在Golang中只有一条消息被其他地方的频道阻塞？

Question

I'm trying to create a channel which used to make sure that everything is ready,

so I can continue with the process, an example would be like this: <kbd>playground</kbd>

package main

import (
    "fmt"
)

// done sends the channel a "okay" status.
func done(ok chan<- bool) {
    ok <- true
}

// waiting is a function that waiting for everything's okay.
func waiting(ok <-chan bool) {
    <-ok

    // Do something here
    // when everything's okay...
}

func main() {
    ok := make(chan bool)

    // Send the "ok" status once.
    go done(ok)

    // function A mockup
    waiting(ok)
    // function B mockup
    waiting(ok)

    fmt.Println("All Ok!")
}

and here's the output:

fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan receive]:
main.main()
    /tmp/sandbox709143808/main.go:29 +0xc0

I'm excepted to send the ok <- true once,

then I can use it in multiple places, and get the output like this:

All Ok!

Program exited.

but I'm not sure how to do that, any idea?

Answer 1

You may close the channel isstead of sending a message .Closing will act as if the ok is broadcasted to all listening froutines

Code

package main

import (
    "fmt"
)

// done sends the channel a "okay" status.
func done(ok chan<- bool) {
    close(ok)
}

// waiting is a function that waits for everything's okay.
func waiting(ok <-chan bool) {
    <-ok

    // Do something here
    // when everything's okay...
}

func main() {
    ok := make(chan bool)

    // Send the "ok" status once.
    go done(ok)
    //go done(ok)

    // function A mockup
    waiting(ok)
    // function B mockup
    waiting(ok)

    fmt.Println("All Ok!")
}

Here is the play link play.golang