java.lang.Integer.numberOfLeadingZeros()方法實例

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值的二進制補碼錶示一比特。

它返回32，如果指定的值沒有一個比特在它的2的補碼錶示的，換句話說，如果它等於零。

聲明

以下是**java.lang.Integer.numberOfLeadingZeros()**方法的聲明

public static int numberOfLeadingZeros(int i)

參數

• 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值的二進制補碼錶示法，或32個1位的數量，如果該值爲零。

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)); } }

讓我們來編譯和運行上面的程序，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