@JensG was on the right track.
The calls to gob
to encode the key in binary made up the vast majority of the cost.
When I transitioned to passing in byte arrays the benchmark started getting the results I was expecting.
Thanks for the help!
BenchmarkJenkins 100000000 20.4 ns/op
BenchmarkSHA1 5000000 463 ns/op
BenchmarkSHA256 1000000 1223 ns/op
Benchmark code:
package bloom
import (
"testing"
"crypto/sha1"
"crypto/sha256"
)
func BenchmarkJenkins(b *testing.B) {
j := jenkinsHash{}
for i := 0; i < b.N; i++ {
j.ComputeHash([]byte{byte(i)})
}
}
func BenchmarkSHA1(b *testing.B) {
for i := 0; i < b.N; i++ {
sha1.Sum([]byte{byte(i)})
}
}
func BenchmarkSHA256(b *testing.B) {
for i := 0; i < b.N; i++ {
sha256.Sum256([]byte{byte(i)})
}
}
Altered code:
package bloom
type jenkinsHash struct {
}
// adapted from http://bretmulvey.com/hash/7.html
func (_ jenkinsHash) ComputeHash(data []byte) (uint64, error) {
var a, b, c uint64
a, b = 0x9e3779b9, 0x9e3779b9
c = 0
i := 0
for i = 0; i < len(data)-12; {
a += uint64(data[i]) | uint64(data[i+1]<<8) | uint64(data[i+2]<<16) | uint64(data[i+3]<<24)
i += 4
b += uint64(data[i]) | uint64(data[i+1]<<8) | uint64(data[i+2]<<16) | uint64(data[i+3]<<24)
i += 4
c += uint64(data[i]) | uint64(data[i+1]<<8) | uint64(data[i+2]<<16) | uint64(data[i+3]<<24)
a, b, c = mix(a, b, c)
}
c += uint64(len(data))
if i < len(data) {
a += uint64(data[i])
i++
}
if i < len(data) {
a += uint64(data[i]) << 8
i++
}
if i < len(data) {
a += uint64(data[i]) << 16
i++
}
if i < len(data) {
a += uint64(data[i]) << 24
i++
}
if i < len(data) {
b += uint64(data[i])
i++
}
if i < len(data) {
b += uint64(data[i]) << 8
i++
}
if i < len(data) {
b += uint64(data[i]) << 16
i++
}
if i < len(data) {
b += uint64(data[i]) << 24
i++
}
if i < len(data) {
c += uint64(data[i]) << 8
i++
}
if i < len(data) {
c += uint64(data[i]) << 16
i++
}
if i < len(data) {
c += uint64(data[i]) << 24
i++
}
a, b, c = mix(a, b, c)
return c, nil
}
func mix(a, b, c uint64) (uint64, uint64, uint64) {
a -= b
a -= c
a ^= (c >> 13)
b -= c
b -= a
b ^= (a << 8)
c -= a
c -= b
c ^= (b >> 13)
a -= b
a -= c
a ^= (c >> 12)
b -= c
b -= a
b ^= (a << 16)
c -= a
c -= b
c ^= (b >> 5)
a -= b
a -= c
a ^= (c >> 3)
b -= c
b -= a
b ^= (a << 10)
c -= a
c -= b
c ^= (b >> 15)
return a, b, c
}