Java.lang.Integer.numberOfTrailingZeros() 方法

描述

java.lang.Integer.numberOfTrailingZeros() 返回指定 int 值的二进制补码表示中最低位（"最右边"）一位之后的零位数。 p>

如果指定的值在其二进制补码表示中没有一位，则返回 32，换句话说，如果它等于 0。

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

此方法返回指定 int 值的二进制补码表示中最高位（"最左边"）一位之前的零位数，如果该值等于 0，则返回 32。

异常

NA

示例

下面的例子展示了 java.lang.Integer.numberOfTrailingZeros() 方法的使用。

package com.tutorialspoint;

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面的程序，这将产生下面的结果 −

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