在golang中晃来晃去的goroutine

Question

In the following code

1. What happens to goroutine1?(at the end of program we have three goroutine goroutine1 without any functionality)
2. What happens to channel?(when we make a channel in loop it release the previous channel memory? close it? or something else?)

func main() {
    for i := 1; i <= 3; i += 1 {
        ch := make(chan int)

        // gorutine1
        go func() {
            time.Sleep(3 * time.Second)
            ch <- i
            fmt.Println("gorutine1 end")
        }()

        // gorutine2
        go func() {
            time.Sleep(1 * time.Second)
            ch <- i+1000
            fmt.Println("gorutine2 end")
        }()

        fmt.Println("loop", <-ch)
    }

    time.Sleep(10 * time.Second)
    fmt.Println("main end")
}

Run above code here

Answer 1

For i=1, the loop creates two goroutines, and starts waiting to read from the channel. goroutine2 writes first and terminates. The channel is read, and then i becomes 2. goroutine1 will wait forever, because nobody will read from the channel again. You create a new channel, and do the same thing. When everything is said and done, you have three instances of goroutine1 waiting to write to three different channels.