go routine for range over channels

Question

I have been working in Golang for a long time. But still I am facing this problem though I know the solution to my problem. But never figured out why is it happening.

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Answer 1

The reason of the deadlock is because the main is waiting for the sq return and finish, but the sq is waiting for someone read the chan then it can continue.

I simplified your code by removing layer of sq call, and split one sentence into 2 :

func main() {
    result := sq(gen(3, 4)) // <-- block here, because sq doesn't return
    for n := range result { 
        fmt.Println(n)
    }
    fmt.Println("Process completed")
}

func gen(nums ...int) <-chan int {
    out := make(chan int)
    go func() {
        for _, n := range nums {
            out <- n
        }
        close(out)
    }()
    return out
}

func sq(in <-chan int) <-chan int {
    out := make(chan int)
    for n := range in {
        out <- n * n   // <-- block here, because no one is reading from the chan
    }
    close(out)
    return out
}

In sq method, if you put code in goroutine, then the sq will returned, and main func will not block, and consume the result queue, and the goroutine will continue, then there is no block any more.

func main() {
    result := sq(gen(3, 4)) // will not blcok here, because the sq just start a goroutine and return
    for n := range result {
        fmt.Println(n)
    }
    fmt.Println("Process completed")
}

func gen(nums ...int) <-chan int {
    out := make(chan int)
    go func() {
        for _, n := range nums {
            out <- n
        }
        close(out)
    }()
    return out
}

func sq(in <-chan int) <-chan int {
    out := make(chan int)
    go func() {
        for n := range in {
            out <- n * n // will not block here, because main will continue and read the out chan
        }
        close(out)
    }()
    return out
}