You can try the following examples on the Go Playground. Benchmark results is at the end.
Note: the implementations below are not safe for concurrent use; I intentionally made them like this to be simpler and faster.
Fastest when using only public API (always hashes all input)
The general concept and interface of Go's hash algorithms is the hash.Hash interface. This does not allow you to save the state of the hasher and to return or rewind to the saved state. So using the public hash APIs of the Go standard lib, you always have to calculate the hash from start.
What the public API offers is to reuse an already constructed hasher to calculate the hash of a new input, using the Hash.Reset() method. This is nice so that no (memory) allocations will be needed to calculate multiple hash values. Also you may take advantage of the optional slice that may be passed to Hash.Sum() which is used to append the current hash to. This is nice so that no allocations will be needed to receive the hash results either.
Here's an example that takes advantage of these:
type Cached1 struct {
hasher hash.Hash
result [sha256.Size]byte
}
func NewCached1() *Cached1 {
return &Cached1{hasher: sha256.New()}
}
func (c *Cached1) Sum(data []byte) []byte {
c.hasher.Reset()
c.hasher.Write(data)
return c.hasher.Sum(c.result[:0])
}
Test data
We'll use the following test data:
var fixed = bytes.Repeat([]byte{1}, 76)
var variantA = []byte{1, 1, 1, 1}
var variantB = []byte{2, 2, 2, 2}
var data = append(append([]byte{}, fixed...), variantA...)
var data2 = append(append([]byte{}, fixed...), variantB...)
var c1 = NewCached1()
First let's get authentic results (to verify if our hasher works correctly):
fmt.Printf("%x\n", sha256.Sum256(data))
fmt.Printf("%x\n", sha256.Sum256(data2))
Output:
fb8e69bdfa2ad15be7cc8a346b74e773d059f96cfc92da89e631895422fe966a
10ef52823dad5d1212e8ac83b54c001bfb9a03dc0c7c3c83246fb988aa788c0c
Now let's check our Cached1 hasher:
fmt.Printf("%x\n", c1.Sum(data))
fmt.Printf("%x\n", c1.Sum(data2))
Output is the same:
fb8e69bdfa2ad15be7cc8a346b74e773d059f96cfc92da89e631895422fe966a
10ef52823dad5d1212e8ac83b54c001bfb9a03dc0c7c3c83246fb988aa788c0c
Even faster but may break (in future Go releases): hashes only the last 4 bytes
Now let's see a less flexible solution which truly calculates the hash of the first 76 fixed part only once.
The hasher of the crypto/sha256 package is the unexported sha256.digest type (more precisely a pointer to this type):
// digest represents the partial evaluation of a checksum.
type digest struct {
h [8]uint32
x [chunk]byte
nx int
len uint64
is224 bool // mark if this digest is SHA-224
}
A value of the digest struct type basically holds the current state of the hasher.
What we may do is feed the hasher the fixed, first 76 bytes, and then save this struct value. When we need to caclulate the hash of some 80 bytes data where the first 76 is the same, we use this saved value as a starting point, and then feed the varying last 4 bytes.
Note that it's enough to simply save this struct value as it contains no pointers and no descriptor types like slices and maps. Else we would also have to make a copy of those, but we're "lucky". So this solution would need adjustment if a future implementation of crypto/sha256 would add a pointer or slice field for example.
Since sha256.digest is unexported, we can only use reflection (reflect package) to achieve our goals, which inherently will add some delays to computation.
Example implementation that does this:
type Cached2 struct {
origv reflect.Value
hasherv reflect.Value
hasher hash.Hash
result [sha256.Size]byte
}
func NewCached2(fixed []byte) *Cached2 {
h := sha256.New()
h.Write(fixed)
c := &Cached2{origv: reflect.ValueOf(h).Elem()}
hasherv := reflect.New(c.origv.Type())
c.hasher = hasherv.Interface().(hash.Hash)
c.hasherv = hasherv.Elem()
return c
}
func (c *Cached2) Sum(data []byte) []byte {
// Set state of the fixed hash:
c.hasherv.Set(c.origv)
c.hasher.Write(data)
return c.hasher.Sum(c.result[:0])
}
Testing it:
var c2 = NewCached2(fixed)
fmt.Printf("%x\n", c2.Sum(variantA))
fmt.Printf("%x\n", c2.Sum(variantB))
Output is again the same:
fb8e69bdfa2ad15be7cc8a346b74e773d059f96cfc92da89e631895422fe966a
10ef52823dad5d1212e8ac83b54c001bfb9a03dc0c7c3c83246fb988aa788c0c
So it works.