Exact difference between overriding and hiding

Question

Can anybody tell the working of overriding and hiding in terms of memory and references.

class A
{
    public virtual void Test1() { //Impl 1}
    public virtual void Test2() { //Impl 2}
}
class B  : A
{
    public override void Test1() { //Impl 3}
    public new void Test2() { Impl 4}
}

static Main()
{
    A aa=new B() //This will give memory to B
    aa.Test1(); //What happens in terms of memory when this executes
    aa.Test2(); //-----------------------SAME------------------------
}

Here memory is with class B but in the second statement aa.Test2 class A's method will be called. Why is it? If B has memory then B's method should be called (in my point of view).

Any link / exercise that describes this fundamental very deeply and completely will be a big help.

Answer 1

Take a look at this answer to a different question by Eric Lippert.

To paraphrase (to the limits of my comprehension), these methods go into "slots". A has two slots: one for Test1 and one for Test2.

Since A.Test1 is marked as virtual and B.Test1 is marked as override, B's implementation of Test1 does not create its own slot but overwrites A's implementation. Whether you treat an instance of B as a B or cast it to an A, the same implementation is in that slot, so you always get the result of B.Test1.

By contrast, since B.Test2 is marked new, it creates its own new slot. (As it would if it wasn't marked new but was given a different name.) A's implementation of Test2 is still "there" in its own slot; it's been hidden rather than overwritten. If you treat an instance of B as a B, you get B.Test2; if you cast it to an A, you can't see the new slot, and A.Test2 gets called.