Mark A. Gibbs wrote:

I have a couple of questions about combining smart pointers and STL allocators.

First of all, am I right in assuming that I can use the shared_* templates right out of the box, as long as I define a custom deleter that uses the allocator's deallocate function?

But if that's so, how exactly is it done? Part of my confusion comes from the fact that the STL allocator documentation I've come across isn't exactly clear on the semantics of allocators (I don't have a copy of the standard). For instance, I'm not 100% sure whether or not I have to call construct() after allocate() and destroy() before deallocate().

Yes, you have to. allocate() gives you raw memory with the correct size and alignment, but there is no object in it. To create the object in that memory, you need to construct() it. Similarly, destroy() invokes the destructor and deallocate() only, well, deallocates.

This is what I have so far:

class allocator_helper { public: allocator_helper(allocator_type& a) : a_(a) {}

void operator()(void const* p) const { a_.deallocate(static_cast(p)); }

This should probably be void operator()(allocator_type::pointer p) const { a_.destroy(p); a_.deallocate(p, 1); }

private: allocator_type& a_; };

foo* bar = static_cast(get_allocator().allocate(1));

new(bar) foo(args); or get_allocator().construct(bar, foo(args));

shared_ptr<foo> p_foo(bar, allocator_helper(get_allocator()));

foo* baz = static_cast(get_allocator().allocate(10)); shared_array<foo> p_foo(baz, allocator_helper(get_allocator()));

Now, for scoped_* I would probably have to rewrite them to use the allocator, right? Or is there a way to use allocators with the existing scoped_* templates?

No, scoped_* do not support allocators. Since they match raw pointers in size, they can keep no additional state such as an instance of, or a reference to, an allocator.