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Why does shared_ptr have allocate_shared while unique_ptr does not have allocate_unique?
I would like to make a unique_ptr using my own allocator: do I have to allocate the buffer myself and then assign it to a unique_ptr?
This seems like an obvious idiom.

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do I have to allocate the buffer myself and then assign it to a unique_ptr?

Not just a buffer, a pointer to an object. But the object might need to be destroyed by the allocator, and the memory definitely needs to deallocated by the allocator, so you also need to pass the allocator the the unique_ptr. It doesn't know how to use an allocator, so you need to wrap it in a custom deleter, and that becomes part of the unique_ptr's type.

I think a generic solution would look something like this:

#include <memory>

template<typename Alloc>
struct alloc_deleter
{
  alloc_deleter(const Alloc& a) : a(a) { }

  typedef typename std::allocator_traits<Alloc>::pointer pointer;

  void operator()(pointer p) const
  {
    Alloc aa(a);
    std::allocator_traits<Alloc>::destroy(aa, std::addressof(*p));
    std::allocator_traits<Alloc>::deallocate(aa, p, 1);
  }

private:
  Alloc a;
};

template<typename T, typename Alloc, typename... Args>
auto
allocate_unique(const Alloc& alloc, Args&&... args)
{
  using AT = std::allocator_traits<Alloc>;
  static_assert(std::is_same<typename AT::value_type, std::remove_cv_t<T>>{}(),
                "Allocator has the wrong value_type");

  Alloc a(alloc);
  auto p = AT::allocate(a, 1);
  try {
    AT::construct(a, std::addressof(*p), std::forward<Args>(args)...);
    using D = alloc_deleter<Alloc>;
    return std::unique_ptr<T, D>(p, D(a));
  }
  catch (...)
  {
    AT::deallocate(a, p, 1);
    throw;
  }
}

int main()
{
  std::allocator<int> a;
  auto p = allocate_unique<int>(a, 0);
  return *p;
}

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