package main
import (
"crypto/md5"
"fmt"
)
func main() {
hash := md5.New()
b := []byte("test")
fmt.Printf("%x\n", hash.Sum(b))
hash.Write(b)
fmt.Printf("%x\n", hash.Sum(nil))
}
Output:
*md5.digest74657374d41d8cd98f00b204e9800998ecf8427e
098f6bcd4621d373cade4e832627b4f6
Could someone please explain to me why/how do I get different result for the two print ?
I'm building up on the already good answers. I'm not sure if Sum
is actually the function you want. From the hash.Hash
documentation:
// Sum appends the current hash to b and returns the resulting slice.
// It does not change the underlying hash state.
Sum(b []byte) []byte
This function has a dual use-case, which you seem to mix in an unfortunate way. The use-cases are:
In case you simply want to compute the hash of something, either use md5.Sum(data)
or
digest := md5.New()
digest.Write(data)
hash := digest.Sum(nil)
This code will, according to the excerpt of the documentation above, append the checksum of data
to nil
, resulting in the checksum of data
.
If you want to chain several blocks of hashes, the second use-case of hash.Sum
, you can do it like this:
hashed := make([]byte, 0)
for hasData {
digest.Write(data)
hashed = digest.Sum(hashed)
}
This will append each iteration's hash to the already computed hashes. Probably not what you want.
So, now you should be able to see why your code is failing. If not, take this commented version of your code (On play):
hash := md5.New()
b := []byte("test")
fmt.Printf("%x\n", hash.Sum(b)) // gives 74657374<hash> (74657374 = "test")
fmt.Printf("%x\n", hash.Sum([]byte("AAA"))) // gives 414141<hash> (41 = 'A')
fmt.Printf("%x\n", hash.Sum(nil)) // gives <hash> as append(nil, hash) == hash
fmt.Printf("%x\n", hash.Sum(b)) // gives 74657374<hash> (74657374 = "test")
fmt.Printf("%x\n", hash.Sum([]byte("AAA"))) // gives 414141<hash> (41 = 'A')
hash.Write(b)
fmt.Printf("%x\n", hash.Sum(nil)) // gives a completely different hash since internal bytes changed due to Write()