Go has two packages for random numbers:
crypto/rand, which provides a way to get random bytesmath/rand, which has a nice algorithm for shuffling intsI want to use the Perm algorithm from math/rand, but provide it with high-quality random numbers.
Since the two rand packages are part of the same standard library there should be a way to combine them in a way so that crypto/rand provides a good source of random numbers that is used by math/rand.Perm to generate a permutation.
Here (and on the Playground) is the code I wrote to connect these two packages:
package main
import (
cryptoRand "crypto/rand"
"encoding/binary"
"fmt"
mathRand "math/rand"
)
type cryptoSource struct{}
func (s cryptoSource) Int63() int64 {
bytes := make([]byte, 8, 8)
cryptoRand.Read(bytes)
return int64(binary.BigEndian.Uint64(bytes) >> 1)
}
func (s cryptoSource) Seed(seed int64) {
panic("seed")
}
func main() {
rnd := mathRand.New(&cryptoSource{})
perm := rnd.Perm(52)
fmt.Println(perm)
}
This code works. Ideally I don't want to define the cryptoSource type myself but just stick together the two rand packages so that they work together. So is there a predefined version of this cryptoSource type somewhere?
That's basically what you need to do. It's not often that you need a cryptographically secure source of randomness for the common usage of math/rand, so there's no adaptor provided. You can make the implementation slightly more efficient by allocating the buffer space directly in the struct, rather than allocating a new slice on every call.
type cryptoSource struct {
buf [8]byte
}
func (s *cryptoSource) Int63() int64 {
cryptoRand.Read(s.buf[:])
return int64(binary.BigEndian.Uint64(s.buf[:]) & (1<<63 - 1))
}
Or just make the source a [8]byte itself
type cryptoSource [8]byte
func (s *cryptoSource) Int63() int64 {
cryptoRand.Read(s[:])
return int64(binary.BigEndian.Uint64(s[:]) & (1<<63 - 1))
}