Why does integer division by zero 1/0 give error but floating point 1/0.0 returns “Inf”?

Question

I\'m just curious about this:

When evaluating 1/0 in Java, the following exception occurs:

Exception in thread \"main\" java.la

Answer 1

The IEEE has defined certain standards for floating point numbers which include definitions for "Not a Number" and positive and negative infinity. These do not apply to integers.

See http://steve.hollasch.net/cgindex/coding/ieeefloat.html

The reason for these special cases in basically rounding errors. Floating point numbers are often always truncated because they are never exact. Integers, on the other hand, are always exact.

Answer 2

That's because integers don't have values for +/-Inf, NaN, and don't allow division by 0, while floats do have those special values.

Answer 3

1/0 is integer division, 1/0.0 is floating point division - Floats can represent invalid values, integers can't.

Answer 4

1/0 is a division of two ints, and throws an exception because you can't divide by integer zero. However, 0.0 is a literal of type double, and Java will use a floating-point division. The IEEE floating-point specification has special values for dividing by zero (among other thing), one of these is double.Infinity.

If you're interested in details, the floating-point spec (which is often cryptic) has a page at Wikipedia: http://en.wikipedia.org/wiki/IEEE_754-2008, and its full text can be also read online: http://ieeexplore.ieee.org/xpl/mostRecentIssue.jsp?punumber=4610933.

Answer 5

fetestexcept and feenableexcept

It is worth noting that the underlying CPU hardware can detect the specific case of 0.0 / 0.0, and:

• set some flag bits, which can be queried with ANSI C fetestexcept(FE_ALL_EXCEPT) == FE_INVALID
• raise an exception, if this was enabled with the feenableexcept(FE_INVALID) glibc extension

I could not find if Java exposes that functionality however.

Here is a minimal runnable C example: What is difference between quiet NaN and signaling NaN?