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Is message passing via channels in go guaranteed to be non-blocking?

In order to assess whether go is a possible option for an audio/video application, I would like to know whether message passing in go satisfies any non-blocking progress guarantees (being obstruction-free, lock-free or wait-free). In particular, the following scenarios are relevant:

Single producer single consumer:

Two threads communicate using a shared channel. Thread A only does asynchronous sends, thread B only does asynchronous receives. Suppose the OS scheduler decides to interrupt thread A at the "worst possible moment" for an indefinite amount of time. Is thread B guaranteed to finish a receive operation in a bounded number of cpu cycles or is there a (theoretical) possibility that thread A can put the channel into a state where thread B needs to wait for the OS to resume thread A?

Multiple producers:

Several threads A1, A2, A3, ... communicate with one or more others threads using a shared channel. The threads Ai only do asynchronous sends. Suppose A2, A3, ... are suspended by the OS scheduler at the "worst possible moment" for an indefinite amount of time. Is thread A1 still guaranteed to finish a send operation in a bounded number of cpu cycles? Suppose further that each thread only wants to do one send. If the program is run sufficiently long (with a "malicious" scheduler which potentially starves some threads or interrupts and resumes threads at the "worst possible moment"), is at least one send guaranteed to succeed?

I am not so much interested in typical scenarios here, but rather worst-case guarantees. See Non-blocking algorithm (Wikipedia) for more details on obstruction-, lock- and wait-free algorithms.

like image 845
Tobias Avatar asked Jul 17 '11 19:07

Tobias


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2 Answers

Normal sends and receives are blocking operations by definition. You can do a non-blocking send or receive by using a select statement:

select {
case ch <- msg:
default:
}

(Receiving is very similar; just replace the case statement.)

The send only takes place when there is room in the channel's buffer. Otherwise the default case runs. Note that internally a mutex is still used (if I'm reading the code correctly).

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Evan Shaw Avatar answered Nov 02 '22 18:11

Evan Shaw


The Go Memory Model doesn't require sends and receives to be non-blocking, and the current runtime implementation locks the channel for send and recv. This means, for instance, that it is possible to starve a sending or receiving go-routine if the OS-scheduler interrupts another thread running another go-routine which tries to send or receive on the same channel while it has already acquired the channel's lock.

So the answer is unfortunately no :(

(unless someone reimplement parts of the runtime using non-blocking algorithms).

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Tobias Avatar answered Nov 02 '22 18:11

Tobias