Java 除以零不会抛出 ArithmeticException - 为什么？

Question

Why doesn't this code throw an ArithmeticException ? Take a look:

public class NewClass {

    public static void main(String[] args) {
        // TODO code application logic here
        double tab[] = {1.2, 3.4, 0.0, 5.6};

        try {
            for (int i = 0; i < tab.length; i++) {
                tab[i] = 1.0 / tab[i];
            }
        } catch (ArithmeticException ae) {
            System.out.println("ArithmeticException occured!");
        }
    }
}

I have no idea!

Answer 1

IEEE 754 defines 1.0 / 0.0 as Infinity and -1.0 / 0.0 as -Infinity and 0.0 / 0.0 as NaN.

By the way, floating point values also have -0.0 and so 1.0/ -0.0 is -Infinity .

Integer arithmetic doesn't have any of these values and throws an Exception instead.

To check for all possible values (eg NaN, 0.0, -0.0) which could produce a non finite number you can do the following.

if (Math.abs(tab[i] = 1 / tab[i]) < Double.POSITIVE_INFINITY)
   throw new ArithmeticException("Not finite");

Answer 2

Why can't you just check it yourself and throw an exception if that is what you want.

    try {
        for (int i = 0; i < tab.length; i++) {
            tab[i] = 1.0 / tab[i];

            if (tab[i] == Double.POSITIVE_INFINITY ||
                    tab[i] == Double.NEGATIVE_INFINITY)
                throw new ArithmeticException();
        }
    } catch (ArithmeticException ae) {
        System.out.println("ArithmeticException occured!");
    }

Answer 3

That's because you are dealing with floating point numbers. Division by zero returns Infinity , which is similar to NaN (not a number).

If you want to prevent this, you have to test tab[i] before using it. Then you can throw your own exception, if you really need it.

Answer 4

0.0 is a double literal and this is not considered as absolute zero! No exception because it is considered that the double variable large enough to hold the values representing near infinity!

Answer 5

Java will not throw an exception if you divide by float zero. It will detect a run-time error only if you divide by integer zero not double zero.

If you divide by 0.0, the result will be INFINITY.

Answer 6

When divided by zero

1. If you divide double by 0, JVM will show Infinity .

    public static void main(String [] args){ double a=10.00; System.out.println(a/0); }

   Console: Infinity

2. If you divide int by 0, then JVM will throw Arithmetic Exception .

    public static void main(String [] args){ int a=10; System.out.println(a/0); }

   Console: Exception in thread "main" java.lang.ArithmeticException: / by zero

Answer 7

There is a trick, Arithmetic exceptions only happen when you are playing around with integers and only during / or % operation.

If there is any floating point number in an arithmetic operation, internally all integers will get converted into floating point. This may help you to remember things easily.

Answer 8

This is behaviour of floating point arithmetic is by specification. Excerpt from the specification, § 15.17.2. Division Operator / :

Division of a nonzero finite value by a zero results in a signed infinity. The sign is determined by the rule stated above.