英文:
Using crypto/rand for generating permutations with rand.Perm
问题
Go语言有两个用于生成随机数的包：
crypto/rand，提供了获取随机字节的方法math/rand，具有一个用于对整数进行洗牌的好算法
我想使用math/rand中的Perm算法，但是要提供高质量的随机数。
由于这两个rand包都是标准库的一部分，应该有一种方法将它们结合起来，使得crypto/rand提供一个好的随机数源，供math/rand.Perm用于生成排列。
这里是我编写的连接这两个包的代码（在Playground上也有）：
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)
}
这段代码是有效的。理想情况下，我不想自己定义cryptoSource类型，而是将这两个rand包组合在一起，使它们能够一起工作。所以是否有一个预定义版本的cryptoSource类型存在？
英文:
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 ints
I 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?
答案1
得分: 3
这基本上是你需要做的。通常情况下，对于math/rand的常规用法，你不经常需要一个加密安全的随机源，所以没有提供适配器。你可以通过直接在值中分配缓冲区空间，而不是在每次调用时分配新的切片，使实现稍微更高效。然而，如果读取操作系统的随机源失败，这将需要引发 panic 来防止返回无效的结果。
type cryptoSource [8]byte
func (s *cryptoSource) Int63() int64 {
_, err := cryptoRand.Read(s[:])
if err != nil {
panic(err)
}
return int64(binary.BigEndian.Uint64(s[:]) & (1<<63 - 1))
}
英文:
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 value, rather than allocating a new slice on every call. However in the unlikely event that reading the OS random source fails, this will need to panic to prevent returning invalid results.
type cryptoSource [8]byte
func (s *cryptoSource) Int63() int64 {
_, err := cryptoRand.Read(s[:])
if err != nil {
panic(err)
}
return int64(binary.BigEndian.Uint64(s[:]) & (1<<63 - 1))
}
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