C++ Syntactical anomaly for instantiation of a derived class - c++

Question

I am working with a GitHub library and ran across a derived class instantiation that perplexes me. In abbreviated form,
class A
{
public:
A() {}
int AFunc(void) { return(1); }
};
class B : public A
{
public:
B(void) : A() {}
int BFunc(void) { return(2); }
};
Within an include file, the class is instantiated as follows:
A &tObject = *(new B());
Sample code then refers to 'tObject' as global variable calling methods from class A and/or B.
For example:
tObject.AFunc();
tObject.BFunc();
So here's the question, is that instantiation legal?
The compiler is only fussing on the call to a service class's method, saying that class A has no such member. That error makes sense to me and I've narrowed the issue to the above explanation.
While I do not have broad compiler experience, I have been programming in C++ for many years. I've never seen such a construct.
Would someone kindly explain how an object declared, in my example, as 'class A' can access methods from the derived class B?
In my experience, I've always declared the derived class as a pointer and then accessed methods from the base or derived class using the '->' construct. Oftentimes, I've stored the derived class as a pointer to the base and then performed a cast to convert when or if I needed access to the derived class's methods.
An insight is highly appreciated.

Answer 1

It cannot. The compiler is right to complain, there is no way this is valid. Remember that C++ is a static language, which means that the compiler will try to find a function named BFunc in A, which it cannot, as there is no such function.
This might be a compiler extension of some sort, but anyways, this isn't legal standard C++. Most probably, the author wanted to make BFunc a virtual method in A, which would have made the access legal.
Would someone kindly explain how an object declared, in my example, as 'class A' can access methods from the derived class B?
As explained, this cannot be.
I've always declared the derived class as a pointer and then accessed methods from the base or derived class using the '->' construct.
You can also do this with references, not just with pointers. Although this is done less often than pointers, so this might explain why you haven't encountered this yet.
Oftentimes, I've stored the derived class as a pointer to the base and then performed a cast to convert when or if I needed access to the derived class's methods.
Exactly, this is the correct way to access the derived class members. As then the compiler will know the type of the object and can actually find BFunc and call it. Now, if the type is not really a B, then you have undefined behavior, but yes, this is what one should do.
Also, please get your terminology right:
the class is instantiated as follows
If there are no templates involved, then there is no instantiation happening. The only thing you are doing here is declaring or more specifically defining a variable named tObject.

Answer 2

// The declaration of the reference as a reference to the base class is not a problem, and is actually performed in some STL implementations.
class A
{
};
class B : public A
{
public:
void f1() {}
};
int main()
{
A * a = new B; // You know its OK using pointers
A & a2 = *(new B); // Also OK, a2 is a reference for a place in memory which is-a-kind-of A
// a2.f1(); // Will not compile - compiler only knows for sure its of type A
((B&) a2).f1(); // This will work. There is really a B there
return 0;
}