Java Long numberOfTrailingZeros()方法

java.lang.Long.numberOfTrailingZeros() 方法返回零位以下的最低阶(“最右”)数指定long值的二进制补码表示一比特。它返回64，如果指定的值没有一个比特的补码表示，换句话说，如果它等于零。

1 语法

public static int numberOfTrailingZeros(long i)

2 参数

i ：这是long 值。

3 返回值

此方法返回零位以下的最低阶(“最右”)数指定long值的二进制补码表示法，或64如果是否等于零。

4 示例 

package com.yiidian;

/**
 * 一点教程网： http://www.yiidian.com
 */
/**
 * Java Long numberOfTrailingZeros()方法
 */
import java.lang.*;

public class LongDemo {

   public static void main(String[] args) {

     long l = 210;
     System.out.println("Number = " + l);
    
     /* returns the string representation of the unsigned long value 
     represented by the argument in binary (base 2) */
     System.out.println("Binary = " + Long.toBinaryString(l));

     /* returns a long value with at most a single one-bit, in the position
     of the lowest-order ("rightmost") one-bit in the specified int value.*/
     System.out.println("Lowest one bit = " + Long.lowestOneBit(l));
     
     /*returns the number of zero bits preceding the highest-order 
     ("leftmost")one-bit */
     System.out.print("Number of leading zeros = ");
     System.out.println(Long.numberOfLeadingZeros(l));
     
     /* returns the number of zero bits following the lowest-order 
     ("rightmost") one-bit */
     System.out.print("Number of trailing zeros = ");
     System.out.println(Long.numberOfTrailingZeros(l));  
   }
}

输出结果为：

Number = 210
Binary = 11010010
Lowest one bit = 2
Number of leading zeros = 56
Number of trailing zeros = 1