I understand that usually, if I wish to access an out-of-scope variable from a Go routine, it is my responsibility to create a copy to be conceptually owned by the Go routine. Is this also true for channels, or are those exempt?
Effective Go #channels explains this with the words "it may seem odd to write req := req but it's a [sic] legal and idiomatic in Go to do this," referring to this code example:
var sem = make(chan int, MaxOutstanding)
// (other code, filling sem, defining process(..), etc., omitted)
func Serve(queue chan *Request) {
for req := range queue {
<-sem
req := req // Create new instance of req for the goroutine.
go func() {
process(req)
sem <- 1
}()
}
}
I happened to have very nearly replicated this example code in my own project (except that I am using a chan struct{} rather than chan int for my semaphore, and declare it local to the Serve func). Staring at it, I am wondering if accessing the same channel from multiple concurrent goroutines is really fine, or if something like sem := sem is called for.
