strictmath_Java StrictMath cbrt（）方法与示例

strictmath

StrictMath类cbrt()方法 (StrictMath Class cbrt() method)

cbrt() method is available in java.lang package.
cbrt()方法在java.lang包中可用。
cbrt() method is used to find the cube root of the given parameter in the method. Here, cbrt stands for cube root.
cbrt()方法用于查找方法中给定参数的立方根。在这里， cbrt代表cube root 。
cbrt() method is a static method so it is accessible with the class name and if we try to access the method with the class object then we will not get an error.
cbrt()方法是一个静态方法，因此可以使用类名进行访问，如果尝试使用类对象访问该方法，则不会收到错误。
In this method, if we pass a positive parameter so it returns the cube root of the given parameter with the same sign(Positive) else if we pass negative parameter so it returns the cube root of the given parameter with the same sign(Negative).
在此方法中，如果传递正参数，则返回给定参数的立方根(正)；否则，传递负参数，使它返回给定参数的立方根(负) 。
cbrt() method does not throw any exception.
cbrt()方法不会引发任何异常。

Syntax:

句法：

    public static double cbrt(double d);

Parameter(s):

参数：

double d – represents a double type value whose cube root to be found.
double d –表示要查找其立方根的double类型值。

Return value:

返回值：

The return type of this method is double – it returns the cube root of given angle.

此方法的返回类型为double-返回给定角度的立方根。

Note:

注意：

If we pass NaN as an argument, method returns the same value (NaN).
如果我们将NaN作为参数传递，则方法将返回相同的值(NaN)。
If we pass zero (0), method returns the same value with the same sign.
如果传递零(0)，则方法将返回具有相同符号的相同值。
If we pass an infinity, method returns the same value with the same sign.
如果我们传递一个无穷大，则方法将返回带有相同符号的相同值。

Example:

例：

// Java program to demonstrate the example
// of  cbrt(double d) method of StrictMath Class.
public class Cbrt {
public static void main(String[] args) {
// variable declarations
double d1 = -0.0;
double d2 = 0.0;
double d3 = -7.0 / 0.0;
double d4 = 7.0 / 0.0;
double d5 = 1000.0;
double d6 = -1000.0;
// Display previous value of d1,d2,d3,d4,d5 and d6
System.out.println("d1: " + d1);
System.out.println("d2: " + d2);
System.out.println("d3: " + d3);
System.out.println("d4: " + d4);
System.out.println("d5: " + d5);
System.out.println("d6: " + d6);
// Here , we will get (-0.0) because we are 
// passing parameter (-0.0) so the cube root is the same
System.out.println("StrictMath.cbrt(d1): " + StrictMath.cbrt(d1));
// Here , we will get (0.0) because we are 
// passing parameter (0.0) so the cube root is the same
System.out.println("StrictMath.cbrt(d2): " + StrictMath.cbrt(d2));
// Here , we will get (-Infinity) because we are 
// passing parameter (-7.0/0.0) so the cube root is (-Infinity)
System.out.println("StrictMath.cbrt(d3): " + StrictMath.cbrt(d3));
// Here , we will get (Infinity) because we are 
// passing parameter (7.0/0.0) so the cube root is (Infinity)
System.out.println("StrictMath.cbrt(d4): " + StrictMath.cbrt(d4));
// Here , we will get (10.0) because we are
// passing parameter (1000.0) so the cube root is 10.0
System.out.println("StrictMath.cbrt(d5): " + StrictMath.cbrt(d5));
// Here , we will get (-10.0) because we are
// passing parameter (-1000.0) so the cube root is (-10.0)
System.out.println("StrictMath.cbrt(d6): " + StrictMath.cbrt(d6));
}
}

Output

输出量

d1: -0.0
d2: 0.0
d3: -Infinity
d4: Infinity
d5: 1000.0
d6: -1000.0
StrictMath.cbrt(d1): -0.0
StrictMath.cbrt(d2): 0.0
StrictMath.cbrt(d3): -Infinity
StrictMath.cbrt(d4): Infinity
StrictMath.cbrt(d5): 10.0
StrictMath.cbrt(d6): -10.0