Java division by zero doesn't throw an ArithmeticException - why?

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!");
        }
    }
}

so how can I change my code to get an ArithmeticException? (I dont want to change the type of an array to int)? — Katie, Jan 03 '13 at 11:30
This is duplicated http://stackoverflow.com/questions/5291606/why-does-1-0-give-error-but-1-0-0-returns-inf — Seba, Jan 03 '13 at 11:30
@assylias: Im not so sure about that, I read somewhere that we shouldnt throw an ArithmerticException — Katie, Jan 03 '13 at 11:32
Katie, you are overthinking. :) assylias solution is just fine. If, for some reason, you don't want to throw ArithmeticException, just throw the exception you want. — gd1, Jan 03 '13 at 11:34
Checking floating value using `==` is potentially error prone. You should always check against sufficiently small number, to rule out rounding errors. — jnovacho, Nov 15 '13 at 11:07
I think jnovacho is referring to assylias's comment. When comparing int values, you can use the `==` operator for comparison perfectly fine. But with floats, `0 == 0.0` will return false. You should use the Double Object: `if (Double.valueOf(0.0).compareTo(tab[i]) == 0)` — lugte098, Jan 24 '14 at 14:40
Possible duplicate of [Why does integer division by zero 1/0 give error but floating point 1/0.0 returns "Inf"?](https://stackoverflow.com/questions/5291606/why-does-integer-division-by-zero-1-0-give-error-but-floating-point-1-0-0-return) — Ciro Santilli OurBigBook.com, Apr 14 '19 at 08:19

score 99 · Answer 1 · edited May 12 '21 at 09:23

99

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 (e.g. 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");

edited May 12 '21 at 09:23

Michael

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answered Jan 03 '13 at 11:27

Peter Lawrey

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PeterLawrey: so how can I change my code to get an ArithmeticException? (I dont want to change the type of an array to int) – Katie Jan 03 '13 at 11:29
2

@Katie you can check `if(tab[i] == 0.0){throw new ArithmeticException();}` before `tab[i] = 1.0 / tab[i];` – codeMan Jan 03 '13 at 11:37
4

BTW `0.0 == -0.0` so you don't have to check for two values. – Peter Lawrey Jan 03 '13 at 12:31
2

Suggested edit queue is full, so here a not broken link to the IEEE 754 https://ieeexplore.ieee.org/document/30711 – Johannes Rabauer Sep 15 '20 at 06:34
3

To check for all special values (i.e. NaN, Infinity, -Infinity), use `if (! Double.isFinite(d))` – Andreas May 12 '21 at 09:27

score 31 · Accepted Answer · answered Jan 03 '13 at 11:36

31

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!");
    }

answered Jan 03 '13 at 11:36

Waqas Ilyas

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13

You need to check for Double.isNaN() as well. – Peter Lawrey Jan 03 '13 at 12:34
5

This doesn't answer the question. – Max Barraclough Jul 27 '20 at 13:08
1

The man asked why he didn't get an error, he didn't ask how do I do it. – hakan hakyemez Apr 17 '21 at 12:15

gd1 · Answer 3 · 2013-01-03T11:31:28.423

26

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.

edited Jan 03 '13 at 11:31

answered Jan 03 '13 at 11:25

gd1

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score 12 · Answer 4 · answered Jan 03 '13 at 11:25

12

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!

answered Jan 03 '13 at 11:25

codeMan

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2

No, 0.0 really is 0 and the IEEE floating point spec defines the result as a special value "positive infinity", not merely a very large value. – Sean Owen May 18 '13 at 15:12

score 12 · Answer 5 · answered Nov 19 '15 at 08:34

12

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.

answered Nov 19 '15 at 08:34

Khalil

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score 8 · Answer 6 · edited Oct 01 '17 at 11:15

8

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

edited Oct 01 '17 at 11:15

Martin Evans

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answered Oct 01 '17 at 10:55

Raman Gupta

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NaN not Infinity :) – San4musa Nov 23 '20 at 01:27
@San4musa That's only for `0/0`. – Unmitigated Aug 08 '21 at 02:41

score 3 · Answer 7 · edited Apr 29 '19 at 12:16

3

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.

edited Apr 29 '19 at 12:16

Adam Horvath

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answered Nov 15 '18 at 18:19

score 0 · Answer 8 · answered Nov 25 '19 at 06:34

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.

Java division by zero doesn't throw an ArithmeticException - why?

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