Why is `make_unique` disallowed?

Question

^{Assume namespace std throughout.}

The C++14 committee draft N3690 defines std::make_unique thus:

[n3690: 20.9.1.4]: unique_ptr creation    [unique.ptr.create]

template <class T, class... Args> unique_ptr<T> make_unique(Args&&... args);

1 Remarks: This function shall not participate in overload resolution unless T is not an array.
2 Returns: unique_ptr<T>(new T(std::forward<Args>(args)...)).

template <class T> unique_ptr<T> make_unique(size_t n);

3 Remarks: This function shall not participate in overload resolution unless T is an array of unknown bound.
4 Returns: unique_ptr<T>(new typename remove_extent<T>::type[n]()).

template <class T, class... Args> unspecified make_unique(Args&&...) = delete;

5 Remarks: This function shall not participate in overload resolution unless T is an array of known bound.

Now, this seems to me to be about as clear as mud, and I think it needs more exposition. But, this editorial comment aside, I believe I've decoded the meanings of each variant:

1. template <class T, class... Args> unique_ptr<T> make_unique(Args&&... args);

   Your bog-standard make_unique for non-array types. Presumably the "remark" indicates that some form of static assertion or SFINAE trick is to prevent the template from being successfully instantiated when T is an array type.

   At a high-level, see it as the smart-pointer equivalent to T* ptr = new T(args);.

2. template <class T> unique_ptr<T> make_unique(size_t n);

   A variant for array types. Creates a dynamically-allocated array of n × Ts, and returns it wrapped in a unique_ptr<T[]>.

   At a high-level, see it as the smart-pointer equivalent to T* ptr = new T[n];.

3. template <class T, class... Args> unspecified make_unique(Args&&...)

   Disallowed. "unspecified" would probably be unique_ptr<T[N]>.

   Would otherwise be the smart-pointer equivalent to something like the invalid T[N]* ptr = new (keep_the_dimension_please) (the_dimension_is_constexpr) T[N];.

First of all, am I correct? And, if so, what's going on with the third function?

• If it's there to disallow programmers from attempting to dynamically-allocate an array while providing constructor arguments for each element (just as new int[5](args) is impossible), then that's already covered by the fact that the first function cannot be instantiated for array types, isn't it?

• If it's there to prevent the addition to the language of a construct like T[N]* ptr = new T[N] (where N is some constexpr) then, well, why? Wouldn't it be completely possible for a unique_ptr<T[N]> to exist that wraps a dynamically-allocated block of N × Ts? Would this be such a bad thing, to the extent that the committee has gone out of its way to disallow its creation using make_unique?

Why is make_unique<T[N]> disallowed?

Answer 1

Quoting from the original proposal:

T[N]

As of N3485, unique_ptr doesn't provide a partial specialization for T[N]. However, users will be strongly tempted to write make_unique<T[N]>(). This is a no-win scenario. Returning unique_ptr<T[N]> would select the primary template for single objects, which is bizarre. Returning unique_ptr<T[]> would be an exception to the otherwise ironclad rule that make_unique<something>() returns unique_ptr<something>. Therefore, this proposal makes T[N] ill-formed here, allowing implementations to emit helpful static_assert messages.

The author of the proposal, Stephan T. Lavavej, illustrates this situation in this video on Core C++ (courtesy of chris), starting from minute 1:01:10 (more or less).

Answer 2

To me the third overload looks superfluous as does not change the fact that the other overloads won't match T[N] and it does not seem to help to generate better error messages. Consider the following implementation:

template< typename T, typename... Args >
typename enable_if< !is_array< T >::value, unique_ptr< T > >::type
make_unique( Args&&... args )
{
  return unique_ptr< T >( new T( forward< Args >( args )... ) );
}

template< typename T >
typename enable_if< is_array< T >::value && extent< T >::value == 0, unique_ptr< T > >::type
make_unique( const size_t n )
{
  using U = typename remove_extent< T >::type;
  return unique_ptr< T >( new U[ n ]() );
}

When you try to call std::make_unique<int[1]>(1), the error message lists both candidates as disabled by enable_if. If you add the third, deleted overload, the error message lists three candidates instead. Also, since it is specified as =delete;, you can not provide a more meaningful error message in the third overload's body, e.g., static_assert(sizeof(T)==0,"array of known bound not allowed for std::make_shared");.

Here's the live example in case you want to play with it.

The fact that the third overload ended up in N3656 and N3797 is probably due to the history of how make_unique was developed over time, but I guess only STL can answer that :)