C++ std::unique_ptr : Why isn't there any size fees with lambdas?

Question

I am reading \"Effective Modern C++\". In the item related to std::unique_ptr it\'s stated that if the custom deleter is a stateless object, then no size fees o

Answer 1

From a unique_ptr implementation:

template>
class unique_ptr
{
public:
   // public interface...

private:

  // using empty base class optimization to save space
  // making unique_ptr with default_delete the same size as pointer

  class _UniquePtrImpl : private deleter_type
  {
  public:
     constexpr _UniquePtrImpl() noexcept = default;

     // some other constructors...

     deleter_type& _Deleter() noexcept
     { return *this; }

     const deleter_type& _Deleter() const noexcept
     { return *this; }

     pointer& _Ptr() noexcept
     { return _MyPtr; }

     const pointer _Ptr() const noexcept
     { return _MyPtr; }

  private:
     pointer   _MyPtr;

  };

  _UniquePtrImpl   _MyImpl;

};

The _UniquePtrImpl class contains the pointer and derives from the deleter_type.

If the deleter happens to be stateless, the base class can be optimized so that it takes no bytes for itself. Then the whole unique_ptr can be the same size as the contained pointer - that is: the same size as an ordinary pointer.