Is there a method to generate a UUID with go language

Question

I have code that looks like this:

u := make([]byte, 16)
_, err := rand.Read(u)
if err != nil {
    return
}

u[8] = (u[8] | 0x80) & 0xBF // what does thi         


        

Answer 1

This library is our standard for uuid generation and parsing:

https://github.com/pborman/uuid

Answer 2

gofrs/uuid is the replacement for satori/go.uuid, which is the most starred UUID package for Go. It supports UUID versions 1-5 and is RFC 4122 and DCE 1.1 compliant.

import "github.com/gofrs/uuid"

// Create a Version 4 UUID, panicking on error
u := uuid.Must(uuid.NewV4())

Answer 3

On Linux, you can read from /proc/sys/kernel/random/uuid:

package main

import "io/ioutil"
import "fmt"

func main() {
    u, _ := ioutil.ReadFile("/proc/sys/kernel/random/uuid")
    fmt.Println(string(u))
}

No external dependencies!

$ go run uuid.go 
3ee995e3-0c96-4e30-ac1e-f7f04fd03e44

Answer 4

You can generate UUIDs using the go-uuid library. This can be installed with:

go get github.com/nu7hatch/gouuid

You can generate random (version 4) UUIDs with:

import "github.com/nu7hatch/gouuid"

...

u, err := uuid.NewV4()

The returned UUID type is a 16 byte array, so you can retrieve the binary value easily. It also provides the standard hex string representation via its String() method.

The code you have also looks like it will also generate a valid version 4 UUID: the bitwise manipulation you perform at the end set the version and variant fields of the UUID to correctly identify it as version 4. This is done to distinguish random UUIDs from ones generated via other algorithms (e.g. version 1 UUIDs based on your MAC address and time).

Answer 5

For Windows, I did recently this:

// +build windows

package main

import (
    "syscall"
    "unsafe"
)

var (
    modrpcrt4 = syscall.NewLazyDLL("rpcrt4.dll")
    procUuidCreate = modrpcrt4.NewProc("UuidCreate")
)

const (
    RPC_S_OK = 0
)

func NewUuid() ([]byte, error) {
    var uuid [16]byte
    rc, _, e := syscall.Syscall(procUuidCreate.Addr(), 1,
             uintptr(unsafe.Pointer(&uuid[0])), 0, 0)
    if int(rc) != RPC_S_OK {
        if e != 0 {
            return nil, error(e)
        } else {
            return nil, syscall.EINVAL
        }
    }
    return uuid[:], nil
}

Answer 6

As part of the uuid spec, if you generate a uuid from random it must contain a "4" as the 13th character and a "8", "9", "a", or "b" in the 17th (source).

// this makes sure that the 13th character is "4"
u[6] = (u[6] | 0x40) & 0x4F
// this makes sure that the 17th is "8", "9", "a", or "b"
u[8] = (u[8] | 0x80) & 0xBF