去通道容量，为什么要比我指定的多一个元素

Question

package main
import "fmt"
import "time"

func main() {
     message := make(chan string ,1) // no buffer
     count := 3

     go func() {
          for i := 1; i <= count; i++ {
               fmt.Println("send message")
               message <- fmt.Sprintf("message %d", i)
          }
     }()

     time.Sleep(time.Second * 3)

     for i := 1; i <= count; i++ {
          fmt.Println(<-message)
     }
}

The output is

send message
send message  [wait for 3 sec]
message 1
send message
message 2
message 3

If I change message := make(chan string ,1) // no buffer to

message := make(chan string ,2) // no buffer

I got

send message
send message
send message [wait 3 sec]
message 1
message 2
message 3

Why 2 buffers channel can store 3 string objects? not 2?

thanks,

Answer 1

It works like this because the buffer holds N messages before it blocks. When the N+1 messages comes in GO will see that it exceeds the capacity you specified and will have to block, waiting for something to take from that channel. When passing a buffer size the sender will always block on N+1 message.

So for instance with size 2 you have an empty buffer:

[][]

Then a message comes in, is put in the buffer:

[m1][]

Then another one, we can keep on going because we have space in the buffer

[m1][m2]

Then another comes in, ups we don't have more space in the buffer so we block

[m1][m2]m3 -> block

Something like this.

The size is basically the number of messages that can be sent to the buffer without blocking.

For the future I recommend http://golang.org/doc/effective_go.html#channels

var sem = make(chan int, MaxOutstanding)

Once MaxOutstanding handlers are executing process, any more will block trying to send into the filled channel buffer, until one of the existing handlers finishes and receives from the buffer.