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I'm revisiting the STM chapter of Marlow's book. There, it is stated that:

When multiple threads block on an MVar, they are guaranteed to be woken up in FIFO order

However, the same can't be done on the STM case:

A transaction can block on an arbitrary condition, so the runtime doesn't know whether any individual transaction will be able to make progress after the TVar is changed; it must run the transaction to find out. Hence, when there are multiple transactions that might be unblocked, we have to run them all; after all, they might all be able to continue now.

What I don't get is why from this it follows that

Because the runtime has to run all the blocked transactions, there is no guarantee that threads will be unblocked in FIFO order ...

I'd expect that even though we have to run all the transactions in an STM block, we can still wake the threads up in a FIFO order. So I guess I'm missing some important details.

Damian Nadales
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    We can certainly _wake_ the threads in FIFO order, but it doesn't guarantee they will _complete_ in the same order. An earlier thread might block _again_, while a later one finishes. – Fyodor Soikin Dec 08 '21 at 10:33

2 Answers2

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The point of STM is to speculate: we try running all the transactions hoping that they do not conflict with one another (or perform a retry). When we do discover a conflict, we allow some transactions to commit, while making the conflicting ones to rollback.

We could run only one transaction, wait for it to complete or to block, and then run another one, and so on, but doing so would amount to use a single "global lock", making the computation completely sequential.

More technically, when threads are waiting on a MVar, those threads will progress on a very simple condition: the MVar becoming non empty. On wake up, a thread will take the value, making it empty. So, at most one thread can perform the take, and there's no point in waking more than one.

By constrast, when threads are waiting because of STM, the condition is much more complex. Suppose they are waiting because they previously performed a retry, so they are waiting for some previously read TVar to be changed. When that happens, we can't really know which thread will block again unless we re-run its transaction. Unlike MVar, it is now possible that waking them all up will cause all of them to complete without conflict, so we try doing just that. In doing so we hope for many (if not all) to complete, and prepare to rollback again for those that do not.

Consider this concrete STM example:

Thread 1:
read X
if X is zero, retry
compute f(X)
write Y = f(X)

Thread 2:
read X
if X is zero, retry
compute g(X)
write Z = g(X)

Assume at the beginning we have "X=Y=Z=0". We run both threads, but they retry. Then a third thread writes X=1. We can wake both, and there's no point in doing so in FIFO order, since both will complete. We can, and should, compute f(X) and g(X) in parallel.

Now, if thread 2 wrote on Y instead of Z at the end, there would be a conflict. In this case, running the transactions in sequence would be more efficient (since otherwise we compute f(X) or g(X) twice). However, in the general case, it's hard to predict if there will be a conflict or not. STM does not attempt to, leaving to the programmer to optimize their code.

chi
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When a bunch of threads are waiting to take from an MVar and someone fills it, only one of the waiting threads can run. That's the whole point. The one that wakes up will (presumably) wake up one more thread after it's done with its task and fills the MVar in turn. So having a well-specified order in which the waiters will get their turn makes sense.

TVars are not MVars. They don't have the empty/reserved vs full/available dichotomy. They're just values that can be accessed and/or changed by transactions. Threads don't "block waiting for the TVar"; they just read the current value for the TVar and then decide that they cannot make progress with that value (and so retry).

The runtime system knows which TVars the thread acessed, so it knows not to wake that thread up again unless at least one of the TVars has changed; with pure computation the thread's decision to retry or not should be a pure function of all of the TVars it read, so we know it can't make progress until one of them changed.

But when you have a bunch of threads waiting in retry that have all read the same TVar and it changes, the runtime system doesn't want to just wake one up. There's no reason to believe that only one of them will be able to make progress with the new value. STM transations are supposed to operate in parallel, optimistically. We want to wake them all up and let them try to run. If they conflict we'll find that out later, by the normal means of detecting contention between STM transactions; we can't tell that from the fact that they all wanted to read the same TVar.

We could potentially wake them all up at the same time and (if there are too many for the available compute units) try to make sure the scheduler starts running them in FIFO order. Maybe that's even done; I have no idea. But I would guess they are treated exactly the same as any other threads that have become available to be scheduled and no special action is taken. They're "supposed" to all be running in parallel, so it's not supposed to matter which one gets out of the gate first.

Ben
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