C# 从工厂返回通用接口(interface)

Question

我实现了一个车辆服务，负责为汽车和卡车等车辆提供服务:

public interface IVehicleService
{
    void ServiceVehicle(Vehicle vehicle);   
}

public class CarService : IVehicleService
{
    void ServiceVehicle(Vehicle vehicle)
    {
        if(!(vehicle is Car))
            throw new Exception("This service only services cars")

       //logic to service the car goes here
    }
}

我还有一个车辆服务工厂，负责根据传入工厂方法的车辆类型创建车辆服务:

public class VehicleServiceFactory 
{
    public IVehicleService GetVehicleService(Vehicle vehicle)
    {
        if(vehicle is Car)
        {
            return new CarService();
        }

        if(vehicle is Truck)
        {
            return new TruckService();
        }

        throw new NotSupportedException("Vehicle not supported");
    }

}

我遇到的主要问题是 CarService.ServiceVehicle 方法。它正在接受 Vehicle，而理想情况下它应该接受 Car，因为它知道它只会为汽车提供服务。所以我决定更新此实现以改用泛型:

public interface IVehicleService<T> where T : Vehicle
{
    void ServiceVehicle(T vehicle); 
}

public class CarService : IVehicleService<Car>
{
    void ServiceVehicle(Car vehicle)
    {
        //this is better as we no longer need to check if vehicle is a car

        //logic to service the car goes here 
    }
}

我遇到的问题是如何更新 VehicleServiceFactory 以返回车辆服务的通用版本。我尝试了以下方法，但它导致编译错误，因为它无法将 CarService 转换为通用返回类型 IVEhicleService:

public class VehicleServiceFactory 
{
    public IVehicleService<T> GetVehicleService<T>(T vehicle) where T : Vehicle
    {
        if(vehicle is Car)
        {
            return new CarService();
        }

        if(vehicle is Truck)
        {
            return new TruckService();
        }

        throw new NotSupportedException("Vehicle not supported");
    }

}

如有任何建议，我们将不胜感激。

Answer 1

简单地将服务转换为接口(interface):

return new CarService() as IVehicleService<T>;

您知道 T 是汽车，但编译器不知道，它不够智能，无法遵循方法逻辑，也不是本应如此；就编译器所知，T 可以是任何东西，只要它是 Vehicle。您需要告诉编译器，“嘿，我知道我在做什么，T 和 Car 实际上是同一类型。”