Wrapping intellisense hints

See more:

C++

I gave up trying to solve this myself, so any help will be appreciated.
Let's say, we have the following declaration:

class A
{
  virtual void() foo() {....};
};

typedef A B;

Then, somewhere in code we have this:

B b;
b.foo();

Now, if you are using MSVC++, Intellisense hint (when you hover your mouse over "b.foo();" call ) will resolve it as "void A::foo()", not "void B::foo()". I.e. if you typedef anything, Intellisense totally ignore your def, and resolves to the very bone of the declaration.

Same will happen if you declare B as descendant of A:

class B: public A
{
};

Intellisense hint will still show "void A::foo()" until you override foo() in B one way or another.

And here is my problem. I'm writing a templated class, which itself is descendant of another templated class, etc, etc. Final class name become very, very long (>200 symbols). Of course, it is then typedef-ed, but - and this is my problem - Intellisense doesn't care about any typedef, and just unwraps the type to this crazy 200 symbols.

My question is this:
is there ANY way at all to force Intellisense to use typedef instead of unwrapped def?

If this is not possible, is it any way to have Intellisense at least display B::foo() instead of A::foo() in the above example, provided that there is no way I can override foo() in B?

My ultimate goal is this:

template<class X1, class X2<X1>, ...., class X10000, typename Q=X1::nested_type> // joking, but you got an idea
class Base : public AnotherTemplatedBase<.....>
{
  typedef typename AnotherTemplatedBase<.....>::very_deeply_nested_type  data_type;
  data_type foo(yet_another_deep_data_type& src) {...}
}

typedef Base< and here are end-user params> UserClass;

....
UserClass uc;
auto x = uc.foo(y);
// and now I want Intellisense to show "data_type UserClass::foo(another_data_type& src)" instead of crazy long unwrapped base name.

Is this even doable? May be in new MSVC++ releases?...