channel | 易学教程

Do buffered channels maintain order?

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问题 In Go, do buffered channels have any order guarantee? For example: you have two goroutines A & B that share a channel. A pushes data onto the channel while B reads from it. Are you guranted that B will read data In the same order that A put it into the channel? I understand that if there are multiple producers or consumers the order may be non-deterministic, but I'm specifically asking about having just 1 producer and 1 consumer. 回答1: You can see the idea of channel illustrated in "The Nature

How to collect values from N goroutines executed in a specific order?

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Below is a struct of type Stuff. It has three ints. A Number , its Double and its Power . Let's pretend that calculating the double and power of a given list of ints is an expensive computation. type Stuff struct { Number int Double int Power int } func main() { nums := []int{2, 3, 4} // given numbers stuff := []Stuff{} // struct of stuff with transformed ints double := make(chan int) power := make(chan int) for _, i := range nums { go doubleNumber(i, double) go powerNumber(i, power) } // How do I get the values back in the right order? fmt.Println(stuff) } func doubleNumber(i int, c chan int)

throw: all goroutines are asleep - deadlock

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问题 Given the following simple Go program package main import ( "fmt" ) func total(ch chan int) { res := 0 for iter := range ch { res += iter } ch <- res } func main() { ch := make(chan int) go total(ch) ch <- 1 ch <- 2 ch <- 3 fmt.Println("Total is ", <-ch) } I am wondering if someone can enlighten me as to why I get throw: all goroutines are asleep - deadlock! thank you 回答1: As you never close the ch channel, the range loop will never finish. You can't send back the result on the same channel.

RabbitMQ and relationship between channel and connection

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The RabbitMQ Java client has the following concepts: Connection - a connection to a RabbitMQ server instance Channel - ??? Consumer thread pool - a pool of threads that consume messages off the RabbitMQ server queues Queue - a structure that holds messages in FIFO order I'm trying to understand the relationship, and more importantly , the associations between them. I'm still not quite sure what a Channel is, other than the fact that this is the structure that you publish and consume from, and that it is created from an open connection. If someone could explain to me what the "Channel"

Is it OK to leave a channel open?

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问题 Is it OK to leave a Go channel open forever (never close the channel) if I never check for its state? Will it lead to memory leaks? Is the following code OK? func (requestCh chan<- Request) GetResponse(data RequestData) Response { reply := make(chan Response) requestCh <- Request{data: data, replyCh: reply} return <-reply } 回答1: It's OK to leave a Go channel open forever and never close it. When the channel is no longer used, it will be garbage collected. Note that it is only necessary to

How does select work when multiple channels are involved?

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问题 I found when using select on multiple non buffered channels like select { case <- chana: case <- chanb: } Even when both channels have data, but when processing this select, the call that falls in case chana and case chanb is not balanced. package main import ( "fmt" _ "net/http/pprof" "sync" "time" ) func main() { chana := make(chan int) chanb := make(chan int) go func() { for i := 0; i < 1000; i++ { chana <- 100 * i } }() go func() { for i := 0; i < 1000; i++ { chanb <- i } }() time.Sleep

Message channels one or many?

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I need to handle emails from about 30 addresses. I implement this in a way where all emails going to one DirectChannel and after to Receiver . In Receiver I can understand from what address is message comes, to do this I create CustomMessageSource that wraps javax.mail.Message to my own type that contains javax.mail.Message and some Enum . Looks like this is not a good decision, cause I can use @Transformer , but how can I use it if I have only 1 channel? That was the first question. Second question: Should I use ONE channel and ONE receiver for all that addresses? Or better to have channel

anonymous struct and empty struct

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问题 http://play.golang.org/p/vhaKi5uVmm package main import "fmt" var battle = make(chan string) func warrior(name string, done chan struct{}) { select { case opponent := <-battle: fmt.Printf("%s beat %s\n", name, opponent) case battle <- name: // I lost :-( } done <- struct{}{} } func main() { done := make(chan struct{}) langs := []string{"Go", "C", "C++", "Java", "Perl", "Python"} for _, l := range langs { go warrior(l, done) } for _ = range langs { <-done } } [1st Question] done <- struct{}{}

How do I use static lifetimes with threads?

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问题 I'm currently struggling with lifetimes in Rust (1.0), especially when it comes to passing structs via channels. How would I get this simple example to compile: use std::sync::mpsc::{Receiver, Sender}; use std::sync::mpsc; use std::thread::spawn; use std::io; use std::io::prelude::*; struct Message<'a> { text: &'a str, } fn main() { let (tx, rx): (Sender<Message>, Receiver<Message>) = mpsc::channel(); let _handle_receive = spawn(move || { for message in rx.iter() { println!("{}", message.text

How does a non initialized channel behave?

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问题 I have a struct that holds a channel that is not initialized. When I write into it, the routine is blocking as expected but the reader is never notified that something is in the pipe. I am surprised that there is no error and I am wondering what is doing Go. In the example bellow, neither the message pushed nor got it are printed. (Uncomment the intialization and it will work like a charm) type MyStruct struct { over chan bool } func main() { nonInitialized := &MyStruct{} // nonInitialized