如何设置显式实现的接口的属性？

Question

I've this code snippet:

public interface Imy
{
    int X { get; set; }
}

public class MyImpl : Imy
{
    private int _x;
    int Imy.X
    {
        get => _x;
        set => _x = value;
    }
}

class Program
{
    static void Main(string[] args)
    {
        var o = new MyImpl();
        o.Imy.X = 3;//error
        o.X = 3;//error
    }
}

I just wish to assign value to X, but get 2 compilation errors. How to fix it?

Answer 1

When you implement the interface explicitly , you need to cast the variable to the interface:

((Imy)o).X = 3;

o is of type MyImpl in your code. You need to cast it to Imy explicitly to use the interface properties.

---

Alternativly, you could declare o as Imy :

Imy o = new MyImpl();
o.X = 3;