For one of my requirement I have to create N number of worker go routines, which will be monitored by one monitoring routine. monitoring routine has to end when all worker routines completes. My code ending in deadlock, please help.
import "fmt" import "sync" import "strconv" func worker(wg *sync.WaitGroup, cs chan string, i int ){ defer wg.Done() cs<-"worker"+strconv.Itoa(i) } func monitorWorker(wg *sync.WaitGroup, cs chan string) { defer wg.Done() for i:= range cs { fmt.Println(i) } } func main() { wg := &sync.WaitGroup{} cs := make(chan string) for i:=0;i<10;i++{ wg.Add(1) go worker(wg,cs,i) } wg.Add(1) go monitorWorker(wg,cs) wg.Wait() }
The WaitGroup type of sync package, is used to wait for the program to finish all goroutines launched from the main function. It uses a counter that specifies the number of goroutines, and Wait blocks the execution of the program until the WaitGroup counter is zero.
A goroutine is a lightweight thread managed by the Go runtime. go f(x, y, z) starts a new goroutine running f(x, y, z) The evaluation of f , x , y , and z happens in the current goroutine and the execution of f happens in the new goroutine.
yourbasic.org/golang. A deadlock happens when a group of goroutines are waiting for each other and none of them is able to proceed.
Your monitorWorker never dies. When all the workers finish, it continues to wait on cs. This deadlocks because nothing else will ever send on cs and therefore wg will never reach 0. A possible fix is to have the monitor close the channel when all workers finish. If the for loop is in main, it will end the loop, return from main, and end the program.
For example: http://play.golang.org/p/nai7XtTMfr
package main import ( "fmt" "strconv" "sync" ) func worker(wg *sync.WaitGroup, cs chan string, i int) { defer wg.Done() cs <- "worker" + strconv.Itoa(i) } func monitorWorker(wg *sync.WaitGroup, cs chan string) { wg.Wait() close(cs) } func main() { wg := &sync.WaitGroup{} cs := make(chan string) for i := 0; i < 10; i++ { wg.Add(1) go worker(wg, cs, i) } go monitorWorker(wg, cs) for i := range cs { fmt.Println(i) } }
Edit: This is an answer to OP's first comment.
Your program has three parts that need to synchronize. First, all of your workers need to send the data. Then your print loop needs to print that data. Then your main function needs to return thereby ending the program. In your example, all the workers send the data, all the data gets printed, but the message is never sent to main that it should return gracefully.
In my example, main does the printing and "monitorWorker" just tells main when it has received every piece of data it needs to print. This way the program ends gracefully and not by deadlock.
If you insist on the print loop being in another goroutine, you can do that. But then an extra communication needs to be sent to main so it returns. In this next example, I use a channel to ensure main ends when all data is printed.
package main import ( "fmt" "strconv" "sync" ) func worker(wg *sync.WaitGroup, cs chan string, i int) { defer wg.Done() cs <- "worker" + strconv.Itoa(i) } func monitorWorker(wg *sync.WaitGroup, cs chan string) { wg.Wait() close(cs) } func printWorker(cs <-chan string, done chan<- bool) { for i := range cs { fmt.Println(i) } done <- true } func main() { wg := &sync.WaitGroup{} cs := make(chan string) for i := 0; i < 10; i++ { wg.Add(1) go worker(wg, cs, i) } go monitorWorker(wg, cs) done := make(chan bool, 1) go printWorker(cs, done) <-done }
When I changed my channel definition from
strChan := make(chan string)
to
strChan := make(chan string, 1)
I was able to fix this error
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