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


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
>> 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?


> 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

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