如何将泛型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。我真的对这是怎么回事有一个抽象的想法，但在寻找答案后，我变得更加困惑。

我首先想使用泛型构造函数创建两个新类，而不是使用它们的函数。一个类应该接受第二个类作为参数来动态地使用方法Attack(((class2.health-class1.damage(，但它必须实现一个IDamagable接口

我可能会犯一些奇怪的错误。很抱歉

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

我建议进行这种转换：

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