并发无法按我期望的那样工作

Question

I am trying to understand why my code does not work, so I managed to recreate the problem in a simpler example.

I expected this code to output the string "broadcasted", but it does not output anything.

package main

import (
    "fmt"
    "time"
)

type hub struct {
    handle    chan []byte
    broadcast chan []byte
}

func (h *hub) init() {
    for {
        select {
        case m := <-h.handle:
            handler(m)
        case _ = <-h.broadcast:
            fmt.Println("broadcasted")
        }
    }
}

var socketHub = hub{
    handle:    make(chan []byte),
    broadcast: make(chan []byte),
}

func main() {

    go socketHub.init()

    m := []byte("this is the message")
    socketHub.handle <- m

    time.Sleep(time.Second * 2)
}

func handler(m []byte) {
    // Do some stuff.
    socketHub.broadcast <- m
}

Why does this not work?

Answer 1

Your broadcast channel is unbuffered. This means that:

1. You send the message to the handle channel: main goroutine blocks until...
2. In the goroutine, the select case gets the message...
3. And calls (in the goroutine) handler, which sends the message to the broadcast channel, blocking until...

And that's it: your child goroutine is blocking, waiting on itself to pick the message. Concurrently, your main goroutine sleeps, then reaches the end of main, exits and terminates the program.

You can 'solve' it in several ways:

• Make your broadcast channel buffered: this way, goroutine sends the message, which succeeds immediately, and returns in the for loop, selecting it back and printing as intended
• Make your send a (new) goroutine, either in handler, or by calling go handler(m) in your loop
• Have two different goroutines listen to handler and broadcast

Which one you choose depends on the exact problem you try to solve: in this tiny example, it's hard to find a 'best' one.