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
> On 07/18/2012 06:57 AM, Larry Evans wrote:
>> On 07/18/12 01:29, Vicente J. Botet Escriba wrote:
>>> In the example from the doc
>>> typedef mpl::vector<
>>> 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
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
-- View this message in context: http://boost.2283326.n4.nabble.com/TypeErasure-Forward-constructors-and-binded-types-tp4633181p4633215.html Sent from the Boost - Dev mailing list archive at Nabble.com.