Re: [boost] [Root Pointer] Benchmark

Subject: Re: [boost] [Root Pointer] Benchmark
From: Gavin Lambert (gavinl_at_[hidden])
Date: 2016-03-15 20:42:39

On 16/03/2016 13:31, Phil Bouchard wrote:
> On 03/15/2016 07:25 PM, Glen Fernandes wrote:
>>
>> You mentioned a while back that you preferred:  p.reset(new
>> X(args...)); because it was intuitive, and in your opinion more
>> intuitive than: p = f<X>(args...);
>>
>> Nobody would have any issue with that. But people would certainly have
>> issue with: Y::Z b(a); p = new (b) Q(b, args...); compared to  p =
>> g<X>(a, args...);
>
> Well first if you have a as an r-value:
> p = g<X>(a, args...)

That's an lvalue, not an rvalue.

> That means the allocator parameter must be constant:
> g<X>(Allocator const &, arg...)
>
> If it's constant it won't be able to change its internal state.  I
> thought that was the purpose of having instantiable allocators?

There's no need for (and no reason for) the allocator to be a const 
reference.

> I can write a wrapper function for:
> p1 = new (a1) node(a1, 1, 'a');
>
> But not for the declaration of the allocator:
> boost::node<U, Allocator<U> >::allocator_type a1(n1, m1);
>
> Why?  Because you allocate node<U>s, not Us.  I could write some
> node_traits<> helper:
>
> template <typename T, template <typename> class A>
>      {
>          typedef typename boost::node<T, A<T> >::allocator_type
> allocator_type;
>      };
>
> So that the declaration becomes:
> boost::node_traits<U, Allocator>::allocator_type a1(n1, m1);
>
> But that's all I can do.

Admittedly I'm not familiar with the changes to allocators in C++11 yet, 
but C++03 allocators support accepting an allocator (even a stateful 
one) of one type and then using it to make allocations of a different 
type, via rebind and a templated constructor.  std::list and friends use 
that functionality extensively.