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。

声明

以下是java.lang.Long.numberOfTrailingZeros()方法的声明

public static int numberOfTrailingZeros(long i)

参数

• i -- 这是long 值。

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。

异常

• NA

例子

下面的例子显示java.lang.Long.numberOfTrailingZeros()方法的使用。

package com.yiibai;

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));  
   }
}  

让我们来编译和运行上面的程序，这将产生以下结果：

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