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>

src/math/bits/bits.go

@@ -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 ---
 

src/math/bits/bits_impl.go

-// 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)])
-}