double d=1.0/0.0;
output is Infinity
double d=1/0;
output is ArithmeticException.
What is difference between between these two? What is meaning of Infinity here?
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Christian Neverdal
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user3593753
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3 Answers
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The first case is treated as a division on double and the later as a division on int and hence the ArthimeticException.
Here is what infinity means
http://docs.oracle.com/javase/7/docs/api/java/lang/Double.html#POSITIVE_INFINITY
The division of doubles and floats is as per the IEEE 754 standards for floating point match which shouldnt throw an exception.
Kalyan Chavali
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ok, but how i implemented dividing by zero with doubles? – Kristijan Iliev Jul 08 '15 at 11:15
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1Question seems to be "why dividing by double zero doesn't throw arithmetic exception, but dividing by integer zero does?". – Pshemo Jul 08 '15 at 11:17
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2To complement this answer. Check this tutorial on data representation that is very ilustrative: http://www.ntu.edu.sg/home/ehchua/programming/java/DataRepresentation.html – Fernando Miguélez Jul 08 '15 at 11:18
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Mathematically, division by zero is undefined, although it can be loosely be regarded as being infinity. (With a little more rigour, it's a number that is greater than x for any value of x.)
An IEEE754 floating point double (used by Java) has a representation of infinity. That is the result of 1.0 / 0.0. In that sense, 1.0 / 0.0 is calculable since it takes place in floating point arithmetic.
An integral type has no representation of infinity, so an exception is thrown. 1 / 0 is computed in integer arithmetic.
Bathsheba
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floating point number have a way to code "infinity". So infinity is a valid value for a double variable. Integer variables do not have this option. So an exception is thrown instead.
Matthias
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