为什么列表< T>不上协接口MyInterface的<有效;出T> [英] Why is List<T> not valid on an covariant interface MyInterface<out T>
问题描述
后续问题前一个问题,这已被确定为一个协方的问题。以这一步,如果我修改 IFactory的
如下:
Follow up question to a previous question, this has been identified as a co-variance issue. Taking this one step further, if I modify IFactory
as follows:
class Program
{
static void Main(string[] args)
{
IFactory<IProduct> factory = new Factory();
}
}
class Factory : IFactory<Product>
{
}
class Product : IProduct
{
}
interface IFactory<out T> where T : IProduct
{
List<T> MakeStuff();
}
interface IProduct
{
}
我得到:
无效差异:类型参数T必须是在Sandbox.IFactory.MakeStuff目不暇接有效的()。 T是协变的。
Invalid variance: The type parameter T must be invariantly valid on Sandbox.IFactory.MakeStuff(). T is covariant.
这是为什么不目不暇接有效吗?如何/应该这样解决?
Why is this not invariantly valid? How can/should this be resolved?
推荐答案
的 @克雷格的回答是正确的。要解决,将其更改为:
@Craig's answer is correct. To resolve, change it to:
IEnumerable<T> MakeStuff()
编辑:至于为什么,看的的IEnumerable< T>接口:
public interface IEnumerable<out T> : IEnumerable
注意的&IList的LT; T>接口没有out关键字。方差支持在接口和委托,而不是类泛型类型参数,所以它并不适用于列出< T>
Note that the IList<T> Interface does not have the out keyword. Variance is supported for generic type parameters in interfaces and delegates, not classes, so it doesn't apply to List<T>.
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