Java Integer numberOfLeadingZeros()方法

java.lang.Integer.numberOfLeadingZeros() 方法返回零位的最高位(“最左侧”)之前的数中指定的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的补码表示的，换句话说，如果它等于零。

1 语法

public static int numberOfLeadingZeros(int i)

2 参数

i ：这是int值。

3 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。

4 示例 

package com.yiidian;

/**
 * 一点教程网： http://www.yiidian.com
 */
/**
 * Java Integer numberOfLeadingZeros()方法
 */
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));
   }
}

输出结果为：

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