Why does 1/0 give error but 1/0.0 returns “Inf”?

问题:

I'm just curious about this:

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

Exception in thread "main" java.lang.ArithmeticException: / by zero at Foo.main(Foo.java:3)

But 1/0.0 is evaluated to Infinity.

public class Foo {
    public static void main (String[] args) {
        System.out.println(1/0.0);
    }
}

Why does this happen?

回答1:

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.

回答2:

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.

回答3:

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

回答4:

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.