观看后: https://www.youtube.com/watch?v=Bme_RiT9CK4 我正在尝试使用C#和reactX复制示例。总之,这是一个货币转换器。用户指定他们想要的货币的数量和类型。当前汇率是从服务器检索的,将花费可变的时间来返回。由于服务器时间的可变性,较旧的货币请求有可能在较新的货币请求之后出现。该警告被用作使事物具有反应性的动机,从而避免出现竞争状况错误。我可以通过在Task上使用Wait()代表货币汇率来避免反应式代码中的竞争状况,但这会阻塞GUI。我应该如何处理呢?也许我应该取消其他任务?以下是我的代码,它不会阻塞但存在竞争条件问题。
public CurrencyWindow()
{
InitializeComponent();
var amount = Observable.FromEventPattern<TextChangedEventArgs>(txtAmount, "TextChanged").Select(t=>int.Parse(txtAmount.Text));
var yen = Observable.FromEventPattern<RoutedEventArgs>(rdoYen, "Checked").Select(t => "Yen");
var dollar = Observable.FromEventPattern<RoutedEventArgs>(rdoDollar, "Checked").Select(t => "Dollar");
var pound = Observable.FromEventPattern<RoutedEventArgs>(rdoPound, "Checked").Select(t => "Pound");
// merge the currency selection into one stream.
var currencyType = yen.Merge(dollar).Merge(pound);
// Lets now get a stream of currency rates.
var exchangeRate = currencyType.Select(s => GetExchangeRate(s));
amount.CombineLatest(exchangeRate, async (i, d) => i * await d).Subscribe(async v =>
{
var val = await v;
txtCost.Text = val.ToString();
});
}
private Task<double> GetExchangeRate(string currency)
{
double exchange = 0;
switch(currency)
{
case "Yen":
exchange = 1;
break;
case "Pound":
exchange = 2;
break;
case "Dollar":
exchange = 3;
break;
}
return Task.Factory.StartNew(() =>
{
System.Threading.Thread.Sleep((int)exchange * 3000);
return exchange;
});
}
答案 0 :(得分:0)
使用此代码,看看它是否有效:
var exchangeRate = currencyType
.Select(s => Observable.FromAsync(() => GetExchangeRate(s)))
.Switch();
amount
.CombineLatest(exchangeRate, (i, d) => i * d)
.Subscribe(v =>
{
var val = v;
txtCost.Text = val.ToString();
});
此代码将确保只有最新生成的currencyType
才会返回汇率。如果使用了新的货币,则之前所有对GetExchangeRate
的机上通话都将被忽略(即使它们是在当前货币之后输入的)。