1. WIKI教程
2. Java.lang包
3. static int numberOfLeadingZeros(int i)

static int numberOfLeadingZeros(int i)

描述 (Description)

java.lang.Integer.numberOfLeadingZeros()方法返回指定int值的二进制补码二进制表示中最高位（“最左侧”）一位之前的零位数。

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，如果它等于零。

声明 (Declaration)

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

public static int numberOfLeadingZeros(int i)

参数 (Parameters)

i - 这是int值。

返回值 (Return Value)

此方法返回指定int值的二进制补码二进制表示中最高位（“最左侧”）一位之前的零位数，如果该值等于零，则返回32位。

异常 (Exception)

NA

例子 (Example)

以下示例显示了java.lang.Integer.numberOfLeadingZeros（）方法的用法。

package com.iowiki;
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));
   }
}

让我们编译并运行上面的程序，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