IMHO, this is so that the sender of ths signal to your channel doesn't gets blocked.
A helpful blog to understand it: https://www.rudderstack.com/blog/implementing-graceful-shutdown-in-go/ Hope it helps! On Thursday 30 July 2015 at 02:42:50 UTC+5:30 Bakul Shah wrote: > On Wed, 29 Jul 2015 01:39:50 PDT Yesudeep Mangalapilly <yesu...@gmail.com> > wrote: > > --- code --- > > package main > > > > import "fmt" > > import "time" > > > > func worker(done chan bool) { > > time.Sleep(time.Second) > > done <- true > > } > > > > func main() { > > done := make(chan bool, 1) // What is the significance of using 1 as > > capacity here? > > go worker(done) > > <-done > > } > > --- end code --- > > > > Why use 1 to create a buffered channel? Can we not signal "done" using a > > regular non-buffered channel? > > In general, synchronize only when you have to. Here the main > thread wants to know when the worker thread terminates but the > worker thread doesn't care when the main thread gets around to > reading from "done". Using a 1 deep buffer channel exactly > captures this usage pattern. An unbuffered channel would make > the worker thread "rendezvous" with the main thread, which is > unnecessary. > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/golang-nuts/6ef0468a-fd7d-4bf8-833b-a7696512aa4dn%40googlegroups.com.
