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)

This may also be implementation-dependent, so you may want to include which compiler you're talking about as well. — Kevin Anderson, Nov 19 '12 at 19:09
I'm not sure how `std::tuple` relates to `std::unique_ptr` in your question. Could you clarify? — Cameron, Nov 19 '12 at 19:09

score 62 · Accepted Answer · answered Nov 19 '12 at 19:23

62

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.

answered Nov 19 '12 at 19:23

chill

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To rephrase a little: when instantiated by itself, a class must have non-zero size, but when instatiated as a base class, zero size is allowed. – Jerry Coffin Nov 19 '12 at 19:58
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Not really true: I instantiated with my own defined deleter, and got an `unique_ptr` with the size of a pointer. All it needs is the deleter to have a "natural size" of zero. – lvella Jul 06 '18 at 18:26

bames53 · Answer 2 · 2012-11-20T06:50:28.863

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.

Thanks for the answer, voted +1 I accepted chill as he has less score and also provided a valid answer. — piotr, Nov 19 '12 at 19:48

How can std::unique_ptr have no size overhead?

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