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/* 
 * Repackaged from org.apache.lucene.util.OpenBitSet (Lucene).
 * svn rev. 1479633, https://svn.apache.org/repos/asf/lucene/dev/trunk 
 * 
 * Minor changes in class hierarchy, removed serialization.
 */


/**
 * 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 com.carrotsearch.hppc;

/**  
 * A variety of high efficiency bit twiddling routines.
 */
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. */
  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. */
  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. */
   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. */
  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. */
  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*/
  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*/
   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;
  }
}