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Question

I'm trying with Go channels and confused with below function example from go blog:

func gen(nums []int) <-chan int {
    out := make(chan int)
    go func() {
        for _, n := range nums {
            out <- n
        }
        close(out)
    }()
    fmt.Println("return statement is called ")
    return out
}

Main :

func main() {
    c := make(chan int)
    c = gen([]int{2, 3, 4, 5})

    // Consume the output.
    // Print 2,3,4,5
    fmt.Println(<-c)
    fmt.Println(<-c)
    fmt.Println(<-c)
    fmt.Println(<-c)
}

Complete Code: http://play.golang.org/p/Qh30wzo4m0

My Doubts:

1. My understanding was, once return is called the function will be terminated and the channel inside that function has no more life.

2. The return statement is called only once. But the content of the out channel is read many times. In this case what is the actual flow of execution?

(I'm new to concurrent programming.)

Answer 1

out := make(chan int)

This is not a buffered channel, which means the out <- n will block until someone somewhere reads that channel (the fmt.Println(<-c) calls)
(See also "do golang channels maintain order")

So the return at the end of the gen() function doesn't mean the literal go func() is terminated (since it is still waiting for readers to consume the content of the out channel).

But main function getting out channel as return from gen() function.
How it is possible to get it after gen() is terminated?

The fact that gen() terminates has no effect on its returned value (the out channel): the goal of "gen()" is to "generate" that out channel.

main can use out (as the returned value of gen()) long after gen() terminates.

The literal go func within gen() still runs, even if gen() is terminated.