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/*-
* Copyright (c) 2008-2012 WiredTiger, Inc.
* All rights reserved.
*
* See the file LICENSE for redistribution information.
*/
#include "wt_internal.h"
#ifdef __WIREDTIGER_UNUSED__
/*
* __wt_nlpo2_round --
* Round up to the next-largest power-of-two for a 32-bit unsigned value.
*
* In 12 operations, this code computes the next highest power of 2 for a 32-bit
* integer. The result may be expressed by the formula 1U << (lg(v - 1) + 1).
* Note that in the edge case where v is 0, it returns 0, which isn't a power of
* 2; you might append the expression v += (v == 0) to remedy this if it
* matters. It would be faster by 2 operations to use the formula and the
* log base 2 method that uses a lookup table, but in some situations, lookup
* tables are not suitable, so the above code may be best. (On a Athlon XP 2100+
* I've found the above shift-left and then OR code is as fast as using a single
* BSR assembly language instruction, which scans in reverse to find the highest
* set bit.) It works by copying the highest set bit to all of the lower bits,
* and then adding one, which results in carries that set all of the lower bits
* to 0 and one bit beyond the highest set bit to 1. If the original number was
* a power of 2, then the decrement will reduce it to one less, so that we round
* up to the same original value. Devised by Sean Anderson, September 14, 2001.
* Pete Hart pointed me to a couple newsgroup posts by him and William Lewis in
* February of 1997, where they arrive at the same algorithm.
* http://graphics.stanford.edu/~seander/bithacks.html
* Sean Eron Anderson, seander@cs.stanford.edu
*/
uint32_t
__wt_nlpo2_round(uint32_t v)
{
v--; /* If v is a power-of-two, return it. */
v |= v >> 1;
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
return (v + 1);
}
uint32_t
__wt_nlpo2(uint32_t v)
{
v |= v >> 1;
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
return (v + 1);
}
#endif /* __WIREDTIGER_UNUSED__ */
/*
* __wt_ispo2 --
* Return if a number is a power-of-two.
*/
int
__wt_ispo2(uint32_t v)
{
/*
* Only numbers that are powers of two will satisfy the relationship
* (v & (v - 1) == 0).
*
* However n must be positive, this returns 0 as a power of 2; to fix
* that, use: (! (v & (v - 1)) && v)
*/
return ((v & (v - 1)) == 0);
}
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