for-each loop

class IsbnVerifier {

    boolean isValid(String stringToVerify) {
        int sum = 0;
        int pos = 0;
        
        for (var chr: stringToVerify.toCharArray()) {
            if (Character.isDigit(chr)) {
                sum += (chr - '0') * (10 - pos);
                pos++;
                continue;
            }
            if (chr == '-')
                continue;
            if (chr == 'X' && pos == 9) {
                sum += 10;
                pos++;
                continue;
            }
            return false;
        }
        
        return pos == 10 && sum % 11 == 0;
    }

}

This variant of validating as you go begins with initializing its sum and pos variables to 0.

A for-each loop is used to iterate the characters of the string to be verified.

A series of if statements begins by checking if the char is a digit, which is the most likely condition. If so, it gets the digit value of the char by subtracting the ASCII value of 0 from its own ASCII value. So, if the char is a 0, then subtracting the ASCII value of 0 from itself results in 0. If the char is a 1, then subtracting the ASCII value of 0 from the ASCII value of 1 results in 1, and so on.

Another way to convert the `char` to a digit is to use the built-in
[`Character.digit()`](https://docs.oracle.com/javase/8/docs/api/java/lang/Character.html#isDigit-char-)
method, which also works for [Unicode](https://docs.oracle.com/javase/tutorial/i18n/text/unicode.html) `char`s.
Since for this exercise all of the `chars` are ASCII, simple ASCII math will do. 

The digit is multiplied by 10 minus the position, with the position starting at 0. So, for the position furthest to the left, the digit is multiplied by 10 - 0 (10). For the position furthest to the right of a string whose length is 10, the digit is multiplied by 10 - 9 (1). The result of the multiplication is added to the sum variable.

The pos variable is incremented and the loop continues.

If the char is not a digit but a dash, the loop continues without incrementing the pos variable.

If the char is an X and the position is 9 (position 9 being at the end of a string whose length is 10), then 10 is added to the sum, pos is incremented, and the loop continues.

If none of those legal conditions apply, then the char is illegal and the function immediately returns false.

After the for-each is done, the pos variable is checked if the position is 10. If not, then the && operator short circuits and returns false from the function. If it is true, the function then returns if the sum is evenly divisible by 11.