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High Performance Primitive Collections: data structures (maps, sets, lists, stacks, queues) generated for combinations of object and primitive types to conserve JVM memory and speed up execution.
/*
* 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;
}
}