Re: [boost] [TypeErasure] Forward constructors and binded types

Subject: Re: [boost] [TypeErasure] Forward constructors and binded types
From: Vicente Botet (vicente.botet_at_[hidden])
Date: 2012-07-18 12:52:17

Steven Watanabe-4 wrote
> 
> AMDG
> 
> On 07/18/2012 06:57 AM, Larry Evans wrote:
>> On 07/18/12 01:29, Vicente J. Botet Escriba wrote:
>>> Hi,,
>>>
>>> In the example from the doc
>> 
>> http://steven_watanabe.users.sourceforge.net/type_erasure/libs/type_erasure/doc/html/boost_typeerasure/multi.html
>> 
>>>
>>>     typedef mpl::vector<
>>>         copy_constructible<_a>,
>>>         copy_constructible<_b>,
>>>         copy_constructible<_c>,
>>>         constructible<_a(const _b&, const _c&)>
>>>     > construct;
>>>
>>>     std::vector<double> vec;
>>>     int i = 10;
>>>     double d = 2.5;
>>>     tuple&lt;construct, _a&amp;, _b, _c&gt; t(vec, i, d);
>>>     any&lt;construct, _a&gt; v(get<1>(t), get<2>(t));
>>>
>>> I don't see which type is constructed and stored in v and how this type
>>> is given. That is I don't see how _a is binded. Could you help me?
>>>
>> 
>> I'd guess that a std::vector<double>&, more specifically the
>> vec is stored as the _a& in the tuple&lt;construct, _&amp;a, _b, _c&gt;.
>> Hmm...  Now I'm not sure anymore.  Maybe it's a new std::vector<double>
>> that's stored and the std::vector<double> CTOR args are
>> 10 and double.  IOW the equivalent of:
>> 
>>   std::vector<double> v_a(10,2.5)
>> 
>> is stored in v?
>> 
> 
> Yes.
> 
>> Could the documentation on this be clarified a bit?
>> 
> 
> What needs to be clarified?  This is the
> only reasonable behavior:
> The arguments obviously match
> constructible<_a(const _b&, const _c&)> right?
> Therefore this constructor is the one called.
> _a was bound to std::vector<double>,
> so this is the type created.
> 
> In Christ,
> Steven Watanabe
> 

My question is how the bind of _a to std::vector<double> if the t variable
is not passed in the constructor of any? I'm sure I'm missing something
evident.

Vicente

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