Does Java's % operator ever overflow?

Question

In C and C++, the behavior of INT_MIN % -1 seems to be undefined / platform-dependent as per Shafik's post.

In Java, does the % operator ever overflow?

Consider this piece of code:

public class Test {
    public static void main(String[] args) {
        // setup variables:
        byte b = Byte.MIN_VALUE % (-1);
        short s = Short.MIN_VALUE % (-1);
        int i = Integer.MIN_VALUE % (-1);
        long l = Long.MIN_VALUE % (-1);

        // my machine prints "0" for all:
        System.out.println(b);
        System.out.println(s);
        System.out.println(i);
        System.out.println(l);
    }
}

Is there a platform-independent guarantee that the above results are 0?

Answer 1

Look at JLS section 15.17.3 it says:

In C and C++, the remainder operator accepts only integral operands, but in the Java programming language, it also accepts floating-point operands.

The remainder operation for operands that are integers after binary numeric promotion (§5.6.2) produces a result value such that (a/b)*b+(a%b) is equal to a. This identity holds even in the special case that the dividend is the negative integer of largest possible magnitude for its type and the divisor is -1 (the remainder is 0). It follows from this rule that the result of the remainder operation can be negative only if the dividend is negative, and can be positive only if the dividend is positive;