Java Double isFinite()用法及代码示例

Java Double 类的 isFinite() 方法如果传递的参数是有限浮点数，则返回 true，否则返回 false(在 NaN 和无穷大参数的情况下)。

用法

public static boolean isFinite(double d)

参数

d-这是要测试的双精度数。

返回值

当有限浮点值作为参数传递时，isFinite() 方法返回真，否则返回假。

例子1

public class Double_isFiniteMethodExample1 {

    public static void main(String[] args) {

        double d1 = 99991.22;

        Double d2 = new Double(d1/0);

         if(Double.isFinite(d2)){

             System.out.println("Value returned is:" +Double.isFinite(d2));

         }

         else{

             System.out.println("Value returned is:"+Double.isFinite(d2));

         }

    }



}

输出：

Value returned is:false

例子2

public class Double_isFiniteMethodExample2 {

    public static void main(String[] args) {

        // will return true

        Double d1 = new Double (0X7f800000);

        System.out.println("Returned result = "+ Double.isFinite(d1));

        //will return true

        Double d2 =new Double (76E-43);

        System.out.println("Returned result = "+ Double.isFinite(d2));

        //will return false

        Double d3 =new Double (d1/0);

        System.out.println("Returned result = "+ Double.isFinite(d3));

        //will return false

       Double d4 =new Double (-d1/0);

        System.out.println("Returned result = "+ Double.isFinite(d4));

        

    }

}

输出：

Returned result = true
Returned result = true
Returned result = false
Returned result = false

相关用法

注：本文由纯净天空筛选整理自 Java Double isFinite() method。非经特殊声明，原始代码版权归原作者所有，本译文未经允许或授权，请勿转载或复制。