如何防止扫描程序多次运行

Question

例如

var subject = new Subject<int>();
var test = subject.Scan(0, (x, y) => {
    Console.WriteLine("scan");
    return x + 1;
});
test.Subscribe(x => Console.WriteLine("subscribe1"));
//test.Subscribe(x => Console.WriteLine("subscribe2"));
Observable.Range(0, 1).Subscribe(subject);
Console.WriteLine("done");
Console.Read();

输出为

scan
subscribe1
done

但是如果你取消第二个订阅的注释，输出是

scan
subscribe1
scan
subscribe2
done

为什么扫描运行两次，我如何防止它?所以输出应该是

scan
subscribe1
subscribe2
done

我使用Subject来累积不同的Observables。然后我使用扫描方法来更新模型，然后我有不同的地方，我需要订阅模型更新。也许没有使用主题有更好的解决方案?

Answer 1

尝试使用Observable.Publish获得IConnectableObservable<T>。

var subject = new Subject<int>();
var test = subject
    .Scan(0, (x, y) => {
        Console.WriteLine("scan");
        return x + 1;
    })
    .Publish();
test.Subscribe(x => Console.WriteLine("subscribe1"));
test.Subscribe(x => Console.WriteLine("subscribe2"));
test.Connect();
Observable.Range(0, 1).Subscribe(subject);
Console.WriteLine("done");
Console.Read();

输出:

scan
subscribe1
subscribe2
done

Publish将冷的Scan观测对象转变为热观测对象，并在Connect被调用时开始发光。

Answer 2

你看到的问题是Subject是一个热可观察对象，而Scan每次订阅它都会创建一个新的冷可观察对象。

尝试在被摄体前移动扫描

var subject = new Subject<int>();
subject.Subscribe(x => Console.WriteLine("subscribe1"));
subject.Subscribe(x => Console.WriteLine("subscribe2"));
Observable.Range(0, 1).Scan(0, (x, y) => {
    Console.WriteLine("scan");
    return x + 1;
}).Subscribe(subject);
Console.WriteLine("done");
Console.Read();

您也可以不使用Subject:

var test = Observable.Range(0, 1).Scan(0, (x, y) => {
    Console.WriteLine("scan");
    return x + 1;
}).Publish();
test.Subscribe(x => Console.WriteLine("subscribe1"));
test.Subscribe(x => Console.WriteLine("subscribe2"));
test.Connect();
Console.WriteLine("done");
Console.Read();

热观测与冷观测