On 7/18/12 5:44 PM, Steven Watanabe wrote:
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/...
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:
Just spelling out some thoughts I had while reading the docu: Even though, it would help to have this information in the docu. The way it is written now feels a bit like playing Jeopardy, you present solutions and the reader tries to figure out the problem. additionally I do not quite gasp on what I could use this feature for. the above is much more complex as using any<...> v = std::vector<double>(10, 2.5); what am I buying with it? And couldn't another syntax like any<...> v(binding_of(vec), (int)10, (double)2.5); be made to work, and be much more comfortable? regards Fabio