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math/bits: move left-over functionality from bits_impl.go to bits.go
Removes an extra function call for TrailingZeroes and thus may increase chances for inlining. Change-Id: Iefd8d4402dc89b64baf4e5c865eb3dadade623af Reviewed-on: https://go-review.googlesource.com/37613 Run-TryBot: Robert Griesemer <gri@golang.org> Reviewed-by: Brad Fitzpatrick <bradfitz@golang.org>
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@ -8,6 +8,8 @@
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// functions for the predeclared unsigned integer types.
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// functions for the predeclared unsigned integer types.
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package bits
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package bits
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const uintSize = 32 << (^uint(0) >> 32 & 1) // 32 or 64
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// UintSize is the size of a uint in bits.
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// UintSize is the size of a uint in bits.
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const UintSize = uintSize
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const UintSize = uintSize
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@ -30,20 +32,72 @@ func LeadingZeros64(x uint64) int { return 64 - Len64(x) }
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// --- TrailingZeros ---
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// --- TrailingZeros ---
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// See http://supertech.csail.mit.edu/papers/debruijn.pdf
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const deBruijn32 = 0x077CB531
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var deBruijn32tab = [32]byte{
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0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
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31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9,
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}
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const deBruijn64 = 0x03f79d71b4ca8b09
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var deBruijn64tab = [64]byte{
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0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4,
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62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5,
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63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11,
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54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6,
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}
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// TrailingZeros returns the number of trailing zero bits in x; the result is UintSize for x == 0.
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// TrailingZeros returns the number of trailing zero bits in x; the result is UintSize for x == 0.
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func TrailingZeros(x uint) int { return ntz(x) }
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func TrailingZeros(x uint) int {
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if UintSize == 32 {
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return TrailingZeros32(uint32(x))
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}
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return TrailingZeros64(uint64(x))
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}
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// TrailingZeros8 returns the number of trailing zero bits in x; the result is 8 for x == 0.
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// TrailingZeros8 returns the number of trailing zero bits in x; the result is 8 for x == 0.
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func TrailingZeros8(x uint8) int { return int(ntz8tab[x]) }
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func TrailingZeros8(x uint8) int {
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return int(ntz8tab[x])
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}
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// TrailingZeros16 returns the number of trailing zero bits in x; the result is 16 for x == 0.
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// TrailingZeros16 returns the number of trailing zero bits in x; the result is 16 for x == 0.
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func TrailingZeros16(x uint16) int { return ntz16(x) }
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func TrailingZeros16(x uint16) (n int) {
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if x == 0 {
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return 16
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}
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// see comment in TrailingZeros64
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return int(deBruijn32tab[uint32(x&-x)*deBruijn32>>(32-5)])
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}
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// TrailingZeros32 returns the number of trailing zero bits in x; the result is 32 for x == 0.
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// TrailingZeros32 returns the number of trailing zero bits in x; the result is 32 for x == 0.
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func TrailingZeros32(x uint32) int { return ntz32(x) }
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func TrailingZeros32(x uint32) int {
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if x == 0 {
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return 32
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}
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// see comment in TrailingZeros64
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return int(deBruijn32tab[(x&-x)*deBruijn32>>(32-5)])
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}
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// TrailingZeros64 returns the number of trailing zero bits in x; the result is 64 for x == 0.
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// TrailingZeros64 returns the number of trailing zero bits in x; the result is 64 for x == 0.
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func TrailingZeros64(x uint64) int { return ntz64(x) }
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func TrailingZeros64(x uint64) int {
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if x == 0 {
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return 64
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}
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// If popcount is fast, replace code below with return popcount(^x & (x - 1)).
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//
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// x & -x leaves only the right-most bit set in the word. Let k be the
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// index of that bit. Since only a single bit is set, the value is two
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// to the power of k. Multiplying by a power of two is equivalent to
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// left shifting, in this case by k bits. The de Bruijn (64 bit) constant
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// is such that all six bit, consecutive substrings are distinct.
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// Therefore, if we have a left shifted version of this constant we can
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// find by how many bits it was shifted by looking at which six bit
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// substring ended up at the top of the word.
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// (Knuth, volume 4, section 7.3.1)
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return int(deBruijn64tab[(x&-x)*deBruijn64>>(64-6)])
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}
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// --- OnesCount ---
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// --- OnesCount ---
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@ -1,67 +0,0 @@
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// Copyright 2017 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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// This file provides basic implementations of the bits functions.
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package bits
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const uintSize = 32 << (^uint(0) >> 32 & 1) // 32 or 64
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func ntz(x uint) (n int) {
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if UintSize == 32 {
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return ntz32(uint32(x))
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}
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return ntz64(uint64(x))
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}
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// See http://supertech.csail.mit.edu/papers/debruijn.pdf
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const deBruijn32 = 0x077CB531
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var deBruijn32tab = [32]byte{
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0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
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31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9,
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}
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func ntz16(x uint16) (n int) {
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if x == 0 {
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return 16
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}
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// see comment in ntz64
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return int(deBruijn32tab[uint32(x&-x)*deBruijn32>>(32-5)])
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}
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func ntz32(x uint32) int {
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if x == 0 {
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return 32
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}
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// see comment in ntz64
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return int(deBruijn32tab[(x&-x)*deBruijn32>>(32-5)])
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}
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const deBruijn64 = 0x03f79d71b4ca8b09
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var deBruijn64tab = [64]byte{
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0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4,
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62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5,
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63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11,
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54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6,
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}
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func ntz64(x uint64) int {
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if x == 0 {
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return 64
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}
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// If popcount is fast, replace code below with return popcount(^x & (x - 1)).
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//
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// x & -x leaves only the right-most bit set in the word. Let k be the
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// index of that bit. Since only a single bit is set, the value is two
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// to the power of k. Multiplying by a power of two is equivalent to
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// left shifting, in this case by k bits. The de Bruijn (64 bit) constant
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// is such that all six bit, consecutive substrings are distinct.
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// Therefore, if we have a left shifted version of this constant we can
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// find by how many bits it was shifted by looking at which six bit
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// substring ended up at the top of the word.
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// (Knuth, volume 4, section 7.3.1)
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return int(deBruijn64tab[(x&-x)*deBruijn64>>(64-6)])
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}
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