DSP

浮点运算 NaN Infinaty

2019-07-13 16:26发布

站内文章 / DSP
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JLS 4.2.4 p40 1 gradual underflow：精度不能满足，导致下溢。 2 round toward zero：从浮点强转成整型的时候向零看趋近，找到最接近的整数。 3 0.0/0.0 is Not-a-Number。 4 比较运算符 <.<=,>,>=返回false，只要有一个操作数是NaN。
5 对于== 运算符，只要有一个操作数是NaN,返回false。
6 对于 != 运算符，只要有一个操作数是NaN，返回true。
7 当且仅当x 是NaN的时候，x != x 为 true 
8 如果x或者y是NaN，(x < y) == !(x >= y ) 返回false   package arkblue.lang; public class FloatTest { public static void main(String[] args) { // An example of overflow: double d = 1e308; System.out.print("overflow produces infinity: "); System.out.println(d + "*10==" + d * 10); // An example of gradual underflow: d = 1e-305 * Math.PI; System.out.print("gradual underflow: " + d + " "); for (int i = 0; i < 4; i++) 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)); System.out.println(1.0 / -0); // Infinity System.out.println(1.0D / -0); // Infinity System.out.println(1.0 / 0); // Infinity System.out.println(1.0D / 0); // Infinity System.out.println(0.0 / 0); // NaN System.out.println((1.0D / 0 * 0) != (0.0D / 0)); // true System.out.println((1.0D / 0) != (0.0D / 0)); // true System.out.println((1.0D / 0 * 0) == (0.0D / 0)); // false System.out.println((1.0D / 0 * 0) > (0.0D / 0)); // false System.out.println((1.0D / 0 * 0) >= (0.0D / 0)); // false System.out.println((1.0D / 0 * 0) < (0.0D / 0)); // false System.out.println((1.0D / 0 * 0) <= (0.0D / 0)); // false double x = Double.NaN; double y = 0.0; System.out.println((x < y) == !(x >= y)); } }   结果：   overflow produces infinity: 1.0E308*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 Infinity Infinity Infinity Infinity NaN true true false false false false false      

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