[英]Covariant use of generic Lazy class in C#
Assuming that this applies: 假设这适用:
public class Cat : Animal { }
and assuming that I have a method: 并假设我有一个方法:
public void Feed(Animal animal) { ... }
And I can call it this way: 我可以这样称呼它:
var animal = new Cat();
Feed(animal);
How can I get this working when Feed
is refactored to only support Lazy<Animal>
as parameter? 当
Feed
被重构为仅支持Lazy<Animal>
作为参数时,如何才能使其工作? I'd like to pass in my var lazyAnimal = new Lazy<Cat>();
我想传入我的
var lazyAnimal = new Lazy<Cat>();
somehow. 不知何故。
This obviously doesnt work: 这显然不起作用:
var lazyAnimal = new Lazy<Cat>();
Feed(lazyAnimal);
You can't, basically - not directly, anyway. 你不能,基本上 - 不管是直接的。
Lazy<T>
is a class, so doesn't suppose generic variance. Lazy<T>
是一个类,所以不要假设泛型方差。
You could create your own ILazy<out T>
interface, implement it using Lazy<T>
, and then make Feed
take ILazy<Animal>
instead. 您可以创建自己的
ILazy<out T>
界面, 使用 Lazy<T>
实现它,然后使Feed
改为使用ILazy<Animal>
。 The implementation would be trivial. 实施将是微不足道的。
Alternatively, you could make Feed
generic: 或者,您可以使
Feed
通用:
public void Feed<T>(Lazy<T> animal) where T : Animal
Or Servy's suggestion of taking Func<Animal>
would work too, and you could call it using a lambda expression for the Lazy<T>
: 或者Servy关于服用
Func<Animal>
的建议也会起作用,你可以使用lambda表达式为Lazy<T>
调用它:
Feed(() => lazyAnimal.Value);
Well, you won't be able to use it exactly as is. 好吧,你将无法完全按原样使用它。 Probably the simplest refactor would be to have it accept a
Func<Animal>
instead of a Lazy
. 可能最简单的重构就是让它接受一个
Func<Animal>
而不是一个Lazy
。 Then you could pass in a lambda that fetches the value of a Lazy
. 然后你可以传入一个获取
Lazy
值的lambda。 Func
is covariant with respect to it's return type. Func
对于它的返回类型是协变的。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.