为什么任务< T>不共变体？ [英] Why is Task<T> not co-variant?

问题描述

class ResultBase {}
class Result : ResultBase {}

Task<ResultBase> GetResult() {
    return Task.FromResult(new Result());
}

编译器告诉我，它不能隐式转换任务< ;结果> 到任务< ResultBase> 。有人能解释这是为什么？我本来期望协方差，让我写这样的代码。

The compiler tells me that it cannot implicitly convert Task<Result> to Task<ResultBase>. Can someone explain why this is? I would have expected co-variance to enable me to write the code in this way.

推荐答案

据的人谁可能是在知道 ...

的理由是协方差的好处是$抵销b $ b杂乱的缺点（即每个人都必须做出是否要在他们的代码中的每一个
座使用任务或ITask一个
号决定）。

The justification is that the advantage of covariance is outweighed by the disadvantage of clutter (i.e. everyone would have to make a decision about whether to use Task or ITask in every single place in their code).

这听起来好像还没有一个非常引人注目的动机无论哪种方式。 ITask<出T> 将需要大量新的过载，可能相当引擎盖下一个位（我不能证明如何实际基类中实现或如何特别是相比天真的实现），但更多的方式在这些形式 LINQ 式的扩展方法。

It sounds to me like there is not a very compelling motivation either way. ITask<out T> would require a lot of new overloads, probably quite a bit under the hood (I cannot attest to how the actual base class is implemented or how special it is compared to a naive implementation) but way more in the form of these linq-like extension methods.

某些别有用心的人取得了良好的一点 - 最好将时间花在类 ES协变和逆变。我不知道有多难那会，但是这听起来像一个更好地利用时间给我。

Somebody else made a good point - the time would be better spent making classes covariant and contravariant. I don't know how hard that would be, but that sounds like a better use of time to me.

在另一方面，有人指出，这将是非常酷有一个真正的收益回报率像功能可在异步方法。我的意思是，没有戏法。

On the other hand, somebody mentioned that it would be very cool to have a real yield return like feature available in an async method. I mean, without sleight of hand.

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