入门教程：频道和缓冲频道

Question

Why does not the channel c buffered out when the second value is sent via another go routine and the first value sent isn't received?

package main
import "fmt"

func sum(s []int, c chan int) {
    sum := 0
    for _, v := range s {
        sum += v
    }
    c <- sum // send sum to c   
}

func main() {
    s := []int{7, 2, 8, -9, 4, 0}
    c := make(chan int)
    go sum(s[:len(s)/2], c)
    go sum(s[len(s)/2:], c)
    x, y :=  <-c ,<-c// receive from c
    fmt.Println(x,y ,x+y)
}

What I was expecting is an error-

fatal error: all goroutines are asleep - deadlock!

This occurs when there is a block when the buffer is full. Since channel c has size 1, sending the second value should have given the above error.

What is happening in above code ?

Answer 1

Just because a write can’t succeed immediately, you won’t get the “deadlock” error so long as there’s some other goroutine that can run.

Let’s imagine a scheduling model where a go function immediately starts the goroutine and makes as much forward progress as it can before yielding to someone else. Then this will happen:

1. The program will call sum() for the first half of the list, calculate the sum, and try to write it to the channel, but since there is no listener, it will block.
2. The program will call sum() for the second half of the list, calculate the sum, and try to write it to the channel, but since there is no listener, it will block.
3. main() will try to read from the channel, wakes up one of the previous goroutines, and gets the value from it.
4. main() will try to read from the channel, wakes up the other blocked goroutine, and gets the value from it.
5. Nobody’s blocked on channel input or output any more and all of the goroutines (including main()) can run to completion.

You can do the same exercise if you pretend go just schedules something in the background and continues running the main goroutine. The important thing is that as soon as there’s a paired read and write on the same channel, both will proceed.