Double Integer value overflow

Question

I am trying to understand the behavior of the code below which I wrote to experiment on calculation overflow.

public static void main(String[] args) {

    System.out.println(getSomeValue());
    System.out.println(getFixedSomeValue());
}

private static double getSomeValue() {
    return (2500000 - 0) * 250000 * (200 + 310);
}

private static double getFixedSomeValue() {
    return (double) (2500000 - 0) * 250000 * (200 + 310);
}

output:

-9.9787264E8
3.1875E14

What I have understood is: It can be because of Integer overflow as:

Double.MAX_VALUE = 1.7976931348623157E308
Integer.MAX_VALUE = 2147483647

I don't understand why the values differ? When the return type of method is double, shouldn't it automatically cast it to double?

Answer 1

(2500000 - 0) * 250000 * (200 + 310) is an expression comprised of int literal and numeric operators, so it is evaluated using integer operators and the result is an int, and therefore overflows (since it is limited to Integer.MAX_VALUE). The overflown result is converted to double before the method returns, but that's too late.

When you add (double) at the beginning, you cast (2500000 - 0) to double and avoid the numeric overflow, since now all the operators are floating point operators that result in double value. This is similar to writing

return (2500000.0 - 0) * 250000 * (200 + 310)
               ^ decimal point = double literal rather than int