c# - 使用实现所需返回类型的返回类型实现接口(interface)

Question

为什么我不能这样做？

IHasOperatingSystem {
IOperatingSystem OperatingSystem { get; }
} 

Computer<T> : IHasOperatingSystem where T : IOperatingSystem {
public T OperatingSystem { get; }
}

它告诉我类型应该是 IOperatingSystem，但是如果 T 实现了 IOperatingSystem，那还不够吗？

另外，我意识到这个问题的标题可能有点令人困惑，但我想不出更好的表达方式。

Answer 1

It's telling me that the type should be IOperatingSystem, but if T implements IOperatingSystem, shouldn't that be sufficient?

不，这不是 C# 的工作方式。为了实现接口(interface)或重写方法，参数类型和返回类型必须完全匹配。从 C# 5 规范的第 13.4.4 节:

For purposes of interface mapping, a class member A matches an interface member B when:

• A and B are methods, and the name, type, and formal parameter lists of A and B are identical.
• ...

(这里的“类型”应该读作“返回类型”。)

现在你可以让你的IHasOperatingSystem类型泛型，当然:

public interface IHasOperatingSystem<T> where T : IOperatingSystem
{
    T OperatingSystem { get; }
}

public class Computer<T> : IHasOperatingSystem<T> where T : IOperatingSystem
{
    public T OperatingSystem { get { ... } }
}

或者，您可以在 Computer<T> 中使用显式接口(interface)实现。类(class):

public interface IHasOperatingSystem
{
    IOperatingSystem OperatingSystem { get; }
} 

public class Computer<T> : IHasOperatingSystem where T : IOperatingSystem
{
    // Explicit interface implementation...
    IHasOperatingSystem.OperatingSystem IOperatingSystem
    {
        // Delegate to the public property
        get { return OperatingSystem; }
    }

    public T OperatingSystem { get { ... } };
}