Golang随机数发生器如何正确播种

Question

I am trying to generate a random string in Go and here is the code I have written so far:

package main

import (
    "bytes"
    "fmt"
    "math/rand"
    "time"
)

func main() {
    fmt.Println(randomString(10))
}

func randomString(l int) string {
    var result bytes.Buffer
    var temp string
    for i := 0; i < l; {
        if string(randInt(65, 90)) != temp {
            temp = string(randInt(65, 90))
            result.WriteString(temp)
            i++
        }
    }
    return result.String()
}

func randInt(min int, max int) int {
    rand.Seed(time.Now().UTC().UnixNano())
    return min + rand.Intn(max-min)
}

My implementation is very slow. Seeding using time brings the same random number for a certain time, so the loop iterates again and again. How can I improve my code?

Answer 1

Each time you set the same seed, you get the same sequence. So of course if you're setting the seed to the time in a fast loop, you'll probably call it with the same seed many times.

In your case, as you're calling your randInt function until you have a different value, you're waiting for the time (as returned by Nano) to change.

As for all pseudo-random libraries, you have to set the seed only once, for example when initializing your program unless you specifically need to reproduce a given sequence (which is usually only done for debugging and unit testing).

After that you simply call Intn to get the next random integer.

Move the rand.Seed(time.Now().UTC().UnixNano()) line from the randInt function to the start of the main and everything will be faster.

Note also that I think you can simplify your string building:

package main

import (
    "fmt"
    "math/rand"
    "time"
)

func main() {
    rand.Seed(time.Now().UTC().UnixNano())
    fmt.Println(randomString(10))
}

func randomString(l int) string {
    bytes := make([]byte, l)
    for i := 0; i < l; i++ {
        bytes[i] = byte(randInt(65, 90))
    }
    return string(bytes)
}

func randInt(min int, max int) int {
    return min + rand.Intn(max-min)
}