一般来说,我建议使用一个基础类。如果不行的话,你可以这样做:
public interface IAnimal { }
public interface INoisyAnimal : IAnimal {
string MakeSound();
}
public static class AnimalExtensions {
public static string MakeSound(this IAnimal someAnimal) {
if (someAnimal is INoisyAnimal) {
return (someAnimal as INoisyAnimal).MakeSound();
}
else {
return "Unknown Noise";
}
}
}
public class Dog : INoisyAnimal {
public string MakeSound() {
return "Bark";
}
}
public class Porcupine : IAnimal { }
这使得每个
IAnimal
看起来像一个
INoisyAnimal
即使它并不是真正的
INoisyAnimal
。例如:
IAnimal dog = new Dog();
IAnimal porcupine = new Porcupine();
Console.WriteLine(dog.MakeSound()); // bark
Console.WriteLine(porcupine.MakeSound()); // Unknown Noise
然而,这仍然不是接口的实际实现。请注意,尽管外观如此,但没有实际实现。
Console.WriteLine(porcupine is INoisyAnimal)
另一种选择可能是创建一个包装器,在需要新功能时扩展您的基类:
public class NoisyAnimalWrapper : INoisyAnimal {
private readonly IAnimal animal;
public NoisyAnimalWrapper(IAnimal animal) {
this.animal = animal;
}
public string MakeSound() {
return "Unknown Noise";
}
}
public static class AnimalExtensions {
public static INoisyAnimal Noisy(this IAnimal someAnimal) {
return someAnimal as INoisyAnimal ??
new NoisyAnimalWrapper(someAnimal);
}
}
当需要时,您可以从任何IAnimal
创建一个INoisyAnimal
:
INoisyAnimal dog = new Dog();
INoisyAnimal porcupine = new Porcupine().Noisy();
Console.WriteLine(dog.MakeSound()); // bark
Console.WriteLine(porcupine.MakeSound()); // Unknown Noise
你还可以将包装器设置为通用的(例如,NoisyAnimal<T> where T : IAnimal, new
),并完全摒弃扩展方法。根据你实际的使用情况,这可能比以前的选项更可取。