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Update #47 - Singleplayer lag fixes

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lax1dude
2025-01-19 13:26:27 -08:00
parent 3f5ee57068
commit 1f0d593a8c
2052 changed files with 133581 additions and 2339 deletions
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sources/main/java/com/carrotsearch/hppc/BitUtil.java Normal file
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/*
* HPPC
*
* Copyright (C) 2010-2024 Carrot Search s.c. and contributors
* All rights reserved.
*
* Refer to the full license file "LICENSE.txt":
* https://github.com/carrotsearch/hppc/blob/master/LICENSE.txt
*/
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 &amp; ~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;
}
}