Question

Possible Duplicate:
Overload resolution and virtual methods

If i call this , why object method is being called?

Classes.Class2 c = new Classes.Class2();
c.GetJ(1);

public class Class1
{
   public virtual void GetJ(int j)
   {

   }
}

class Class2:Class1
{
    public override void GetJ(int j)
    {
       int j3 = 8;
    }

     public void GetJ(object j)
     {
        int j1 = 82;
     }
}

Answer 1

See the C# 4.0 Specification ( https://www.microsoft.com/en-us/download/details.aspx?id=7029 )

7.4 Member lookup

First, a set of accessible members named N is determined:

• If T is a type parameter, then the set is the union of the sets of accessible members named N in each of the types specified as a primary constraint or secondary constraint (§10.1.5) for T, along with the set of accessible members named N in object.

• Otherwise, the set consists of all accessible (§3.5) members named N in T, including inherited members and the accessible members named N in object. If T is a constructed type, the set of members is obtained by substituting type arguments as described in §10.3.2. Members that include an override modifier are excluded from the set.

I don’t understand that behavior, but this is in the specification, so it is correct, even if a bit disturbing (and I have a C++ background)…

Edit:

This is indeed a duplicate question (as correctly discovered by Anders Abel in his comment).

See the original Overload resolution and virtual methods for an explanation for this behavior.