C# - 为什么我不能创建通用接口(interface)类型的 IEnumerable？

Question

例如，我有一个制作水果冰沙的应用程序。我可以做一个IEnumerable苹果冰沙列表，并且有效。

我希望能够制作一份通用的冰沙列表 IEnumerable<ISmoothie<IFruit>>万一我决定在生产线上添加另一种水果，比如橙子。当我尝试这样做时，代码将无法编译并且出现错误:

Cannot implicitly convert type 'System.Collections.Generic.List>' to 'System.Collections.Generic.IEnumerable>'. An explicit conversion exists (are you missing a cast?)

public class Program
{
    public static void Main()
    {
        IEnumerable<ISmoothie<Apple>> appleSmoothies = new List<ISmoothie<Apple>>();   // I Can specifically make Apple Smoothies!
        IEnumerable<ISmoothie<IFruit>> genericSmoothies = new List<Smoothie<Apple>>(); // Does Not Compile - What if I want generic Smoothies? 
        Console.WriteLine("Hello World");
    }

    public class Apple : IApple
    {
        // Concrete Implimentation of an Apple
    }

    public interface IApple : IFruit
    {
        // Specific Interface for Apples
    }

    public interface IFruit
    {
        // Basic Fruit Interface for all Fruits
    }

    public class Smoothie<T> : ISmoothie<T> where T : IFruit
    {
        // Generic Fruit Smoothie
        public List<T> Ingredients {get; set;}
        public int Size {get; set;}
        public void Drink()
        {
            // Drink Logic
        }
    }

    public interface ISmoothie<T> where T : IFruit
    {
        List<T> Ingredients {get; set;}
        int Size {get; set;}
        void Drink();
    }
}

Answer 1

您需要将out 参数添加到您的界面以指定Covariance

public interface ISmoothie<out T> where T : IFruit
{
}

---

out (generic modifier) (C# Reference)

For generic type parameters, the out keyword specifies that the type parameter is covariant. You can use the out keyword in generic interfaces and delegates.

Covariance enables you to use a more derived type than that specified by the generic parameter. This allows for implicit conversion of classes that implement covariant interfaces and implicit conversion of delegate types. Covariance and contravariance are supported for reference types, but they are not supported for value types.

更新

使用out参数时有限制，

public class Smoothie<T> : ISmoothie<T> where T : IFruit
{
   // Generic Fruit Smoothie
   public T Type { get; set; } // we can do this
}

public interface ISmoothie<out T> where T : IFruit
{
   T Type { get; set; } // compiler error CS1961 Invalid variance: 
}

out T 表示类型 T 被限制为仅作为返回(出站)值出现在泛型类、接口(interface)或方法的方法中