Java Java.lang.Long.numberOfTrailingZeros()用法及代码示例

描述

这个java.lang.Long.numberOfTrailingZeros() 方法返回指定 long 值的二进制补码表示中最低阶 ("rightmost") one-bit 之后的零位数。如果指定值在其二进制补码表示中没有 one-bits，换句话说，如果它等于 0，则返回 64。

声明

以下是声明java.lang.Long.numberOfTrailingZeros()方法

public static int numberOfTrailingZeros(long i)

参数

i─ 这是长期价值。

返回值

此方法返回指定 long 值的二进制补码表示中最低阶 ("rightmost") one-bit 之后的零位数，如果该值等于零，则返回 64。

异常

NA

示例

下面的例子展示了 java.lang.Long.numberOfTrailingZeros() 方法的用法。

package com.tutorialspoint;

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));  
   }
}

让我们编译并运行上面的程序，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