/**
     * A constant holding the minimum value an {@code int} can
     * have, -2<sup>31</sup>.
     */
    @Native public static final int   MIN_VALUE = 0x80000000;
    /**
     * A constant holding the maximum value an {@code int} can
     * have, 2<sup>31</sup>-1.
     */
    @Native public static final int   MAX_VALUE = 0x7fffffff;
    /**
     * The {@code Class} instance representing the primitive type
     * {@code int}.
     *
     * @since   JDK1.1
     */
    @SuppressWarnings("unchecked")
    public static final Class<Integer>  TYPE = (Class<Integer>) Class.getPrimitiveClass("int");
    /**
     * All possible chars for representing a number as a String
     */
    final static char[] digits = {
        '0' , '1' , '2' , '3' , '4' , '5' ,
        '6' , '7' , '8' , '9' , 'a' , 'b' ,
        'c' , 'd' , 'e' , 'f' , 'g' , 'h' ,
        'i' , 'j' , 'k' , 'l' , 'm' , 'n' ,
        'o' , 'p' , 'q' , 'r' , 's' , 't' ,
        'u' , 'v' , 'w' , 'x' , 'y' , 'z'
    };

   /**
     * Cache to support the object identity semantics of autoboxing for values between
     * -128 and 127 (inclusive) as required by JLS.
     *
     * The cache is initialized on first usage.  The size of the cache
     * may be controlled by the {@code -XX:AutoBoxCacheMax=<size>} option.
     * During VM initialization, java.lang.Integer.IntegerCache.high property
     * may be set and saved in the private system properties in the
     * sun.misc.VM class.
     */
  private static class IntegerCache {
        static final int low = -128;
        static final int high;
        static final Integer cache[];
        static {
            // high value may be configured by property
            int h = 127;
            String integerCacheHighPropValue =
                sun.misc.VM.getSavedProperty("java.lang.Integer.IntegerCache.high");
            if (integerCacheHighPropValue != null) {
                try {
                    int i = parseInt(integerCacheHighPropValue);
                    i = Math.max(i, 127);
                    // Maximum array size is Integer.MAX_VALUE
                    h = Math.min(i, Integer.MAX_VALUE - (-low) -1);
                } catch( NumberFormatException nfe) {
                    // If the property cannot be parsed into an int, ignore it.
                }
            }
            high = h;
            cache = new Integer[(high - low) + 1];
            int j = low;
            for(int k = 0; k < cache.length; k++)
                cache[k] = new Integer(j++);
            // range [-128, 127] must be interned (JLS7 5.1.7)
            assert IntegerCache.high >= 127;
        }
        private IntegerCache() {}
    }

 public Xxx XxxValue() {
        return (Xxx)value;
    }

    /**
     * Returns the value of this {@code Integer} as an
     * {@code int}.
     */
    public int intValue() {
        return value;
    }

public boolean equals(Object obj) {
        if (obj instanceof Integer) {
            return value == ((Integer)obj).intValue();
        }
        return false;
    }

   @Override
    public int hashCode() {
        return Integer.hashCode(value);
    }

    /**
     * Returns a hash code for a {@code int} value; compatible with
     * {@code Integer.hashCode()}.
     *
     * @param value the value to hash
     * @since 1.8
     *
     * @return a hash code value for a {@code int} value.
     */
    public static int hashCode(int value) {
        return value;
    }

/**
     * Returns an {@code int} value with at most a single one-bit, in the
     * position of the highest-order ("leftmost") one-bit in the specified
     * {@code int} value.  Returns zero if the specified value has no
     * one-bits in its two's complement binary representation, that is, if it
     * is equal to zero.
     *
     * @param i the value whose highest one bit is to be computed
     * @return an {@code int} value with a single one-bit, in the position
     *     of the highest-order one-bit in the specified value, or zero if
     *     the specified value is itself equal to zero.
     * @since 1.5
     */
    public static int highestOneBit(int i) {
        // HD, Figure 3-1
        i |= (i >>  1);
        i |= (i >>  2);
        i |= (i >>  4);
        i |= (i >>  8);
        i |= (i >> 16);//保证把最高位的1及其后面所有位变成1
        return i - (i >>> 1);//减去，把（最高位的1变成0）了后面所有位变成1 这样返回的就是最高位的权值了
    }

  /**
     * Returns an {@code int} value with at most a single one-bit, in the
     * position of the lowest-order ("rightmost") one-bit in the specified
     * {@code int} value.  Returns zero if the specified value has no
     * one-bits in its two's complement binary representation, that is, if it
     * is equal to zero.
     *
     * @param i the value whose lowest one bit is to be computed
     * @return an {@code int} value with a single one-bit, in the position
     *     of the lowest-order one-bit in the specified value, or zero if
     *     the specified value is itself equal to zero.
     * @since 1.5
     */
    public static int lowestOneBit(int i) {
        // HD, Section 2-1
        return i & -i;
    }

package testlang;
/**
 * @author CSDN@日星月云
 * @date 2022/10/24 22:32
 */
public class testInteger {
    public static void main(String[] args) {
        Integer i=new Integer(15);
        System.out.println(i.toString());//15
        System.out.println(Integer.lowestOneBit(i));//1 传入一个int参数i，返回其二进制最低位1的权值
        System.out.println(Integer.highestOneBit(i));//8 传入一个int参数i，返回其二进制最高位1的权值
    }
}

/**
     * Returns the number of zero bits preceding the highest-order
     * ("leftmost") one-bit in the two's complement binary representation
     * of the specified {@code int} value.  Returns 32 if the
     * specified value has no one-bits in its two's complement representation,
     * in other words if it is equal to zero.
     *
     * <p>Note that this method is closely related to the logarithm base 2.
     * For all positive {@code int} values x:
     * <ul>
     * <li>floor(log<sub>2</sub>(x)) = {@code 31 - numberOfLeadingZeros(x)}
     * <li>ceil(log<sub>2</sub>(x)) = {@code 32 - numberOfLeadingZeros(x - 1)}
     * </ul>
     *
     * @param i the value whose number of leading zeros is to be computed
     * @return the number of zero bits preceding the highest-order
     *     ("leftmost") one-bit in the two's complement binary representation
     *     of the specified {@code int} value, or 32 if the value
     *     is equal to zero.
     * @since 1.5
     */
    public static int numberOfLeadingZeros(int i) {
        // HD, Figure 5-6
        if (i == 0)
            return 32;
        int n = 1;
        if (i >>> 16 == 0) { n += 16; i <<= 16; }
        if (i >>> 24 == 0) { n +=  8; i <<=  8; }
        if (i >>> 28 == 0) { n +=  4; i <<=  4; }
        if (i >>> 30 == 0) { n +=  2; i <<=  2; }
        n -= i >>> 31;
        return n;
    }

  /**
     * Returns the number of zero bits following the lowest-order ("rightmost")
     * one-bit in the two's complement binary representation of the specified
     * {@code int} value.  Returns 32 if the specified value has no
     * one-bits in its two's complement representation, in other words if it is
     * equal to zero.
     *
     * @param i the value whose number of trailing zeros is to be computed
     * @return the number of zero bits following the lowest-order ("rightmost")
     *     one-bit in the two's complement binary representation of the
     *     specified {@code int} value, or 32 if the value is equal
     *     to zero.
     * @since 1.5
     */
    public static int numberOfTrailingZeros(int i) {
        // HD, Figure 5-14
        int y;
        if (i == 0) return 32;
        int n = 31;
        y = i <<16; if (y != 0) { n = n -16; i = y; }
        y = i << 8; if (y != 0) { n = n - 8; i = y; }
        y = i << 4; if (y != 0) { n = n - 4; i = y; }
        y = i << 2; if (y != 0) { n = n - 2; i = y; }
        return n - ((i << 1) >>> 31);
    }

 public static int bitCount(int i) {
        // HD, Figure 5-2
        i = i - ((i >>> 1) & 0x55555555);
        i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
        i = (i + (i >>> 4)) & 0x0f0f0f0f;
        i = i + (i >>> 8);
        i = i + (i >>> 16);
        return i & 0x3f;
    }

/**
     * Returns the value obtained by rotating the two's complement binary
     * representation of the specified {@code int} value left by the
     * specified number of bits.  (Bits shifted out of the left hand, or
     * high-order, side reenter on the right, or low-order.)
     *
     * <p>Note that left rotation with a negative distance is equivalent to
     * right rotation: {@code rotateLeft(val, -distance) == rotateRight(val,
     * distance)}.  Note also that rotation by any multiple of 32 is a
     * no-op, so all but the last five bits of the rotation distance can be
     * ignored, even if the distance is negative: {@code rotateLeft(val,
     * distance) == rotateLeft(val, distance & 0x1F)}.
     *
     * @param i the value whose bits are to be rotated left
     * @param distance the number of bit positions to rotate left
     * @return the value obtained by rotating the two's complement binary
     *     representation of the specified {@code int} value left by the
     *     specified number of bits.
     * @since 1.5
     */
    public static int rotateLeft(int i, int distance) {
        return (i << distance) | (i >>> -distance);
    }

    /**
     * Returns the value obtained by rotating the two's complement binary
     * representation of the specified {@code int} value right by the
     * specified number of bits.  (Bits shifted out of the right hand, or
     * low-order, side reenter on the left, or high-order.)
     *
     * <p>Note that right rotation with a negative distance is equivalent to
     * left rotation: {@code rotateRight(val, -distance) == rotateLeft(val,
     * distance)}.  Note also that rotation by any multiple of 32 is a
     * no-op, so all but the last five bits of the rotation distance can be
     * ignored, even if the distance is negative: {@code rotateRight(val,
     * distance) == rotateRight(val, distance & 0x1F)}.
     *
     * @param i the value whose bits are to be rotated right
     * @param distance the number of bit positions to rotate right
     * @return the value obtained by rotating the two's complement binary
     *     representation of the specified {@code int} value right by the
     *     specified number of bits.
     * @since 1.5
     */
    public static int rotateRight(int i, int distance) {
        return (i >>> distance) | (i << -distance);
    }

  /**
     * Returns the value obtained by reversing the order of the bits in the
     * two's complement binary representation of the specified {@code int}
     * value.
     *
     * @param i the value to be reversed
     * @return the value obtained by reversing order of the bits in the
     *     specified {@code int} value.
     * @since 1.5
     */
    public static int reverse(int i) {
        // HD, Figure 7-1
        i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
        i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
        i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
        i = (i << 24) | ((i & 0xff00) << 8) |
            ((i >>> 8) & 0xff00) | (i >>> 24);
        return i;
    }

    /**
     * Returns the signum function of the specified {@code int} value.  (The
     * return value is -1 if the specified value is negative; 0 if the
     * specified value is zero; and 1 if the specified value is positive.)
     *
     * @param i the value whose signum is to be computed
     * @return the signum function of the specified {@code int} value.
     * @since 1.5
     */
    public static int signum(int i) {
        // HD, Section 2-7
        return (i >> 31) | (-i >>> 31);
    }

    /**
     * Returns the value obtained by reversing the order of the bytes in the
     * two's complement representation of the specified {@code int} value.
     *
     * @param i the value whose bytes are to be reversed
     * @return the value obtained by reversing the bytes in the specified
     *     {@code int} value.
     * @since 1.5
     */
    public static int reverseBytes(int i) {
        return ((i >>> 24)           ) |
               ((i >>   8) &   0xFF00) |
               ((i <<   8) & 0xFF0000) |
               ((i << 24));
    }

package testlang;
/**
 * @author CSDN@日星月云
 * @date 2022/10/24 22:32
 */
public class testInteger {
    public static void main(String[] args) {
        Integer i=new Integer(15);
        System.out.println(i.toString());//15
        // 传入一个int参数i，返回其二进制最低位1的权值
        System.out.println(Integer.lowestOneBit(i));//1
        // 传入一个int参数i，返回其二进制最高位1的权值
        System.out.println(Integer.highestOneBit(i));//8
        //在指定 int 值的二进制补码表示形式中最高位（最左边）的 1 位之前，返回零位的数量
        System.out.println(Integer.numberOfLeadingZeros(i));//28 32位，15=1111,4个一，前面还有28个零
        //返回指定的 int 值的二进制补码表示形式中最低（“最右边”）的为 1 的位后面的零位个数。
        System.out.println(Integer.numberOfTrailingZeros(i));//0   最右边为1后面的零的个数是0
        //返回指定 int 值的二进制补码表示形式的 1 位的数量。此函数有时用于人口普查。
        System.out.println(Integer.bitCount(i));//4   4个一
        //返回根据指定的位数循环左移指定的 int 值的二进制补码表示形式而得到的值。（位是从左边（即高位）移出，从右边（即低位）再进入）
        System.out.println(Integer.rotateLeft(i,1));//30  循环左移  11110  15*2=30
        //返回根据指定的位数循环右移指定的 int 值的二进制补码表示形式而得到的值。（位是从右边（即低位）移出，从左边（即高位）再进入）
        System.out.println(Integer.rotateRight(i,1));//-2147483641 循环右移  1(28个0)111 二进制补码
        //原码是 -(28个1)001
        System.out.println(Integer.toBinaryString(Integer.rotateRight(i,1)));//10000000000000000000000000000111
        //返回通过反转指定 int 值的二进制补码表示形式中位的顺序而获得的值。
        System.out.println(Integer.reverse(i));//-268435456  翻转   1111(28个0)二进制补码
        //原码是 -001(28个0)
        System.out.println(-Math.pow(2,28));//2.68435456E8
        //返回指定 int 值的符号函数。（如果指定值为负，则返回 －1；如果指定值为零，则返回 0；如果指定的值为正，则返回 1。）
        System.out.println(Integer.signum(i));//1
    }
}