Java StrictMath floor()方法

java.lang.StrictMath.floor() 是一个内置的方法，它返回最大的双倍值，小于或等于给定的参数，并且等于整数值。

语法:

public static double floor(double num)

参数： 该方法接受一个参数 num ，该参数为双数类型。
返回值： 该方法返回最大的值，该值最接近正无穷大，小于或等于参数，并且等于一个整数。
示例

Input: num = 9.6Output: 9.0Input: num = -7.8Output: -8.0

下面的程序说明了java.lang.StrictMath.floor()方法：
程序1

// Java program to illustrate the//java.lang.StrictMath.floor() import java.lang.*; public class Geeks { public static void main(String[] args) {     double num1 = 7.8, num2 = 1.4 ;     double fValue = StrictMath.floor(num1);    System.out.println("The floor value of "+                             num1+" = " + fValue);     fValue = StrictMath.floor(num2);    System.out.println("The floor value of "+                             num2+" = " + fValue);}}

输出

The floor value of 7.8 = 7.0The floor value of 1.4 = 1.0

程序2

// Java program to illustrate the//java.lang.StrictMath.floor() import java.lang.*; public class Geeks { public static void main(String[] args) {     double num1 = -7.8, num2 = -1.4 ,num3 = 0.1 ;     double fValue = StrictMath.floor(num1);    System.out.println("The floor value of "+                             num1+" = " + fValue);     fValue = StrictMath.floor(num2);    System.out.println("The floor value of "+                             num2+" = " + fValue);     fValue = StrictMath.floor(num3);    System.out.println("The floor value of "+                             num3+" = " + fValue); }}

输出

The floor value of -7.8 = -8.0The floor value of -1.4 = -2.0The floor value of 0.1 = 0.0