我一直在尝试在 Go 中使用汇编语言，并且编写了一个Hamming Weight函数作为练习。
我基于这个 SO 答案构建了一个原生 Go 版本，而汇编版本基于AMD 的这个文档（第 180 页）。在对这两个函数进行基准测试后，我发现原生 Go 版本比汇编版本快大约 1.5 倍 - 2 倍，尽管手写汇编版本几乎与go tool 6g -S popcount.go.
输出自go test -bench=.
PASS
BenchmarkPopCount 100000000 19.4 ns/op
BenchmarkPopCount_g 200000000 8.97 ns/op
ok popcount 4.777s
popcount.go
package popcount
func popCount(i uint32) uint32 // Defined in popcount_amd64.s
func popCount_g(i uint32) uint32 {
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24
}
popcount_test.go
package popcount
import "testing"
func TestPopcount(t *testing.T) {
for i := uint32(0); i < uint32(100); i++ {
if popCount(i) != popCount_g(i) {
t.Fatalf("failed on input = %v", i)
}
}
}
func BenchmarkPopCount(b *testing.B) {
for i := 0; i < b.N; i++ {
popCount(uint32(i))
}
}
func BenchmarkPopCount_g(b *testing.B) {
for i := 0; i < b.N; i++ {
popCount_g(uint32(i))
}
}
popcount_amd64.s
// func popCount(i uint32) uint32
TEXT ·popCount(SB),$0
MOVL i+0(FP), BP // i
MOVL BP, BX // i
SHRL $1, BX // i >> 1
ANDL $0x055555555, BX // (i >> 1) & 0x55555555
SUBL BX, BP // w = i - ((i >> 1) & 0x55555555)
MOVL BP, AX // w
SHRL $2, BP // w >> 2
ANDL $0x033333333, AX // w & 0x33333333
ANDL $0x033333333, BP // (w >> 2) & 0x33333333
ADDL BP, AX // x = (w & 0x33333333) + ((w >> 2) & 0x33333333)
MOVL AX, BX // x
SHRL $4, BX // x >> 4
ADDL AX, BX // x + (x >> 4)
ANDL $0x00F0F0F0F, BX // y = (x + (x >> 4) & 0x0F0F0F0F)
IMULL $0x001010101, BX // y * 0x01010101
SHRL $24, BX // population count = (y * 0x01010101) >> 24
MOVL BX, toReturn+8(FP) // Store result.
RET // return
输出自go tool 6g -S popcount.go
"".popCount_g t=1 size=64 value=0 args=0x10 locals=0
000000 00000 (popcount.go:5) TEXT "".popCount_g+0(SB),4,$0-16
000000 00000 (popcount.go:5) NOP ,
000000 00000 (popcount.go:5) NOP ,
000000 00000 (popcount.go:5) MOVL "".i+8(FP),BP
0x0004 00004 (popcount.go:5) FUNCDATA $2,gclocals┬À9308e7ef08d2cc2f72ae1228688dacf9+0(SB)
0x0004 00004 (popcount.go:5) FUNCDATA $3,gclocals┬À3280bececceccd33cb74587feedb1f9f+0(SB)
0x0004 00004 (popcount.go:6) MOVL BP,BX
0x0006 00006 (popcount.go:6) SHRL $1,BX
0x0008 00008 (popcount.go:6) ANDL $1431655765,BX
0x000e 00014 (popcount.go:6) SUBL BX,BP
0x0010 00016 (popcount.go:7) MOVL BP,AX
0x0012 00018 (popcount.go:7) ANDL $858993459,AX
0x0017 00023 (popcount.go:7) SHRL $2,BP
0x001a 00026 (popcount.go:7) ANDL $858993459,BP
0x0020 00032 (popcount.go:7) ADDL BP,AX
0x0022 00034 (popcount.go:8) MOVL AX,BX
0x0024 00036 (popcount.go:8) SHRL $4,BX
0x0027 00039 (popcount.go:8) ADDL AX,BX
0x0029 00041 (popcount.go:8) ANDL $252645135,BX
0x002f 00047 (popcount.go:8) IMULL $16843009,BX
0x0035 00053 (popcount.go:8) SHRL $24,BX
0x0038 00056 (popcount.go:8) MOVL BX,"".~r1+16(FP)
0x003c 00060 (popcount.go:8) RET ,
我从这里知道这些FUNCDATA行包含垃圾收集器的信息，但除此之外我没有看到任何明显的差异。
是什么导致这两个函数之间的速度差异如此之大？