linux文件删除原理

1 sync.Once说明

sync.Once用于执行只需执行一次的函数功能。

2 例子

func main() {
    o sync.Once

    for {
        o.Do(func() {
            fmt.Println("sync once")
        })

        time.Sleep(1 * time.Second)
        fmt.Println("not once")
    }
}

3 源码数据结构

type Once struct {
    m    Mutex
    done uint32
}

done uint32，标记位，若已执行过，置为1；
m Mutex，锁

4 源码实现

func (o *Once) Do(f func()) {
    if atomic.LoadUint32(&o.done) == 1 {
        return
    }
    // Slow-path.
    o.m.Lock()
    defer o.m.Unlock()
    if o.done == 0 {
        defer atomic.StoreUint32(&o.done, 1)
        f()
    }
}

• if atomic.LoadUint32(&o.done) == 1是if o.done == 1的线程安全写法，如果o.done等于0才会执行后续逻辑，atomic.LoadUint32是原子变量取值函数。
• defer atomic.StoreUint32(&o.done, 1)是o.done = 1的线程安全写法，当然取值的时候也要用atomic.LoadUint32才有效果。如果写的时候用atomic.StoreUint32，取值的时候直接用if o.done == 1，则无法保证原子性。

5 错误实现

Note: Here is an incorrect implementation of Do:

if atomic.CompareAndSwapUint32(&o.done, 0, 1) {
    f()
}

Do guarantees that when it returns, f has finished. This implementation would not implement that guarantee:
given two simultaneous calls, the winner of the cas would call f, and the second would return immediately, without waiting for the first's call to f to complete. This is why the slow-path falls back to a mutex, and why the atomic.StoreUint32 must be delayed until after f returns. 此时如果第二个直接返回的协程立即访问到尚未初始化完成的变量(第一个协程调用f可能还未初始化完成)，则会出问题。

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