channel

Is it possible to let function funcWithNonChanResult have the following interface: func funcWithNonChanResult() int { And if I want it to use function funcWithChanResult with the interface: func funcWithChanResult() chan int { In other words, can I somehow convert chan int to int ? Or I must have chan int result type in all the functions which use funcWithChanResult ? Currently, I tried these methods: result = funcWithChanResult() // cannot use funcWithChanResult() (type chan int) as type int in assignment result <- funcWithChanResult() // invalid operation: result <- funcWithChanResult()

Recently I use notification channel to support android O. But the problem is I cannot change the sound Uri dynamically. Our app have notification sound setting which user can change app notification sound as they want. But as you know, Android now do not allow developer to update notification channel before user reinstall app. There I consider several possible solutions which is not looks good. User ringtone manager to play ringtone instead of setSound. But when user disable notification in app setting, still ringtone will not stop playing. (This will be bad user experience) Delete

Is there a way to terminate the subscription to a particular channel for any particular consumer from the server side (controller) so that disconnected callback in my coffee script file can be invoked? http://api.rubyonrails.org/classes/ActionCable/Channel/Base.html#class-ActionCable::Channel::Base-label-Rejecting+subscription+requests class ChatChannel < ApplicationCable::Channel def subscribed @room = Chat::Room[params[:room_number]] reject unless current_user.can_access?(@room) end end Before calling reject you can also inform the subscriber of the reject's reason: class ChatChannel <

I'm trying to call a WCF service (hosted in a Windows Service, not IIS) and am getting the following error: The maximum message size quota for incoming messages has been exceeded for the remote channel. See the server logs for more details. I have tried increasing the MaxReceivedMessageSize and the ReaderQuotas to their maximum values without any luck. I have also turned on logging and checked the messsage size that's getting "sent." It's definitely nowhere near the maximum. We're talking about sending an object that serialized into 372KB XML. Two questions: Does anyone know what "server logs"

I'm working through the examples at tour.golang.org, and I've encountered this code I don't really understand: package main import "fmt" func fibonacci(c, quit chan int) { x, y := 0, 1 for { select { case c <- x: // case: send x to channel c? x, y = y, x+y case <-quit: // case: receive from channel quit? fmt.Println("quit") return } } } func main() { c := make(chan int) quit := make(chan int) go func() { // when does this get called? for i := 0; i < 10; i++ { fmt.Println(<-c) } quit <- 0 }() fibonacci(c, quit) } I understand the basics of how channels work, but what I don't get is how the

前言 你如果熟悉go ，fiber （对于fiber可以简单理解成为轻量级线程）和 channel 就对应go 的goroutines 和channel,在go语言中用法如下： package main import "fmt" func sum(s []int, c chan int) { //方法执行体 sum := 0 for _, v := range s { sum += v } c <- sum // send sum to c } func main() { //main 函数 s := []int{7, 2, 8, -9, 4, 0} // 构建一个数组 c := make(chan int) //新建一个channel go sum(s[:len(s)/2], c) // 新起一个协程，然后在协程中执行方法,相当于new 一个fiber go sum(s[len(s)/2:], c) // new 一个fiber 。。 x, y := <-c, <-c // receive from c // 通过channel来传递消息 fmt.Println(x, y, x+y) } fiber example： Channel<Object> objectChannel = Channels.newChannel(0); // 0 ，receive() block

Does order in which the Goroutines block on a channel determine the order they will unblock? I'm not concerned with the order of the messages that are sent (they're guaranteed to be ordered), but the order of the Goroutines that'll unblock. Imagine a empty Channel ch shared between multiple Goroutines (1, 2, and 3), with each Goroutine trying to receive a message on ch . Since ch is empty, each Goroutine will block. When I send a message to ch , will Goroutine 1 unblock first? Or could 2 or 3 possibly receive the first message? (Or vice-versa, with the Goroutines trying to send) I have a

One of Go's slogans is Do not communicate by sharing memory; instead, share memory by communicating . I am wondering whether Go allows two different Go-compiled binaries running on the same machine to communicate with one another (i.e. client-server), and how fast that would be in comparison to boost::interprocess in C++? All the examples I've seen so far only illustrate communication between same-program routines. A simple Go example (with separate client and sever code) would be much appreciated! One of the first things I thought of when I read this was Stackless Python. The channels in Go

I am trying to understand netty buffers and watermarks. As a test case, I have a netty server which writes to a client, the client is blocked (essentially has a sleep of 10 seconds between each read) Under normal I/O TCP sender would be throttled (sending is slowed down because of flow control) if the receiver is blocked, this is not the case here. The sender seems to keep writing and flushing data on each send. Where is this data being written? Is there going to be Flow control in the netty's flush() as well? See: https://en.wikipedia.org/wiki/Transmission_Control_Protocol#Flow_control Is it