Are data members allocated in the same memory space as their objects in C++?

Question

Say I\'ve got a class like this:

class Test
{
  int x;
  SomeClass s;
}

And I instantiate it like this:

Test* t = new Test;         


        

Answer 1

Each time you "instantiate" an object/symbol using a new (we are speaking C++ here), a new memory zone will be allocated for this object. If not, it will be put on the "local" memory zone.

The problem is that I have no standard definition for "local" memory zone.

An example

This means that, for example:

struct A
{
   A()
   {
      c = new C() ;
   }

   B b ;
   C * c ;
}

void doSomething()
{
   A aa00 ;
   A * aa01 = new A() ;
}

The object aa00 is allocated on the stack.

As aa00::b is allocated on a "local" memory according to aa00, aa00::b is allocated inside the memory range allocated by the new aa01 instruction. Thus, aa00::b is also allocated on stack.

But aa00::c is a pointer, allocated with new, so the object designed by aa00::c is on the heap.

Now, the tricky example: aa01 is allocated via a new, and as such, on the heap.

In that case, as aa01::b is allocated on a "local" memory according to aa01, aa00::b is allocated inside the memory range allocated by the new aa01 instruction. Thus, aa00::b is on the heap, "inside" the memory already allocated for aa01.

As aa01::c is a pointer, allocated with new, the object designed by aa01::c is on the heap, in another memory range than the one allocated for aa01.

Conclusion

So, the point of the game is:
1 - What's the "local" memory of the studied object: Stack of Heap?
2 - if the object is allocated through new, then it is outside this local memory, i.e., it is elsewhere on the heap
3 - if the object is allocated "without new", then it is inside the local memory.
4 - If the "local" memory is on the stack, then the object allocated without new is on the stack, too.
5 - If the "local" memory is on the heap, then the object allocated without new is on the heap, too, but still inside the local memory.

Sorry, I have no better vocabulary to express those concepts.