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

Subject: Re: [boost] [TypeErasure] Forward constructors and binded types
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2012-07-18 11:44:10

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<construct, _a&, _b, _c> t(vec, i, d);
>>     any<construct, _a> 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<construct, _&a, _b, _c>.
> 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