了解去通道僵局

Question

package main

import (
    "fmt"
    "time"
)

func main() {
    p := producer()
    for c := range p {
        fmt.Println(c)
    }
}

func producer() <-chan string {
    ch := make(chan string)
    go func() {
        for i := 0; i < 5; i++ {
            ch <- fmt.Sprint("hello", i)
            time.Sleep(1 * time.Second)
        }
        // commented the below to show the issue
        // close(ch)
    }()
    return ch
}

Running the above code will print 5 messages and then give a "all go routines are a sleep - deadlock error". I understand that if I close the channel the error is gone.

The thing I would like to understand is how does go runtime know that the code will be waiting infinitely on the channel and that there is nothing else that will be sending data into the channel.

Now if I add an additional go routine to the main() function.. it does not throw any error and keeps waiting on the channel.

go func() {
        for {
            time.Sleep(2 * time.Millisecond)
        }
    }()

So does this mean.. the go runtime is just looking for presence of a running go routine that could potentially send data into the channel and hence not throwing the deadlock error ?

Answer 1

The runtime panics with the "all go routines are a sleep - deadlock error" error when all goroutines are blocked on channel and mutex operations.

The sleeping goroutine does not block on one of these operations. There is no deadlock and therefore no panic.