Java division by zero doesnt throw an ArithmeticException - why?

问题

Why this code doesn\'t 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!

回答1:

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!");
    }

回答2:

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

BTW Floating point also has -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, including 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");

回答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.

回答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!

回答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.

回答6:

When divided by zero

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

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

回答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.

回答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.

来源：https://stackoverflow.com/questions/14137989/java-division-by-zero-doesnt-throw-an-arithmeticexception-why

标签

java

arithmeticexception