我如何将通用接口作为方法参数传递？

Question

所以我目前正在试验泛型，我刚刚偶然发现了一个令人震惊的问题。我真的对它如何工作有一个抽象的想法，但在搜索答案后我变得更加困惑。

我首先想使用通用构造函数创建 2 个新的 classes，而不是使用它们的函数。一个 class 应该接受第二个 class 作为参数来使用方法 Attack() 和动态（class2.health - class1.damage），但它必须实现一个 IDamagable 接口。

我可能会犯奇怪的错误。对不起。

public class GenericsIdeas : MonoBehaviour
{

public interface IDamagable<T>
{
    void TakeDamage(T dmg);
}

public class Enemy<T1, T2> : IDamagable<T1>
{
    T1 damage;
    T2 health;

    public void Attack(IDamagable d)
    {
        d.TakeDamage((dynamic)damage);
    }

    public void TakeDamage(T1 damage)
    {
        dynamic dmg = (dynamic)damage;
        dynamic hp = (dynamic)health;
        health -= dmg;
        Debug.Log("Enemy health: " + health);
    }

    public Enemy(T1 t1, T2 t2)
    {
        damage = t1;
        health = t2;
    }
}

public class Player<T1, T2> : IDamagable<T1>
{
    T1 damage;
    T2 health;

    public void Attack(IDamagable d)
    {
        d.TakeDamage((dynamic)damage);
    }

    public void TakeDamage(T1 damage)
    {
        dynamic dmg = (dynamic)damage;
        dynamic hp = (dynamic)health;
         health -= dmg;
        Debug.Log("Player health: " + health);
    }

    public Player(T1 t1, T2 t2)
    {
        damage = t1;
        health = t2;
    }
}

void Start()
{
    Player<int, int> player = new Player<int, int>(6, 10);
    Enemy<float, float> enemy = new Enemy<float, float>(4.32f, 20.61f);
    player.Attack(enemy.IDamagable); //???
}
}

Answer 1

我会建议这种转变：

public interface IDamagable
{
    void TakeDamage<T>(T dmg);
}

我如何将通用接口作为方法参数传递？

How could I pass generic interface as a method parameter?

c#

generics

interface

class