为什么我们需要在go例程中运行未缓冲的通道，但是缓冲通道无需go例程也可以工作？

Question

I was going through tutorials on gobyexample. I noticed the author has shown example of gochannel using go routine but in buggered go channel he is directly sending messages to channel .

I tried on my local system to run unbuffered channel without go routine but it is throwing "fatal error: all goroutines are asleep - deadlock!" but buffered channels are working fine without go routine

func channelDemo() {
    message := make(chan string)
    // go func() {
    //  message <- "Hello"
    // }()
    message <- "Hello"
    msg := <-message

    fmt.Println("msg", msg)
}
func channelBufferingDemo() {
    messages := make(chan string, 3)

    messages <- "Buffered"
    messages <- "channel"
    fmt.Println(<-messages)
    fmt.Println(<-messages)
}

Answer 1

A channel send will succeed only if the channel can accept the input, that is, either there's a listener for that channel, or there is available buffer in the channel. Otherwise goroutine will go to sleep until one of those become true: either someone starts listening, or someone reads from the channel and there is now buffer space.

With a channel with no buffer, the only way you can write to it is if someone is listening to it. If there is only one goroutine and if you write to the channel, all goroutines will be asleep.

With a channel with buffer size 3 and one goroutine, you can write to if 3 times without reading from it. The 4th write will put all goroutines to sleep.