具有不同线程安全模式的System.Lazy <T>

Question

.NET 4.0的System.Lazy <T>类通过枚举LazyThreadSafetyMode提供了三种线程安全模式，我将其总结为：

• LazyThreadSafetyMode.None - 不是线程安全的。

• LazyThreadSafetyMode.ExecutionAndPublication - 只有一个线程同时将尝试创造潜在价值。创建成功后，所有等待线程将获得相同的值。如果在创建过程中发生未处理的异常，则将在每个等待的线程上将其重新抛出，在每次后续访问基础值的尝试中将对其进行缓存和重新抛出。

• LazyThreadSafetyMode.PublicationOnly - 多个并发线程将尝试创造潜在价值，但第一个成功将决定传递给所有线程值。如果在创建过程中发生未处理的异常，则不会对其进行缓存，并且并发访问和随后尝试访问基础值的尝试将重试创建，并且可能会成功。

我想要一个延迟初始化的值，该值遵循略有不同的线程安全规则，即：

只有一个并发线程将尝试创建基础值。创建成功后，所有等待线程将获得相同的值。如果在创建过程中发生未处理的异常，它将在每个等待的线程上重新抛出，但不会被缓存，并且随后尝试访问基础值的尝试将重试创建，并且可能会成功。

因此，与LazyThreadSafetyMode.ExecutionAndPublication的主要区别在于，如果创建时“先行尝试”失败，则可以在以后重新尝试。

是否存在提供这些语义的现有（.NET 4.0）类，还是我必须自己滚动？如果我自己动手，是否存在一种聪明的方法来重用实现中的现有Lazy <T>以避免显式的锁定/同步？

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注意：对于一个用例，想象一下“创建”可能是昂贵的并且容易出现间歇性错误，例如涉及从远程服务器获取大量数据。我不想进行多次并发尝试来获取数据，因为它们很可能全部失败或全部成功。但是，如果它们失败了，我希望稍后再试。

Answer 1

我尝试的是达林更新后的答案版本，但没有我指出的比赛条件...警告，我不确定这最终是否完全摆脱了比赛条件。private static int waiters = 0;private static volatile Lazy<object> lazy = new Lazy<object>(GetValueFromSomewhere);public static object Value{&nbsp; &nbsp; get&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; Lazy<object> currLazy = lazy;&nbsp; &nbsp; &nbsp; &nbsp; if (currLazy.IsValueCreated)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return currLazy.Value;&nbsp; &nbsp; &nbsp; &nbsp; Interlocked.Increment(ref waiters);&nbsp; &nbsp; &nbsp; &nbsp; try&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return lazy.Value;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // just leave "waiters" at whatever it is... no harm in it.&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; catch&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (Interlocked.Decrement(ref waiters) == 0)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lazy = new Lazy<object>(GetValueFromSomewhere);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; throw;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}更新：我以为发布此消息后发现了比赛情况。该行为实际上应该是可以接受的，只要您对一个罕见的情况感到满意，即Lazy<T>在另一个线程已经从成功的快速返回之后，某个线程抛出了一个从慢速观察到的异常Lazy<T>（将来的请求将全部成功）。waiters = 0t1：一直运行到Interlocked.Decrement（waiters= 1）之前t2：进入并运行到Interlocked.Increment（waiters= 1）之前t1：进行Interlocked.Decrement并准备覆盖（waiters= 0）t2：一直运行到Interlocked.Decrement（waiters= 1）之前t1：lazy用新的覆盖（称为lazy1）（waiters= 1）t3：进入并在lazy1（waiters= 2）处阻止t2：是否执行Interlocked.Decrement（waiters= 1）t3：从lazy1（waiters现在不相关）获取并返回值t2：抛出异常我无法提出一系列导致比“该线程在另一个线程产生成功结果之后引发异常”更糟糕的事情的事件。Update2：声明lazy为volatile确保所有读者立即都能看到受保护的覆盖。有些人（包括我自己在内）看到volatile并立即想到“好吧，可能是使用不正确”，他们通常是正确的。这就是我在这里使用它的原因：在上面示例中的事件序列中，t3仍然可以读取旧的，lazy而不是lazy1如果它位于lazy.Valuet1修改lazy为包含的那一刻之前lazy1。 volatile防止这种情况，以便下次尝试可以立即开始。我还提醒自己，为什么我脑子里有这样的话：“低锁并发编程很难，只需使用C＃lock语句！！！” 我一直在写原始答案。Update3：只是更改了Update2中的一些文本，指出了volatile必要的实际情况- Interlocked此处使用的操作显然是在当今重要的CPU架构上全屏实现的，而不是我最初只是假设的半屏实现的，因此volatile保护的范围比我原来想象的要窄得多。

Answer 2

只有一个并发线程将尝试创建基础值。创建成功后，所有等待线程将获得相同的值。如果在创建过程中发生未处理的异常，它将在每个等待的线程上重新抛出，但不会被缓存，并且随后尝试访问基础值的尝试将重试创建，并且可能会成功。由于Lazy不支持该功能，因此您可以尝试自行滚动：private static object syncRoot = new object();private static object value = null;public static object Value{&nbsp; &nbsp; get&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; if (value == null)&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lock (syncRoot)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (value == null)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // Only one concurrent thread will attempt to create the underlying value.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // And if `GetTheValueFromSomewhere` throws an exception, then the value field&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // will not be assigned to anything and later access&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // to the Value property will retry. As far as the exception&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // is concerned it will obviously be propagated&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // to the consumer of the Value getter&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; value = GetTheValueFromSomewhere();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return value;&nbsp; &nbsp; }}更新：为了满足您对传播到所有等待读取器线程的相同异常的要求：private static Lazy<object> lazy = new Lazy<object>(GetTheValueFromSomewhere);public static object Value{&nbsp; &nbsp; get&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; try&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return lazy.Value;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; catch&nbsp; &nbsp; &nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // We recreate the lazy field so that subsequent readers&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // don't just get a cached exception but rather attempt&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // to call the GetTheValueFromSomewhere() expensive method&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // in order to calculate the value again&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lazy = new Lazy<object>(GetTheValueFromSomewhere);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // Re-throw the exception so that all blocked reader threads&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // will get this exact same exception thrown.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; throw;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}