java.lang.Integer.numberOfLeadingZeros()方法範例

java.lang.Integer.numberOfLeadingZeros() 方法返回零位的最高位(“最左側”)之前的數中指定的int值的二進位制二補數表示一位元。

它返回32，如果指定的值沒有一個位元在它的2的二補數表示的，換句話說，如果它等於零。

宣告

以下是java.lang.Integer.numberOfLeadingZeros()方法的宣告

public static int numberOfLeadingZeros(int i)

引數

• i -- 這是int值。

返回值

此方法返回零位的最高位(“最左側”)前在指定的int值的二進位制二補數表示法，或32個1位的數量，如果該值為零。

異常

• NA

例子

下面的例子顯示java.lang.Integer.numberOfLeadingZeros()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));
   }
}

讓我們來編譯和執行上面的程式，這將產生以下結果：

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