function with goroutine returned will kill its goroutine?

[Kenshin Wang]

package main

import "fmt"

func main() {

ca := createCounter(2)

cb := createCounter(102)

for i := 0; i < 5; i++ {

a := <-ca

fmt.Printf("A %d \nB %d \n", a, <-cb)

}

}

func createCounter(start int) chan int {

next := make(chan int)

go func(i int) {

for {

next <- i

i++

}

}(start)

return next

}

the output is:

A 2

B 102

A 3

B 103

A 4

B 104

A 5

B 105

A 6

B 106

 

could anyone can explain why the output is this?

especially, i don't understand when the goroutine will be destroyed,i think when the createCounter() function returned the goroutine will be killed, but it seems not ...

thank you

 

[Jan Mercl]

On Mon, Jul 4, 2016 at 5:08 PM Kenshin Wang <kenshi...@gmail.com> wrote:

        go func(i int) {
                for {

                        next <- i

                        i++

                }

        }(start)

> could anyone can explain why the output is this?

A goroutine dies when it returns. A goroutine with an endless loop is immortal (modulo unhandled panics).

--

-j

[Christoph Berger]

In addition to that, all goroutines exit when the main function exits. In your code, this happens when i == 5. This explains the output that you get. Both goroutines are able to produce five numbers until the main loop finishes and the main function exits.

[Paul Borman]

You probably should say all goroutines are terminated when main exits (or os.Exit is called or the program abnormally terminates), lest someone complain that their defer functions were not run.

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[Christoph Berger]

Good point, thanks. Yes, when the process exits, the goroutines end immediately without running their defer functions.