (the part which does not works to work then do this).

class Main {
    public static void main(String[] args) {
        System.out.println(D5PSum(10));
    }

    private static int D5PSum(int number) {
        String n = Integer.toString(number);
        int sum = 0;
        for (int i = 0; i < n.length(); i++) {

            sum += Integer.parseInt(n.charAt(i)+"");


        }
        return sum;
    }
}

To get sum of the digits of a string str:

int sum = str.chars().map(Character::getNumericValue).sum();