如何限制方法中傳遞的參數

Question

我的問題在示例代碼中。 我如何限制開發人員傳遞真實參數。 我嘗試了一些有關泛型的方法，但無法解決。

這里重要的是我想限制編譯時間。 所以我知道如何在運行時中進行預防。

namespace TheLiving
{
    public interface IFood
    {
        int Protein { get; set; }
        int Carbohydrate { get; set; }
    }

    public interface IMeat : IFood
    {
        int Nitrogen { get; set; }
    }

    public interface IVegetable : IFood
    {
        int Vitamin { get; set; }
    }


    public class Veal : IMeat
    {
        public int Protein { get; set; }
        public int Carbohydrate { get; set; }
        public int Nitrogen { get; set; }
    }

    public class Spinach : IVegetable
    {
        public int Protein { get; set; }
        public int Carbohydrate { get; set; }
        public int Vitamin { get; set; }
    }

    public interface IEating
    {
        void Eat(IFood food);
    }

    public class Lion : IEating
    {
        public int Protein { get; set; }
        public int Carbohydrate { get; set; }
        public int Nitrogen { get; set; }


        //But lion is eating only Meat. So any developer can pass vegatable to lion for eating. 
        //May be god is not a good developer. So i want restrict him on Compile Time!! for passing only Meat. :)
        //The important thing here is i want restrict on Compile Time not RunTime!!!
        public void Eat(IFood food)
        {
            Protein = food.Protein;
            Carbohydrate = food.Carbohydrate;
            //Nitrogen = ?? //So i know that i can cast and validate food but i want ensure this on DesignTime!!
        }
    }

    public class Sheep : IEating
    {
        public int Protein { get; set; }
        public int Carbohydrate { get; set; }
        public int Vitamin { get; set; }

        public void Eat(IFood food)
        {
            Protein = food.Protein;
            Carbohydrate = food.Carbohydrate;
            //Vitamin = food.??
        }
    }
}

Answer 1

我認為您必須具有IHerbivore ， ICarnivore和IOmnivore才能在設計時允許它。

public interface IHerbivore
{
    void Eat(IVegetable food);
}

public interface ICarnivore
{
    void Eat(IMeat food);
}

public interface IOmnivore : IHerbivore, ICarnivore
{
}

這樣，您的獅子可以成為ICarnivore ，只能吃肉

Answer 2

或者，您可以更改界面...

public interface IEating
{
    bool Eat(IFood food); //return wheater the eater eats the food or not
}

像這樣實現獅子：

public class Lion : IEating
{
    public int Protein { get; set; }
    public int Carbohydrate { get; set; }
    public int Nitrogen { get; set; }

    public bool Eat(IFood food)
    {
        IMeat meat = food as IMeat;
        if (meat != null)
        {

            Protein = meat.Protein;
            Carbohydrate = meat.Carbohydrate;
            Nitrogen = meat.Nitrogen;
            return true;
        }
        return false;
    }
}

Answer 3

public interface IEating<in T> where T : IFood {

    void Eat(T food);
}

public class Lion : IEating<IMeat>, IEating<IFood> {

    public int Protein { get; set; }
    public int Carbohydrate { get; set; }
    public int Nitrogen { get; set; }

    public void Eat(IMeat food) {

        Protein = food.Protein;
        Carbohydrate = food.Carbohydrate;
        Nitrogen = food.Nitrogen;
    }

    public void Eat(IFood food) {

        var meat = food as IMeat;
        if (meat == null) return;

        Eat(meat);
    }
}

public class Sheep : IEating<IVegetable>, IEating<IFood> {

    public int Protein { get; set; }
    public int Carbohydrate { get; set; }
    public int Vitamin { get; set; }

    public void Eat(IVegetable food) {
        Protein = food.Protein;
        Carbohydrate = food.Carbohydrate;
        Vitamin = food.Vitamin;
    }

    public void Eat(IFood food) {

        var vegetable = food as IVegetable;
        if (vegetable == null) return;

        Eat(vegetable);
    }
}

Answer 4

MS代碼合同呢？ 我發誓它的Visual Studio集成至少在編譯時添加警告。