Why does shared_ptr<T>::use_count() return a long instead of an unsigned type?

Question

shared_ptr observers 20.8.2.2.5 C++14 Final Draft (n4296)

   long use_count() const noexcept; 

Returns: the number of shared_ptr objects, *this included, that share ownership with *this, or 0 when *this is empty.

[Note: use_count() is not necessarily efficient. — end note]

Answer 1

According to this page

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2003/n1450.html

The return type of use_count is signed to avoid pitfalls such as p.use_count() > -1 evaluating to false.

with a reference to

John Lakos, Large-Scale C++ Software Design, section 9.2.2, page 637, Addison-Wesley, July 1996, ISBN 0-201-63362-0.

Basically, it looks like a nanny-grade decision, tailored for freshman year software developers.