What should be the behavior of unique_ptr in this situation?

Question

Say I have the following:

std::unique_ptr<A> pA; pA(new A); 

In this confusing example, what should be the behavior of pA(new A); ?

As far as I can tell, in MSVC2010 void operator()(T*) const; from default_delete is called immediately after returning new and immediately deletes the pointer. If g ++ (4.7.0) gave me an error no match for call (std::unique_ptr<A>)(A*) .

Answer 1

The code should not compile. std::unique_ptr does not overload operator() .

Visual C ++ 2011 Developer Preview correctly rejects code. Visual C ++ 2010 accepts an error code in its implementation of std::unique_ptr .

Answer 2

MSVC uses stateless determiner optimization for unique_ptr , i.e. uses optimization with an empty base class and simply inherits from the deleter. Unfortunately, inheritance is public , so you have access to the overloaded operator() default_delete functor.