Why do C++ classes without member variables occupy space?

Question

I found that both MSVC and GCC compilers allocate at least one byte per each class instance even if the class is a predicate with no member variables (or with just static member variables). The following code illustrates the point.

#include <iostream>

class A
{
public:
   bool operator()(int x) const
   {
      return x>0;
   }
};

class B
{
public:
   static int v;
   static bool check(int x)
   {
      return x>0;
   }
};

int B::v = 0;

void test()
{
   A a;
   B b;
   std::cout << "sizeof(A)=" << sizeof(A) << "\n"
             << "sizeof(a)=" << sizeof(a) << "\n"
             << "sizeof(B)=" << sizeof(B) << "\n"
             << "sizeof(b)=" << sizeof(b) << "\n";
}

int main()
{
   test();
   return 0;
}

Output:

sizeof(A)=1
sizeof(a)=1
sizeof(B)=1
sizeof(b)=1

My question is why does compiler need it? The only reason that I can come up with is ensure that all member var pointers differ so we can distinguish between two members of type A or B by comparing pointers to them. But the cost of this is quite severe when dealing with small-size containers. Considering possible data alignment, we can get up to 16 bytes per class without vars (?!). Suppose we have a custom container that will typically hold a few int values. Then consider an array of such containers (with about 1000000 members). The overhead will be 16*1000000! A typical case where it can happen is a container class with a comparison predicate stored in a member variable. Also, considering that a class instance should always occupy some space, what type of overhead should be expected when calling A()(value) ?

Answer 1

It’s necessary to satisfy an invariant from the C++ standard: every C++ object of the same type needs to have a unique address to be identifiable.

If objects took up no space, then items in an array would share the same address.