3

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?

1
  • Java in nature is platform independent so I guess (not checked) that the answer is YES Commented Sep 10, 2014 at 12:28

1 Answer 1

6

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;

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4 Comments

I was just about to post that ^^ Almost everything is in the JLS, especially for this kind of questions.
sorry @Joffrey i had that topic on my screen when i was reading that question :-$
Haha it's no problem, the point here is to get an answer to the question, no matter who writes it ;)
i do think the same - it's not so important who does it - it's more important WHAT has been submitted!

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