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嵌套在orElse中的STM事务中的验证

[英]Validation in STM transactions nested with orElse

This commentary page describes a lot of the fine details of STM in GHC, but I'd like clarity on a couple points. 该评论页面描述了GHC中STM的许多详细信息,但我想在几点上保持清晰。

First , is a nested transaction invalidated when variables accessed in the parent change? 首先 ,当父级中访问的变量发生更改时,嵌套事务是否无效?

For instance we have in thread A : 例如我们在线程A

takeTMVar a `orElse` takeTMVar b `orElse` takeTMVar c

Say that while A is executing the nested transaction takeTMVar b , another thread B does a putTMVar a () ; 假设当A执行嵌套事务takeTMVar b ,另一个线程B执行putTMVar a () ; can thread A successfully complete its nested transaction, or is it invalidated (this would strike me as wrong)? 线程A可以成功完成其嵌套事务,还是使其无效(这会使我感到错误)?

A second point which I think I understand but wouldn't mind reassurance on: in the case where the entire top-level transaction described above for A is retried and finally blocks, is it correct that A will be awoken when any of a , b , or c change? 我想理解的第二点是,但我不介意放心:在上述针对A所述的整个顶级交易被重试并最终阻塞的情况下,当ab任何a被唤醒时, A是否被唤醒是正确的吗? ,或c变化?

Finally as a bonus , do the semantics of the transaction above change if we (or library authors) change orElse to infixr ? 最后,作为奖励 ,做上述变化事务的语义,如果我们(或库的作者)改变orElseinfixr

I don't think "nested" is the right term to describe this. 我认为“嵌套”是不恰当的描述方式。 These are three alternate transactions; 这是三个备用交易; none is nested within another. 没有一个嵌套在另一个内部。 In particular, exactly one of the three is going to happen and be committed -- but which one happens is not deterministic. 特别是,这三种情况中的一种恰好会发生并被落实-但是发生的哪一种并不确定。 This one sentence should be enough to answer all three questions, but just to be sure, let's carefully say for each: 这句话足以回答所有三个问题,但是请确保为每个问题仔细说一遍:

  1. There's no guarantee. 没有保证。 Maybe takeTMVar b will complete and commit; 也许takeTMVar b将完成并提交; or maybe it will be pre-empted and takeTMVar a will be woken up and complete. 否则它将被抢占,而takeTMVar a将被唤醒并完成。 But they won't both complete, that's for sure. 但是他们不会全部完成,这是肯定的。

  2. Yes, that's correct: all three TMVar s can wake this thread up. 是的,这是正确的:所有三个TMVar都可以唤醒该线程。

  3. The semantics don't change: whenever several of them can commit, the left-most one will. 语义不会改变:只要其中几个可以提交,最左边的一个就会提交。 (In particular, the paper describing STM says, "The orElse function obeys useful laws: it is associative and has unit retry .".) (特别是描述STM的论文说,“ orElse函数遵循有用的定律:它是关联的,并且具有单元retry 。”。)

  4. (from your question in the comments) The semantics of STM on page 8 of the linked paper really does guarantee that the left-most successful transaction is the one that succeeds. (摘自您提出的问题)链接的论文第8页STM的语义确实保证了最成功的事务成功的事务。 So: if thread A is executing takeTMVar b (but has not yet committed) and thread B executes and commits a write to a , and nothing else happens afterwards , you can be sure that thread A will be restarted and return the newly written value from a . 所以:如果线程A正在执行takeTMVar b (但尚未提交)和线程B执行,并承诺一个写a ,并没有什么事情发生后 ,你可以肯定的是线程A将重新启动,并返回来自新写入的值a The "nothing else happens afterwards" part is important: the semantics makes a promise about what happens, but not about how the implementation achieves it; “此后什么也没有发生”部分很重要:语义对发生的事情作出了承诺,但对实现的实现方式不做承诺。 so if, say, another thread took from a immediately (so that the takeTMvar a is still going to retry ), a sufficiently clever implementation is allowed to notice this and not restart thread A from the beginning of the transaction. 所以,如果说,另一个线程从拿了a立即(以便takeTMvar a仍然在不断retry ),足够聪明的实现是允许注意到这一点,而不是重新启动线程A从交易的开始。

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