Re: 5/0.0 = Infinity : NO error is thrown

From:
=?ISO-8859-1?Q?Arne_Vajh=F8j?= <arne@vajhoej.dk>
Newsgroups:
comp.lang.java.programmer
Date:
Sun, 13 Jan 2008 22:33:34 -0500
Message-ID:
<478ad809$0$90276$14726298@news.sunsite.dk>
fullofquestions wrote:

Hello everyone. I have encountered an issue that perhaps is related to
my setup. I am running NetBeans 6.0, jre1.6.04 and at the moment I was
going through a java book to document a whole bunch of topics. Anyway,
I created an extremely simple class to go over exception handling. The
issue is that I can't find a way for the instance to throw a
'java.lang.ArithmeticException: / by zero.' The following file

public class index {
  public static void main( String args[] )
  {
    Scanner scanner = new Scanner( System.in ); // scanner for input
    System.out.println("5/0.0 = " + 5/0.0);
    System.out.println("5/0.0 = " + 5/0);
  }
}

Produces the following output:

5/0.0 = Infinity
Exception in thread "main" java.lang.ArithmeticException: / by zero

What is the difference between dividing by 0(int) and 0.0(double)? If
I have a bunch of calculations involving doubles where I want to check
for div by zero, how shoud I go about it?


There exist a standard for floating point operations IEEE 754.

It says that division by zero return infinite instead of faulting.

Most CPU's follow that standard.

Java had to use the available hardware support for floating point
for obvious performance reasons.

The short answer is that it is just this way.

You can test for infinite.

Integer arithmetic is different from floating point.

Arne

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