java.lang.Integer.numberOfLeadingZeros()%uA0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。
声明
以下是java.lang.Integer.numberOfLeadingZeros()方法的声明
public static int numberOfLeadingZeros(int i)
参数
-
i%uA0-- 这是int值。
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。
异常
-
NA
例子
下面的例子显示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)) } }
让我们来编译和运行上面的程序，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