dart - Dart 2 中与隔离物的双向通信

Question

我正在尝试隔离，我想知道我怎样才能产生其中一些进行繁重的计算，当根隔离询问他们当前的计算值时，他们会“按需”响应它。

据我所知，唯一可以用作新创建的隔离消息的对象是 SendPort，这意味着只有生成的隔离可以与根隔离。我尝试发送一个 <SendPort,ReceivePort> 元组，但由于 ReceivePort 不是 SendPort，它被认为是非法的。

简而言之：

root <-- 隔离好

root <-> 隔离如何？

Answer 1

随着Gunter的评论，我做了这个：

import 'dart:async';
import 'dart:io';
import 'dart:isolate';

Stopwatch stopwatch = new Stopwatch();

main(args) async {
  ReceivePort rPort = new ReceivePort();
  rPort.listen((data) {
    print("<root> $data received");
    if (data is List) {
      String action = data[0];
      if (action == "register") {
        (data[1] as SendPort).send(stopwatch.elapsedMilliseconds);
      }
    }
  });
  stopwatch.start();
  await Isolate.spawn(elIsolate, rPort.sendPort);
  print("isolate spawned in ${stopwatch.elapsedMilliseconds} msecs"); //isolate spawned in 377 msecs
}

void elIsolate(SendPort sPort) {
  ReceivePort rPort = new ReceivePort();
  rPort.listen((data) {
    print("<Isolate> '$data' received"); //<Isolate> '387' received
  });
  sPort.send(["register", rPort.sendPort]);
}

在Kevin的回答中，代码简化为：

import 'dart:async';
import 'dart:io';
import 'dart:isolate';
import 'package:stream_channel/stream_channel.dart';

Stopwatch stopwatch = new Stopwatch();

main(args) async {
  ReceivePort rPort = new ReceivePort();
  IsolateChannel channel = new IsolateChannel.connectReceive(rPort);
  channel.stream.listen((data) {
    print("<root> '$data' received at ${stopwatch.elapsedMilliseconds} msecs"); //<root> 'hello world' received at 1141 msecs
    channel.sink.add(stopwatch.elapsedMilliseconds);
  });
  stopwatch.start();
  await Isolate.spawn(elIsolate, rPort.sendPort);
  print("isolate spawned in ${stopwatch.elapsedMilliseconds} msecs"); //isolate spawned in 1111 msecs
}

void elIsolate(SendPort sPort) {
  IsolateChannel channel = new IsolateChannel.connectSend(sPort);
  channel.stream.listen((data) {
    print("<Isolate> '$data' received");
  });
  channel.sink.add("hello world");
}

Answer 2

查看package:stream_channel中的IsolateChannel。

这应该为您尝试做的事情提供很多帮助。