douxiuyu2028 2013-02-27 15:14
浏览 45
已采纳

sync / atomic.once.go中的两个原子样式代码是否必要?

The code in sync/atomic.once.go is :

func (o *Once) Do(f func()) {
        if atomic.LoadUint32(&o.done) == 1 { //A
        //if o.done == 1 {
            return
        }
        // Slow-path.
        o.m.Lock()
        defer o.m.Unlock()
        if o.done == 0 {
            f()
            atomic.CompareAndSwapUint32(&o.done, 0, 1) //B
            //o.done = 1
        }
    }

I don't think the two 'atomic-style' code A,B above is necessary or useful. I think the lock is enough, and it could be ok if A,B are not atomic style. I must miss something, please be kind to tell me the purpose of the code A,B. Thank you.

  • 写回答

1条回答 默认 最新

  • dqqfuth6736 2013-02-27 15:29
    关注

    The original is correct. The reason is that the Go Memory Model says, that without synchronization (if o.done == 1) the changes to o.done might not be observed at all.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 用verilog实现tanh函数和softplus函数
  • ¥15 求京东批量付款能替代天诚
  • ¥15 slaris 系统断电后,重新开机后一直自动重启
  • ¥15 51寻迹小车定点寻迹
  • ¥15 谁能帮我看看这拒稿理由啥意思啊阿啊
  • ¥15 关于vue2中methods使用call修改this指向的问题
  • ¥15 idea自动补全键位冲突
  • ¥15 请教一下写代码,代码好难
  • ¥15 iis10中如何阻止别人网站重定向到我的网站
  • ¥15 滑块验证码移动速度不一致问题