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

This situation caused of output channel of sq function is not buffered. So sq is waiting until next function will read from output, but if sq is not async, it will not happen (Playground link):

package main

import (
    "fmt"
    "sync"
)

var wg sync.WaitGroup

func main() {
    numsCh := gen(3, 4)
    sqCh := sq(numsCh) // if there is no sq in body - we are locked here until input channel will be closed
    result := sq(sqCh) // but if output channel is not buffered, so `sq` is locked, until next function will read from output channel

    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, 100)
    for n := range in {
        out <- n * n
    }
    close(out)
    return out
}