c# – 是否可以强制显式实现接口(或部分接口)？

public interface MyInterface
{
    int Property1
    {
        get;
    }

    void Method1();
    void Method2();
}

有没有办法强制接口的实现者明确地实现它的一部分？就像是：

public interface MyInterface
{
    int Property1
    {
        get;
    }

    explicit void Method1();
    explicit void Method2();
}

编辑：至于为什么我关心界面是否明确实现;就功能而言,它并不重要,但是使用代码隐藏一些不必要的细节可能会有所帮助.

我正试图在我的系统中模仿多重继承,使用这种模式：

public interface IMovable
{
    MovableComponent MovableComponent
    {
         get;
    }
}

public struct MovableComponent
{
    private Vector2 position;
    private Vector2 velocity;
    private Vector2 acceleration;

    public int Method1() 
    {
        // Implementation
    }

    public int Method2() 
    {
        // Implementation
    }

}

public static IMovableExtensions
{
    public static void Method1(this IMovable movableObject) 
    {
        movableObject.MovableComponent.Method1();
    }

    public static void Method2(this IMovable movableObject) 
    {
        movableObject.MovableComponent.Method2();
    }
}

public class MovableObject : IMovable
{
    private readonly MovableComponent movableComponent = new MovableComponent();

    public MovableComponent MovableComponent 
    {
        get { return movableComponent; }  // Preferably hiddem,all it's methods are available through extension methods.
    } 
}

class Program
{
     static void Main(string[] args)
     {
          MovableObject movableObject = new MovableObject();
          movableObject.Method1();   // Extension method
          movableObject.Method2();   // Extension method
          movableObject.MovableComponent    // Should preferably be hidden.

     }
}

如果MovableComponent属性是显式实现的,那么在大多数情况下,它将对使用该类的任何人隐藏.希望解释不是太可怕.