Golang中的取消模式

Question

Here is a quote from 50 Shades Of Go: Traps, Gotchas and Common mistakes:

You can also use a special cancellation channel to interrupt the workers.

func First(query string, replicas ...Search) Result {  
    c := make(chan Result)
    done := make(chan struct{})
    defer close(done)
    searchReplica := func(i int) { 
        select {
        case c <- replicas[i](query):
        case <- done:
        }
    }
    for i := range replicas {
        go searchReplica(i)
    }

    return <-c
}

As far as understand, it means that we use channel done to interrupt the workers ahead of time without waiting for full execution (in our case execution of replicas[i](query). Therefore, we can receive a result from the fastest worker ("First Wins Pattern") and then cancel the work in all other workers and save the resources.

On the other hand, according to the specification:

For all the cases in the statement, the channel operands of receive operations and the channel and right-hand-side expressions of send statements are evaluated exactly once, in source order, upon entering the "select" statement.

As far as I understand, it means we cannot interrupt the workers, as in any case, all workers will evaluate function replicas[i]query and only then select case <- done and finish their execution.

Could you please point out at the mistake in my reasoning?

Answer 1

Your reasoning is correct, the wording on the site is not completely clear. What this "construct" achieves is that the goroutines will not be left hanging forever, but once the searches finish, the goroutines will end properly. Nothing more is happening there.

In general, you can't interrupt any goroutine from the outside, the goroutine itself has to support some kind of termination (e.g. shutdown channel, context.Context etc.). See cancel a blocking operation in Go.

So yes, in the example you posted, all searches will be launched, concurrently, result of the fastest one will be returned as it arrives, the rest of the goroutines will continue to run as long as their search is finished.

What happens to the rest? The rest will be discarded (case <- done will be chosen, as an unbuffered channel cannot hold any elements, and there will be no one else receiving more from the channel).

You can verify this in this Go Playground example.