java.lang.Integer.numberOfTrailingZeros() 返回零位以下的最低阶(“最右”)的数在指定的int值的二进制补码表示一比特。

它返回32，如果指定的值没有一个比特在它的2的补码表示，换句话说，如果它等于零。

声明

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

public static int numberOfTrailingZeros(int i)

参数

• i -- 这是int值。

返回值

此方法返回零位的最高位(“最左侧”)前在指定的int值的二进制补码表示法，或32个1位的数量，如果该值为零。

异常

• NA

例子

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

package com.yiibai;

import java.lang.*;

public class IntegerDemo {

   public static void main(String[] args) {

     int i = 170;
     System.out.println("Number = " + i);
    
     /* returns the string representation of the unsigned integer value 
     represented by the argument in binary (base 2) */
     System.out.println("Binary = " + Integer.toBinaryString(i));

     // returns the number of one-bits 
     System.out.println("Number of one bits = " + Integer.bitCount(i));

     /* returns an int value with at most a single one-bit, in the position 
     of the highest-order ("leftmost") one-bit in the specified int value */
     System.out.println("Highest one bit = " + Integer.highestOneBit(i));

     /* returns an int 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 = " + Integer.lowestOneBit(i));

     /*returns the number of zero bits preceding the highest-order 
     ("leftmost")one-bit */
     System.out.print("Number of leading zeros = ");
     System.out.println(Integer.numberOfLeadingZeros(i));

     /* returns the number of zero bits following the lowest-order 
     ("rightmost") one-bit */
     System.out.print("Number of trailing zeros = ");
     System.out.println(Integer.numberOfTrailingZeros(i));  
   }
}

让我们编译并运行上述程序，这将产生以下结果：

Number = 170
Binary = 10101010
Number of one bits = 4
Highest one bit = 128
Lowest one bit = 2
Number of leading zeros = 24
Number of trailing zeros = 1