public class Foo : IEquatable<Foo>
{
    public bool Equals(Foo other)
    {
        Type type1 = this.GetType();
        Type type2 = other.GetType();

        if (type1 != type2)
            return false;

        if (type1 == typeof(A))
        {
            A a = (A)this;
            A b = (A)other;

            return a.Equals(b);
        }
        else if (type1 == typeof(B))
        {
            B c = (B)this;
            B d = (B)other;

            return c.Equals(d);
        }
        else
        {
            throw new Exception("Something is wrong");
        }
    }
}

public class A : Foo,IEquatable<A>
{
    public int Number1 { get; set; }
    public int Number2 { get; set; }

    public bool Equals(A other)
    {
        return this.Number1 == other.Number1 && this.Number2 == other.Number2;
    }
}

public class B : Foo,IEquatable<B>
{
    public int Number1 { get; set; }
    public int Number2 { get; set; }
    public int Number3 { get; set; }

    public bool Equals(B other)
    {
        return this.Number1 == other.Number1 && this.Number2 == other.Number2 && this.Number3 == other.Number3;
    }
}

A a = new A();
Foo foo = a;

foo.Equals(another);

作为直接回答您的问题,您似乎执行IEquatable&Foo>总是推迟(具体)子类的IEquatable< self>实现.这看起来像：

(坏代码,仅用于演示)

// You need to specify what you want when this method is called on a 
// vanilla Foo object. I assume here that Foo is abstract. If not,please
// specify desired behavIoUr.
public bool Equals(Foo other)
{
    if (other == null || other.GetType() != GetType())
        return false;

    // You can cache this MethodInfo..
    var equalsMethod = typeof(IEquatable<>).MakeGenericType(GetType())
                                           .GetMethod("Equals");

    return (bool)equalsMethod.Invoke(this,new object[] { other });
}

但是真的不清楚为什么你需要平等的比较来总是“通过”基类的IEquatable< self>实现.

该框架已经具有虚拟Equals方法,这将导致调度相应方法的等式调用.此外,EqualityComparar< T> .Default(大多数集合类型用于进行相等性检查)已经具有智能以适当地选择IEquatable(self)或者object.Equals(object).

只要我能看到,尝试在基类中创建一个只转发请求的相等性的实现,就是没有添加任何值.

没有进一步的解释为什么你需要基类IEquatable<>实施方面,我建议在每种类型上正确实现平等.例如：

public class A : Foo,IEquatable<A>
{
    public int Number1 { get; set; }
    public int Number2 { get; set; }

    public bool Equals(A other)
    {
        return other != null 
            && Number1 == other.Number1
            && Number2 == other.Number2;
    }

    public override bool Equals(object obj)
    {
        return Equals(obj as A);
    }

    public override int GetHashCode()
    {
        return Number1 ^ Number2;
    }
}