go - 我们只能使用没有volatile的RWMutex？

Question

在https://golang.org/ref/mem#tmp_10 , 该程序如下所示不安全，无法保证打印最新消息

type T struct {
    msg string
}

var g *T

func setup() {
    t := new(T)
    t.msg = "hello, world"
    g = t
}

func main() {
    go setup()
    for g == nil {
    }
    print(g.msg)
}

在 JAVA 中，volatile g 可以吗，我们必须使用 rwmutex 来保持在 golang 中打印最新的消息，如下所示？

type T struct {
    msg    string
    rwlock sync.RWMutex
}

var g = &T{}

func setup() {
    g.rwlock.Lock()
    defer g.rwlock.Unlock()
    g.msg = "hello, world"
}

func main() {
    go setup()
    printMsg()
}

func printMsg() {
    g.rwlock.RLock()
    defer g.rwlock.RUnlock()
    print(g.msg)
}

Answer 1

这里有一些其他选项。

忙等等。这个程序在当前版本的 Go 中完成，但规范不保证它。 Run it on the playground .

package main

import (
    "runtime"
    "sync/atomic"
    "unsafe"
)

type T struct {
    msg string
}

var g unsafe.Pointer

func setup() {
    t := new(T)
    t.msg = "hello, world"
    atomic.StorePointer(&g, unsafe.Pointer(t))
}

func main() {
    go setup()
    var t *T
    for {
        runtime.Gosched()

        t = (*T)(atomic.LoadPointer(&g))
        if t != nil {
            break
        }
    }
    print(t.msg)
}

channel 。 Run it on the playground .

func setup(ch chan struct{}) {
    t := new(T)
    t.msg = "hello, world"
    g = t
    close(ch) // signal that the value is set
}

func main() {
    var ch = make(chan struct{})
    go setup(ch)
    <-ch // wait for the value to be set.
    print(g.msg)
}

WaitGroup . Run it on the playground .

var g *T

func setup(wg *sync.WaitGroup) {
    t := new(T)
    t.msg = "hello, world"
    g = t
    wg.Done()
}

func main() {
    var wg sync.WaitGroup
    wg.Add(1)
    go setup(&wg)
    wg.Wait()
    print(g.msg)
}