/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.apache.lucene.util; // from org.apache.solr.util rev 555343

 A variety of high efficiency bit twiddling routines.
@lucene.internal
/**  A variety of high efficiency bit twiddling routines.
 * @lucene.internal
 */
public final class BitUtil {

  private BitUtil() {} // no instance

  // The pop methods used to rely on bit-manipulation tricks for speed but it
  // turns out that it is faster to use the Long.bitCount method (which is an
  // intrinsic since Java 6u18) in a naive loop, see LUCENE-2221

  
Returns the number of set bits in an array of longs. /** Returns the number of set bits in an array of longs. */
  public static long pop_array(long[] arr, int wordOffset, int numWords) {
    long popCount = 0;
    for (int i = wordOffset, end = wordOffset + numWords; i < end; ++i) {
      popCount += Long.bitCount(arr[i]);
    }
    return popCount;
  }

  
Returns the popcount or cardinality of the two sets after an intersection.
 Neither array is modified. /** Returns the popcount or cardinality of the two sets after an intersection.
   *  Neither array is modified. */
  public static long pop_intersect(long[] arr1, long[] arr2, int wordOffset, int numWords) {
    long popCount = 0;
    for (int i = wordOffset, end = wordOffset + numWords; i < end; ++i) {
      popCount += Long.bitCount(arr1[i] & arr2[i]);
    }
    return popCount;
  }

   
Returns the popcount or cardinality of the union of two sets.
 Neither array is modified. /** Returns the popcount or cardinality of the union of two sets.
    *  Neither array is modified. */
   public static long pop_union(long[] arr1, long[] arr2, int wordOffset, int numWords) {
     long popCount = 0;
     for (int i = wordOffset, end = wordOffset + numWords; i < end; ++i) {
       popCount += Long.bitCount(arr1[i] | arr2[i]);
     }
     return popCount;
   }

  
Returns the popcount or cardinality of A & ~B. Neither array is modified. /** Returns the popcount or cardinality of {@code A & ~B}.
   *  Neither array is modified. */
  public static long pop_andnot(long[] arr1, long[] arr2, int wordOffset, int numWords) {
    long popCount = 0;
    for (int i = wordOffset, end = wordOffset + numWords; i < end; ++i) {
      popCount += Long.bitCount(arr1[i] & ~arr2[i]);
    }
    return popCount;
  }

  
Returns the popcount or cardinality of A ^ B
Neither array is modified. /** Returns the popcount or cardinality of A ^ B
    * Neither array is modified. */
  public static long pop_xor(long[] arr1, long[] arr2, int wordOffset, int numWords) {
    long popCount = 0;
    for (int i = wordOffset, end = wordOffset + numWords; i < end; ++i) {
      popCount += Long.bitCount(arr1[i] ^ arr2[i]);
    }
    return popCount;
  }

  
returns the next highest power of two, or the current value if it's already a power of two or zero/** returns the next highest power of two, or the current value if it's already a power of two or zero*/
  public static int nextHighestPowerOfTwo(int v) {
    v--;
    v |= v >> 1;
    v |= v >> 2;
    v |= v >> 4;
    v |= v >> 8;
    v |= v >> 16;
    v++;
    return v;
  }

  
returns the next highest power of two, or the current value if it's already a power of two or zero/** returns the next highest power of two, or the current value if it's already a power of two or zero*/
   public static long nextHighestPowerOfTwo(long v) {
    v--;
    v |= v >> 1;
    v |= v >> 2;
    v |= v >> 4;
    v |= v >> 8;
    v |= v >> 16;
    v |= v >> 32;
    v++;
    return v;
  }

  // magic numbers for bit interleaving
  private static final long MAGIC0 = 0x5555555555555555L;
  private static final long MAGIC1 = 0x3333333333333333L;
  private static final long MAGIC2 = 0x0F0F0F0F0F0F0F0FL;
  private static final long MAGIC3 = 0x00FF00FF00FF00FFL;
  private static final long MAGIC4 = 0x0000FFFF0000FFFFL;
  private static final long MAGIC5 = 0x00000000FFFFFFFFL;
  private static final long MAGIC6 = 0xAAAAAAAAAAAAAAAAL;

  // shift values for bit interleaving
  private static final long SHIFT0 = 1;
  private static final long SHIFT1 = 2;
  private static final long SHIFT2 = 4;
  private static final long SHIFT3 = 8;
  private static final long SHIFT4 = 16;

  
Interleaves the first 32 bits of each long value
Adapted from: http://graphics.stanford.edu/~seander/bithacks.html#InterleaveBMN
/**
   * Interleaves the first 32 bits of each long value
   *
   * Adapted from: http://graphics.stanford.edu/~seander/bithacks.html#InterleaveBMN
   */
  public static long interleave(int even, int odd) {
    long v1 = 0x00000000FFFFFFFFL & even;
    long v2 = 0x00000000FFFFFFFFL & odd;
    v1 = (v1 | (v1 << SHIFT4)) & MAGIC4;
    v1 = (v1 | (v1 << SHIFT3)) & MAGIC3;
    v1 = (v1 | (v1 << SHIFT2)) & MAGIC2;
    v1 = (v1 | (v1 << SHIFT1)) & MAGIC1;
    v1 = (v1 | (v1 << SHIFT0)) & MAGIC0;
    v2 = (v2 | (v2 << SHIFT4)) & MAGIC4;
    v2 = (v2 | (v2 << SHIFT3)) & MAGIC3;
    v2 = (v2 | (v2 << SHIFT2)) & MAGIC2;
    v2 = (v2 | (v2 << SHIFT1)) & MAGIC1;
    v2 = (v2 | (v2 << SHIFT0)) & MAGIC0;

    return (v2<<1) | v1;
  }

  
Extract just the even-bits value as a long from the bit-interleaved value
/**
   * Extract just the even-bits value as a long from the bit-interleaved value
   */
  public static long deinterleave(long b) {
    b &= MAGIC0;
    b = (b ^ (b >>> SHIFT0)) & MAGIC1;
    b = (b ^ (b >>> SHIFT1)) & MAGIC2;
    b = (b ^ (b >>> SHIFT2)) & MAGIC3;
    b = (b ^ (b >>> SHIFT3)) & MAGIC4;
    b = (b ^ (b >>> SHIFT4)) & MAGIC5;
    return b;
  }

  
flip flops odd with even bits
/**
   * flip flops odd with even bits
   */
  public static long flipFlop(final long b) {
    return ((b & MAGIC6) >>> 1) | ((b & MAGIC0) << 1 );
  }

   
Same as zigZagEncode(long) but on integers. /** Same as {@link #zigZagEncode(long)} but on integers. */
   public static int zigZagEncode(int i) {
     return (i >> 31) ^ (i << 1);
   }

   
Zig-zag
encode the provided long. Assuming the input is a signed long whose
absolute value can be stored on n bits, the returned value will
be an unsigned long that can be stored on n+1 bits.
/**
    * <a href="https://developers.google.com/protocol-buffers/docs/encoding#types">Zig-zag</a>
    * encode the provided long. Assuming the input is a signed long whose
    * absolute value can be stored on <tt>n</tt> bits, the returned value will
    * be an unsigned long that can be stored on <tt>n+1</tt> bits.
    */
   public static long zigZagEncode(long l) {
     return (l >> 63) ^ (l << 1);
   }

   
Decode an int previously encoded with zigZagEncode(int). /** Decode an int previously encoded with {@link #zigZagEncode(int)}. */
   public static int zigZagDecode(int i) {
     return ((i >>> 1) ^ -(i & 1));
   }

   
Decode a long previously encoded with zigZagEncode(long). /** Decode a long previously encoded with {@link #zigZagEncode(long)}. */
   public static long zigZagDecode(long l) {
     return ((l >>> 1) ^ -(l & 1));
   }
}