goroutine

问题 From: http://blog.nindalf.com/how-goroutines-work/ As the goroutines are scheduled cooperatively, a goroutine that loops continuously can starve other goroutines on the same thread. Goroutines are cheap and do not cause the thread on which they are multiplexed to block if they are blocked on network input sleeping channel operations or blocking on primitives in the sync package. So given the above, say that you have some code like this that does nothing but loop a random number of times and

问题 I want to stop goroutine execution on timeout. But it seems like it is not working for me. I am using iris framework. type Response struct { data interface{} status bool } func (s *CicService) Find() (interface{}, bool) { ch := make(chan Response, 1) go func() { time.Sleep(10 * time.Second) fmt.Println("test") fmt.Println("test1") ch <- Response{data: "data", status: true} }() select { case <-ch: fmt.Println("Read from ch") res := <-ch return res.data, res.status case <-time.After(50 * time

问题 I have the following setup: func startsMain (){ go main () } fun stopMain (){ //kill main } func main() { //infinite loop } I am creating cucumber steps and I need to be able to start and shut down the application. 回答1: You can kill you infinite loop using select and channels! var quit chan struct{} func startLoop() { quit := make(chan struct{}) go loop() } func stopLoop() { // As mentioned by Kaedys //close(quit) // permits signalling everyone havins such a `case <-quit:` // statement to be

问题 To settle some misunderstandings I have about goroutines, I went to the Go playground and ran this code: package main import ( "fmt" ) func other(done chan bool) { done <- true go func() { for { fmt.Println("Here") } }() } func main() { fmt.Println("Hello, playground") done := make(chan bool) go other(done) <-done fmt.Println("Finished.") } As I expected, Go playground came back with an error: Process took too long . This seems to imply that the goroutine created within other runs forever.

问题 I wonder what is the most down to earth explanation of this famous quote: Don't communicate by sharing memory; share memory by communicating. (R. Pike) In The Go Memory Model I can read this: A send on a channel happens before the corresponding receive from that channel completes. (Golang Spec) There is also a dedicated golang article explaining the quote. And key contribution is a working example also by Andrew G. Well. Sometimes too much talking around .... I have derived from the Memory

问题 I'm writing software in Go that does a lot of parallel computing. I want to collect data from worker threads and I'm not really sure how to do it in a safe way. I know that I could use channels but in my scenario they make it more complicated since I have to somehow synchronize messages (wait until every thread sent something) in the main thread. Scenario The main thread creates n Worker instances and launches their work() method in a goroutine so that the workers each run in their own thread

问题 I've been trying to solve this simple problem I encountered in Golang concurrency. I've been searching all possible solutions, but found nothing specific to my problem(or I might be missed one). Here's my code: package main import ( "fmt" "time" ) func producer(ch chan int, d time.Duration, num int) { for i:=0; i<num; i++ { ch <- i time.Sleep(d) } } func main() { ch := make(chan int) go producer(ch, 100*time.Millisecond, 2) go producer(ch, 200*time.Millisecond, 5) for { fmt.Println(<-ch) }

问题 As i learned from golang docs, if i set runtime.GOMAXPROCS(8) with a cpu of 8 cores (intel i7), then start a goroutine of infinite-loop, other gorutines should not be blocked because there are engough threads and goprocs. But this is not true when using net/http package, an infinite-loop goroutine will block http server after a few invocations. Can anyone help to explain why ? If i comment the line of "go infinite loop", start client after server, client will output 1000 asterisks; but if i

问题 How can other goroutines keep executing whilst invoking a syscall? (when using GOMAXPROCS=1) As far as I'm aware of, when invoking a syscall the thread gives up control until the syscall returns. How can Go achieve this concurrency without creating a system thread per blocking-on-syscall goroutine? From the documentation: Goroutines They're called goroutines because the existing terms—threads, coroutines, processes, and so on—convey inaccurate connotations. A goroutine has a simple model: it

问题 I am trying to print json result from struct I created as following: type Machine struct { m_ip string m_type string m_serial string } and print out m:= &Machine{ m_ip:"test", m_type:"test", m_serial:"test" } m_json:= json.Marshal(m) fmt.Println(m_json) However, result returned just {} Secondly,I tried to changed the first letter of words to Uppercase as follow: type Machine struct{ MachIp string MachType string MachSerial string } and it works! Why doesn't the word with lowercase character