📜  Java中的 StrictMath expm1() 方法及示例

📅  最后修改于: 2022-05-13 01:54:23.526000             🧑  作者: Mango

Java中的 StrictMath expm1() 方法及示例

Java.lang.StrictMath.expm1() 是Java中的一个内置方法,用于返回给定 num 值的指数 e^num -1 该方法产生了四种不同的情况:

  • 当给定参数为 NaN 时,该方法返回 NaN。
  • 当参数是正无穷大时,结果是正无穷大。
  • 当参数为负无穷大时,结果为负无穷大。
  • 对于 0,方法返回 0,其符号与参数相同。

句法 :

public static double expm1(double num)

参数:该方法接受一个double类型的参数num ,指的是要对其进行指数运算的值。
返回值:该方法将返回e num – 1操作的结果。
例子 :

Input: num = (1.0/0.0)
Output: Infinity

Input: 32.2
Output: 9.644557735961714E13

下面的程序说明了Java.lang.StrictMath.expm1() 方法:
方案一:

java
// Java program to illustrate the
// java.lang.StrictMath.expm1()
import java.lang.*;
 
public class Geeks {
 
public static void main(String[] args) {
 
    double num1 = 0.0, num2 = -(1.0/0.0);
        double num3 = (1.0/0.0), num4 = 32.2;
     
    /*It  returns e^num - 1 */
    double eValue = StrictMath.expm1(num1);
    System.out.println("The expm1 Value of "+
                              num1+" = "+eValue);
     
    eValue = StrictMath.expm1(num2);
    System.out.println("The expm1 Value of "+
                              num2+" = "+eValue);
     
    eValue = StrictMath.expm1(num3);
    System.out.println("The expm1 Value of "+
                              num3+" = "+eValue);
     
    eValue = StrictMath.expm1(num4);
    System.out.println("The expm1 Value of "+
                              num4+" = "+eValue);}
}


java
// Java program to illustrate the
// java.lang.StrictMath.expm1()
import java.lang.*;
 
public class Geeks {
 
public static void main(String[] args) {
 
    double num1 = 2.0 , num2 = -51.8;
        double num3 = 61.0, num4 = -32.2;
     
    /*It  returns e^num - 1 */
    double eValue = StrictMath.expm1(num1);
    System.out.println("The expm1 Value of "+
                              num1+" = "+eValue);
     
    eValue = StrictMath.expm1(num2);
    System.out.println("The expm1 Value of "+
                              num2+" = "+eValue);
     
    eValue = StrictMath.expm1(num3);
    System.out.println("The expm1 Value of "+
                              num3+" = "+eValue);
     
    eValue = StrictMath.expm1(num4);
    System.out.println("The expm1 Value of "+
                              num4+" = "+eValue);
}
}


方案二:

Java

// Java program to illustrate the
// java.lang.StrictMath.expm1()
import java.lang.*;
 
public class Geeks {
 
public static void main(String[] args) {
 
    double num1 = 2.0 , num2 = -51.8;
        double num3 = 61.0, num4 = -32.2;
     
    /*It  returns e^num - 1 */
    double eValue = StrictMath.expm1(num1);
    System.out.println("The expm1 Value of "+
                              num1+" = "+eValue);
     
    eValue = StrictMath.expm1(num2);
    System.out.println("The expm1 Value of "+
                              num2+" = "+eValue);
     
    eValue = StrictMath.expm1(num3);
    System.out.println("The expm1 Value of "+
                              num3+" = "+eValue);
     
    eValue = StrictMath.expm1(num4);
    System.out.println("The expm1 Value of "+
                              num4+" = "+eValue);
}
}