Channel Buffering

Posted by yhuang
Public (Editable by Users)

By default Go channels are unbuffered, meaning that they will only accept sends (chan <-) if there is a corresponding receive (<- chan) ready to receive the sent value. Buffered channels accept a limited number of values without a corresponding receiver for those values.

Go
Edit 
package main

import "fmt"

func main() {

    messages := make(chan string, 2)

    messages <- "buffered"
    messages <- "channel"

    fmt.Println(<-messages)
    fmt.Println(<-messages)
}