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259

answers:

4
+3  A: 

It only has a vtable, no data fields.

Daniel Earwicker
+6  A: 

I suppose it's to differentiate it from "empty", which is what you get if compile a class with no members at all. "nearly-empty" seems to mean it hasa vtable and nothing else.

anon
+6  A: 

C++ has something called an "empty base optimization". If a class has no members, it doesn't need to take up space when it's used as a base class. An example of why this is important is std::unary_function<T, U>. It exists to provide you with an easy set of typedefs. Those typedefs should not contribute to the size of your functor class.

If you have a base class with a vtable pointer, this pointer can likely be shared with the derived class. You simply create a vtable for the derived class which adds its own methods after that of the base class.

You can now achieve a similar "no extra overhead" base class. Apparently GCC calls that "nearly empty".

MSalters
Sounds plausible. Do you have a reference to confirm it?
Tobias
+4  A: 

The C++ ABI provides a definition of "nearly empty" classes and an interesting discussion of how they affect vtable construction:

A class that contains a virtual pointer, but no other data except (possibly) virtual bases. In particular, it:

  • has no non-static data members other than zero-width bitfields,
  • has no direct base classes that are not either empty, nearly empty, or virtual,
  • has at most one non-virtual, nearly empty direct base class, and
  • has no proper base class that is empty, not morally virtual, and at an offset other than zero.

I ran across this while researching the effect of nearly empty virtual bases on object size, vtable size, and virtual call overhead.

Trevor Robinson