我试图看看如何使用Rx将多个可观察事件流式传输到一组事件中.但是当我运行下面的代码时,我得到一个例外.那么这是否意味着由于违反Rx语法,多个观察者总是容易出现异常？我的意思是如果这些多个观察者中的两个偶然同时生成一个事件(任何两个可观察者将同时产生一些概率),它应该给出一个例外.
DateTimeOffset start;
object sync = new object();
var subject = new Subject<long>();
var observer = Observer.Create<long>(c =>
{
lock (sync)
{
Console.WriteLine(c);
}
})
;
var observable1 = Observable.Interval(TimeSpan.FromSeconds(2));
var observable2 = Observable.Interval(TimeSpan.FromSeconds(5));
var observable3 = Observable.Never<long>().Timeout
(start = DateTimeOffset.Now.AddSeconds(15),
(new long[] { 1 }).ToObservable());
var observable4 = Observable.Never<long>().Timeout(start);
observable1.Subscribe(observer);
observable2.Subscribe(observer);
observable3.Subscribe(observer);
observable4.Subscribe(observer);
Thread.Sleep(20000);
感谢Gideon的解释.这是我得到的例外.你是对的,这是一个时间的例子.这是一个编码错误.谢谢.
System.TimeoutException: The operation has timed out.
at System.Reactive.Observer.<Create>b__8[T](Exception e)
at System.Reactive.AnonymousObserver`1.Error(Exception exception)
at System.Reactive.AbstractObserver`1.OnError(Exception error)
at System.Reactive.Subjects.Subject`1.OnError(Exception error)
at System.Reactive.AnonymousObservable`1.AutoDetachObserver.Error(Exception e
xception)
at System.Reactive.AbstractObserver`1.OnError(Exception …/* Atleast is it needed in this case?*/
public class Service
{
private static List<String> subscribedFields;
private static List<String> unsubscribedFields;
//---------------------------------
// is this necessary?
//---------------------------------
public static void setFields(List<String> subscribedFields, List<String> unsubscribedFields)
{
Service.subscribedFields = subscribedFields;
Service.unsubscribedFields = unsubscribedFields;
}
public static List<String> getSubscribedFields()
{
return subscribedFields;
}
public static List<String> getUnsubscribedFields()
{
return unsubscribedFields;
}
}
// some other class
public class User{
// Is this not enough to change the lists? Isn't the setter redundant?
Change(Service.getSubscribedFields());
Change(Service.getUnsubscribedFields()); …我是Rx的新手.我想知道是否可以向不同的订阅者发送消息,以便它们在不同的线程上运行？IObserable如何控制呢？简单的Subject实现,据我所知,它在一个线程上一个接一个地调用订阅者.
public class Subsciber : IObserver<int>
{
public void OnNext(int a)
{
// Do something
}
public void OnError(Exception e)
{
// Do something
}
public void OnCompeleted()
{
}
}
public static class Program
{
public void static Main()
{
var observable = new <....SomeClass....>();
var sub1 = new Subscriber();
var sub2 = new Subscriber();
observable.Subscribe(sub1);
observable.Subscribe(sub2);
// some waiting function
}
}
如果我使用Subject作为'SomeClass',那么在sub1的OnNext()完成之前,不会调用sub2的OnNext().如果sub1花了很多时间,我不希望它延迟sub2的接收.有人能告诉我Rx如何为SomeClass提供这种实现.
当我尝试最大化窗口时,orinigal窗口渲染仍然存在,而另一个最大化窗口出现使其变得混乱.
import java.awt.BorderLayout;
import java.awt.FlowLayout;
import java.awt.GridBagConstraints;
import java.awt.GridBagLayout;
import java.awt.GridLayout;
import java.awt.Insets;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.JScrollPane;
import javax.swing.JTabbedPane;
import javax.swing.JTable;
import javax.swing.table.DefaultTableModel;
import javax.swing.*;
import javax.swing.table.TableColumnModel;
/**
* @author ad *
*/
public class Blotter {
private JFrame topFrame;
private JPanel mainContentPanel;
private JList unsubscribedFields;
private JList subscribedFields;
private JButton butSubscribe;
private JButton butUnsubscribe;
private JButton butApply;
private JButton butOk;
private JButton butCancel;
private JPanel panConfirm;
private JPanel panToggle;
private JPanel panBottom;
private …我在windows7上使用dev cpp来编译我的代码.
int d = 0x12;
char* e = (char*)&d;
printf("%d %d\n", sizeof (int), sizeof (float));
printf("%p %p\n", &d, (float*)&d);
printf("%p %p %p %p %p\n", &d, &e[0], &e[1], &e[2], &e[3]);
printf(" %d | %x | %#1x | %#1x | %#1x |%p\n", d, e[0], e[1], e[2], e[3], &e[0]);
getchar();
4 4
0028FF40 0028FF40
0028FF40 0028FF40 0028FF41 0028FF42 0028FF43
18 | 12 | 0 | 0 | 0 |0028FF40
你看,如果我使用%d打印d,它打印4个字节的精细.但是,如果我使用下面的%f,它会在必须打印e的第一个字节的位置显示零.任何人都可以帮助解决为什么会这样 为什么e的内容取决于d的格式？
int d = 0x12;
char* e = (char*)&d;
printf("%d %d\n", sizeof …