Why doesn't Java throw an Exception when dividing by 0.0?

java types integer double divide-by-zero
froadie picture froadie · Mar 4, 2010 · Viewed 50.5k times · Source

I have code to calculate the percentage difference between 2 numbers - (oldNum - newNum) / oldNum * 100; - where both of the numbers are doubles. I expected to have to add some sort of checking / exception handling in case oldNum is 0. However, when I did a test run with values of 0.0 for both oldNum and newNum, execution continued as if nothing had happened and no error was thrown. Running this code with ints would definitely cause an arithmetic division-by-zero exception. Why does Java ignore it when it comes to doubles?

Answer

Michael Borgwardt picture Michael Borgwardt · Mar 4, 2010

Java's float and double types, like pretty much any other language out there (and pretty much any hardware FP unit), implement the IEEE 754 standard for floating point math, which mandates division by zero to return a special "infinity" value. Throwing an exception would actually violate that standard.

Integer arithmetic (implemented as two's complement representation by Java and most other languages and hardware) is different and has no special infinity or NaN values, thus throwing exceptions is a useful behaviour there.

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