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math/bits: move left-over functionality from bits_impl.go to bits.go

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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>
This commit is contained in:
Robert Griesemer 2017-02-28 14:55:00 -08:00
parent 09294ab754
commit 32b41c8dc7
2 changed files with 59 additions and 72 deletions
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64
src/math/bits/bits.go
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@ -8,6 +8,8 @@
// functions for the predeclared unsigned integer types.
package bits
const uintSize = 32 << (^uint(0) >> 32 & 1) // 32 or 64
// UintSize is the size of a uint in bits.
const UintSize = uintSize
@ -30,20 +32,72 @@ func LeadingZeros64(x uint64) int { return 64 - Len64(x) }
// --- TrailingZeros ---
// See http://supertech.csail.mit.edu/papers/debruijn.pdf
const deBruijn32 = 0x077CB531
var deBruijn32tab = [32]byte{
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9,
}
const deBruijn64 = 0x03f79d71b4ca8b09
var deBruijn64tab = [64]byte{
0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4,
62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5,
63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11,
54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6,
}
// TrailingZeros returns the number of trailing zero bits in x; the result is UintSize for x == 0.
func TrailingZeros(x uint) int { return ntz(x) }
func TrailingZeros(x uint) int {
if UintSize == 32 {
return TrailingZeros32(uint32(x))
}
return TrailingZeros64(uint64(x))
}
// TrailingZeros8 returns the number of trailing zero bits in x; the result is 8 for x == 0.
func TrailingZeros8(x uint8) int { return int(ntz8tab[x]) }
func TrailingZeros8(x uint8) int {
return int(ntz8tab[x])
}
// TrailingZeros16 returns the number of trailing zero bits in x; the result is 16 for x == 0.
func TrailingZeros16(x uint16) int { return ntz16(x) }
func TrailingZeros16(x uint16) (n int) {
if x == 0 {
return 16
}
// see comment in TrailingZeros64
return int(deBruijn32tab[uint32(x&-x)*deBruijn32>>(32-5)])
}
// TrailingZeros32 returns the number of trailing zero bits in x; the result is 32 for x == 0.
func TrailingZeros32(x uint32) int { return ntz32(x) }
func TrailingZeros32(x uint32) int {
if x == 0 {
return 32
}
// see comment in TrailingZeros64
return int(deBruijn32tab[(x&-x)*deBruijn32>>(32-5)])
}
// TrailingZeros64 returns the number of trailing zero bits in x; the result is 64 for x == 0.
func TrailingZeros64(x uint64) int { return ntz64(x) }
func TrailingZeros64(x uint64) int {
if x == 0 {
return 64
}
// If popcount is fast, replace code below with return popcount(^x & (x - 1)).
//
// x & -x leaves only the right-most bit set in the word. Let k be the
// index of that bit. Since only a single bit is set, the value is two
// to the power of k. Multiplying by a power of two is equivalent to
// left shifting, in this case by k bits. The de Bruijn (64 bit) constant
// is such that all six bit, consecutive substrings are distinct.
// Therefore, if we have a left shifted version of this constant we can
// find by how many bits it was shifted by looking at which six bit
// substring ended up at the top of the word.
// (Knuth, volume 4, section 7.3.1)
return int(deBruijn64tab[(x&-x)*deBruijn64>>(64-6)])
}
// --- OnesCount ---

67
src/math/bits/bits_impl.go
View File

@ -1,67 +0,0 @@
// Copyright 2017 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
// This file provides basic implementations of the bits functions.
package bits
const uintSize = 32 << (^uint(0) >> 32 & 1) // 32 or 64
func ntz(x uint) (n int) {
if UintSize == 32 {
return ntz32(uint32(x))
}
return ntz64(uint64(x))
}
// See http://supertech.csail.mit.edu/papers/debruijn.pdf
const deBruijn32 = 0x077CB531
var deBruijn32tab = [32]byte{
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9,
}
func ntz16(x uint16) (n int) {
if x == 0 {
return 16
}
// see comment in ntz64
return int(deBruijn32tab[uint32(x&-x)*deBruijn32>>(32-5)])
}
func ntz32(x uint32) int {
if x == 0 {
return 32
}
// see comment in ntz64
return int(deBruijn32tab[(x&-x)*deBruijn32>>(32-5)])
}
const deBruijn64 = 0x03f79d71b4ca8b09
var deBruijn64tab = [64]byte{
0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4,
62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5,
63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11,
54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6,
}
func ntz64(x uint64) int {
if x == 0 {
return 64
}
// If popcount is fast, replace code below with return popcount(^x & (x - 1)).
//
// x & -x leaves only the right-most bit set in the word. Let k be the
// index of that bit. Since only a single bit is set, the value is two
// to the power of k. Multiplying by a power of two is equivalent to
// left shifting, in this case by k bits. The de Bruijn (64 bit) constant
// is such that all six bit, consecutive substrings are distinct.
// Therefore, if we have a left shifted version of this constant we can
// find by how many bits it was shifted by looking at which six bit
// substring ended up at the top of the word.
// (Knuth, volume 4, section 7.3.1)
return int(deBruijn64tab[(x&-x)*deBruijn64>>(64-6)])
}