Java Reference
In-Depth Information
System.out.print(" " + (d /= 100000));
System.out.println();
// An example of NaN:
System.out.print("0.0/0.0 is Not-a-Number: ");
d = 0.0/0.0;
System.out.println(d);
// An example of inexact results and rounding:
System.out.print("inexact results with float:");
for (int i = 0; i < 100; i++) {
float z = 1.0f / i;
if (z * i != 1.0f)
System.out.print(" " + i);
}
System.out.println();
// Another example of inexact results and rounding:
System.out.print("inexact results with double:");
for (int i = 0; i < 100; i++) {
double z = 1.0 / i;
if (z * i != 1.0)
System.out.print(" " + i);
}
System.out.println();
// An example of cast to integer rounding:
System.out.print("cast to int rounds toward 0: ");
d = 12345.6;
System.out.println((int)d + " " + (int)(-d));
}
}
This program produces the output:
overflow produces infinity: 1.0e+308*10==Infinity
gradual underflow: 3.141592653589793E-305
3.1415926535898E-310 3.141592653E-315 3.142E-320 0.0
0.0/0.0 is Not-a-Number: NaN
inexact results with float: 0 41 47 55 61 82 83 94 97
inexact results with double: 0 49 98
cast to int rounds toward 0: 12345 -12345
This example demonstrates, among other things, that gradual underflow can result in
a gradual loss of precision.
The results when
i
is
0
involve division by zero, so that
z
becomes positive infinity,
and
z * 0
is NaN, which is not equal to
1.0
.
