Automatic casting

Question

I have to write program that gets a number n from the user, and then calculates the sum: s = 1/1 + 1/2 + ... + 1/n.

I wrote this code:

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {
        Scanner unos = new Scanner(System.in);
        System.out.println("n=?");
        int n = unos.nextInt();

        double s = 0.0;
        for (int i = 1; i <= n; i++) {
            s = s + (1.0 / i);
        }
        System.out.println("s=" + s);
    }
}

How does Java decide to convert the int value i into double in this statement:

s = s + (1.0 / i);

Answer 1

The rules that govern what type gets converted/promoted to what other type are defined in the Java Language Spec Chapter 5 - Conversions and Promotions.

Specifically for most arithmetic operations, look at the Binary Numeric Promotion section.

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value of a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:

• If either operand is of type double, the other is converted to double.
• Otherwise, if either operand is of type float, the other is converted to float.

In your case, 1.0 is a double, so i is converted to a double (widening conversion). Since s already is a double, no further conversion is necessary.