How can std::unique_ptr have no size overhead?

Question

If the size of an empty class can't be 0, what magic is doing std::tuple so the sizeof of unique_ptr is returning 8 in a 64 bit machine?

In unique_ptr the member is defined as:

  typedef std::tuple<typename _Pointer::type, _Dp>  __tuple_type;                    __tuple_type  _M_t; 

Where _Dp is the deleter class.

Compiler is gcc version 4.7.1 (Debian 4.7.1-7)

Answer 1

The reason is that the typename _Dp = default_delete<_Tp> is an empty class and the tuple template employs empty base class optimization.

If you instantiate the unique_ptr with a non-default delete, you should see the size increase.

Answer 2

unique_ptr as specified can have zero overhead because the only thing needed to implement it is to modify the process of copying/moving a raw pointer; no additional information is necessary. Therefore unique_ptr doesn't need to store anything besides the pointer and can be the same size as a pointer.

As to how your particular implementation, achieves that; Only most derived types need to have a size greater than zero. Empty base classes can take up zero bytes. It's quite common for standard library implementations to take advantage of the so-called 'empty base class' optimization for all kinds of things, from stateless allocators in containers to tuple.