Multiplication of numbers as string

Why is '0' subtracted from n1.charAt(i) and n2.charAt(j)?

Question: Given two numbers represented as strings, return multiplication of the numbers as a string.

public String multiply(String num1, String num2) {
    String n1 = new StringBuilder(num1).reverse().toString();
    String n2 = new StringBuilder(num2).reverse().toString();

    int[] d = new int[num1.length()+num2.length()];

    //multiply each digit and sum at the corresponding positions
    for(int i=0; i<n1.length(); i++){
        for(int j=0; j<n2.length(); j++){
            d[i+j] += (n1.charAt(i)-'0') * (n2.charAt(j)-'0');
        }
    }

    StringBuilder sb = new StringBuilder();

    //calculate each digit
    for(int i=0; i<d.length; i++){
        int mod = d[i]%10;
        int carry = d[i]/10;
        if(i+1<d.length){
            d[i+1] += carry;
        }
        sb.insert(0, mod);
    }

    //remove front 0's
    while(sb.charAt(0) == '0' && sb.length()> 1){
        sb.deleteCharAt(0);
    }

    return sb.toString();
}
Jon Skeet
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quotationmark

Why is '0' subtracted from n1.charAt(i) and n2.charAt(j) ?

Because the number of the Unicode character representing the digit 0 is 48, not 0.

Imagine you wanted to label 'A' => 0, 'B' => 1 etc... then you'd use

n1.charAt(i) - 'A'

etc... but as it happens, we want '0' => 0, '1' => 1 etc, so we subtract '0'. You need to differentiate between "the value, as an int" and "the character, as a char" (where the latter still has a number value, but it's not what you might expect it to be).

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