Question

I’m trying to write a password generator. It requires that characters be in ASCII representation, but I’m trying to use crypto/rand. This provides numbers in big.Int format, though, and I need to convert the relevant lower 8 bits to a form usable in a string. I’ve tried converting from big.Int to uint8 with no luck so far.

Is there a good and easy method to do this? I have seen answers involving using encoding/binary to convert from int64 to [8]uint8, but those seem needlessly complex for my purpose. Any guidance at all would be appreciated :).

Answer 1

If you want to get bytes out of crypto/rand, I’d skip the use of big.Int and rand.Int() entirely and use either rand.Read() or rand.Reader:

package main

import (
    "crypto/rand"
    "fmt"
    "io"
)

// If you want just a byte at a time ...
// (you could change 'byte' to int8 if you prefer)    
func secureRandomByte() byte {
    data := make([]byte, 1)
    if _, err := rand.Read(data); err != nil {
        // handle error
        panic(err)
    }
    return data[0]
}

// If you want to read multiple bytes at a time ...
func genPassword(len int) (string, error) {
    data := make([]byte, len)
    if _, err := io.ReadFull(rand.Reader, data); err != nil {
        // handle error
        return "", err
    }
    for i := range data {
        // XXX quick munge into something printable
        data[i] &= 0x3F
        data[i] += 0x20
    }
    return string(data), nil
}

func main() {
    b := secureRandomByte()
    fmt.Printf("%T = %v\n", b, b)
    fmt.Println(genPassword(16))
}