Infinity (Infinity) and Nan in Java

Source: Internet
Author: User

1, i = = i + 1

A number never equals its own plus 1? Java enforcement requires the use of IEEE 754 floating-point arithmetic operations [IEEE 754], which allows you to represent infinity with a double or float. As we learned in school, Infinity plus 1 is infinitely large.

You can initialize I with any floating-point arithmetic expression that is computed as infinity, for example:

Double i = 1.0/0.0;

However, you'd better be able to use the constants provided to you by the Standard class library:

Double i = double.positive_infinity;

In fact, you do not have to initialize I to infinity to ensure that the loop executes forever. Any floating-point number that is large enough to achieve this, such as:

Double i = 1.0e40;

2, I! = i

A number is always equal to its own? IEEE 754 floating-point arithmetic retains a special value used to denote a number that is not a number [IEEE 754]. This value is Nan ("not a number" abbreviation), and for all floating-point calculations that do not have a good numeric definition, such as 0.0/0.0, the value is it. The specification describes that NaN is notequal to any floating-point value, including itself [JLS].

You can initialize I with any floating-point arithmetic expression that evaluates to Nan, for example:

Double i = 0.0/0.0;

Again, for clarity, you can use the constants provided by the standard class library:

Double i = Double.NaN;

There are other amazing places to NaN. any floating-point operation, as long as one or more of its operands is Nan, the result is Nan. This rule is very reasonable, but it has a strange result. For example, the following program will print false:

Class Test {

public static void Main (string[] args) {

Double i = 0.0/0.0;

System.out.println (I-i = = 0);



In summary, both the float and double types have a special Nan value, which is used to denote a number that is not a number.

3, Nan and any number comparison are returned false

if ((0 > C) | | (0 = = c) | | (0 < C)) {
SYSTEM.OUT.PRINTLN ("NaN compared with 0 are not always false.");
SYSTEM.OUT.PRINTLN ("NaN compared with 0 are always false!");


Double.NaN = = Double.NaN, the result is false. However,

Double A = new double (Double.NaN);

Double b = new double (Double.NaN);]

A.equals (b); //true

4, ()

And when we use the () method to compare two Nan, we get an equal result. Can be verified with the following code:

float Nan=float.nan;
float Anothernan=float.nan;
System.out.println ( (Nan,anothernan));

The Compare () method returns 0, which indicates that the two numbers are equal, and returns-1, which means the first is smaller than the second and returns 1 is the opposite.
The return result of the above statement is 0.
In general, the basic type of compare () method is the same as the effect of using = = directly, but inconsistency on Nan is a benefit and disadvantage, depending on the expectations of the person who uses it. When the semantics of a program require that two Nan should not be considered equal (for example, using Nan to represent two infinity, a friend who has studied higher mathematics remembers that two infinite-looking symbols are the same, but should not be considered equal), use = = to judge; If Nan is seen as insignificant (after all, I only care about the numbers, two is not the number of things to the same class well, use ().

Another floating-point number that is inconsistent on the = = and compare () method is positive 0 and minus 0 (which, of course, is a very persistent problem for computers representing signed numbers), and we (Magnum) certainly know that 0.0f and -0.0f should be equal numbers, but try the following code:

float negzero=-0.0f;
float zero=0.0f;
System.out.println (Zero==negzero);
System.out.println ( (Zero,negzero));

The returned result is true and-1. See, = = think positive 0 and negative 0 are equal, and the Compare () method thinks positive 0 is greater than negative 0. So for a comparison of 0, = = is a better choice.

Infinity (Infinity) and Nan in Java

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