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Subject: Re: [boost] [Root Pointer] Benchmark
From: Phil Bouchard (philippeb8_at_[hidden])
Date: 2016-03-14 20:43:37

On 03/14/2016 08:12 PM, Glen Fernandes wrote:
> Comment out what from where? More simply, if someone wants to use
> root_ptr with a conforming C++ allocator (such as Allocator<T> in that
> example) what must they do?
> e.g. With shared_ptr they would use allocate_shared:
> shared_ptr<int> p1 = allocate_shared<int>(Allocator<int>(state));
> e.g. With vector they would just use the constructor:
> vector<int, Allocator<int> > v1(Allocator<int>(state));
> e.g. With function they would just use the constructor:
> function<void()> f1(allocator_arg, Allocator<void>(state));
> From the code that you showed, it looks like for block_ptr, they would
> need to do:
> nodealloc<int>::PoolType a1(state);
> root_ptr<int> p1 = new(a1) nodealloc<int>(a1);
> But first they need to define the nodealloc class template?! That
> doesn't seem right.

No I just scribbled something quickly but I need to clean up
node_base.hpp and make nodealloc<> generic.

> What happens if they want to pass constructor parameters as well?
> struct T { T(int, char, bool) { } };
> auto p1 = std::allocate_shared<T>(Allocator<T>(state), 5, 'g', false);
> How would they do that with root_ptr? Would they have to define a new
> nodealloc class template to do it?

No like I was saying nodealloc<> could be used for other instantiated
allocators as well.

And I could well easily define function wrappers as well to something
like "std::allocate_node<>" but ideally I would prefer not to because I
find it redundant if the placement operator new works correctly.

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