异步迭代器任务<IEnumerable<T>&
时间：2023-08-28问题描述
我正在尝试实现一个返回迭代器的异步函数.思路如下:
I’m trying to implement an asynchronous function that returns an iterator. The idea is the following:
private async Task<IEnumerable<char>> TestAsync(string testString)
{
foreach (char c in testString.ToCharArray())
{
// do other work
yield return c;
}
}
但是，由于 Task 不是迭代器接口类型，因此有错误消息指出该函数不能是迭代器块.有解决办法吗?
However, there is an error message that the function cannot be an iterator block because Task<IEnumerable<char>> is not an iterator interface type. Is there a solution?
推荐答案
听起来你可能真正想要的是 IObservable<T>，有点像基于推送的异步 IEnumerable<T>.查看 Reactive Extensions，又名 Rx(在 MIT 许可的代码)(无从属关系)了解大量方法与 IObservable<T> 一起工作，使其像 LINQ-to-Objects 等一样工作.
It sounds like what you may really be looking for is something like IObservable<T>, which is sort of like a push-based asynchronous IEnumerable<T>. See Reactive Extensions, a.k.a. Rx (code licensed under MIT) (no affiliation) for a huge host of methods that work with IObservable<T> to make it work like LINQ-to-Objects and more.
IEnumerable<T> 的问题在于，没有什么可以真正使枚举本身异步.如果您不想添加对 Rx 的依赖(这确实是 IObservable<T> 大放异彩的原因)，这个替代方案可能适合您:
The problem with IEnumerable<T> is that there's nothing that really makes the enumeration itself asynchronous. If you don't want to add a dependency on Rx (which is really what makes IObservable<T> shine), this alternative might work for you:
public async Task<IEnumerable<char>> TestAsync(string testString)
{
return GetChars(testString);
}
private static IEnumerable<char> GetChars(string testString)
{
foreach (char c in testString.ToCharArray())
{
// do other work
yield return c;
}
}
虽然我想指出的是，在不知道异步实际上做什么的情况下，可能有更好的方法来实现您的目标.您发布的所有代码实际上都不会异步执行任何操作，而且我真的不知道 //do other work 中的任何内容是否是异步的(在这种情况下，这不是您底层的解决方案问题虽然它会让你的代码编译).
though I'd like to point out that without knowing what's actually being done asynchronously, there may be a much better way to accomplish your goals. None of the code you posted will actually do anything asynchronously, and I don't really know if anything in // do other work is asynchronous (in which case, this isn't a solution to your underlying problem though it will make your code compile).
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