On Thu, Sep 4, 2008 at 4:51 PM, Joaquin M Lopez Munoz <joaquin@tid.es> wrote:
Because identity<T> operates at run time as a functor taking objects of type T:
identity<std::string> i; std::string str="hello"; std::cout<<i(str)<<std::endl; // prints hello
Thank you very much for the response. Now I understand that identity<> is not an MPL construct. I was reading through the identity.hpp file and I think I understand it now. It can be used to convert chained pointers and smart pointers to references. For example, int ****i; identity<int> get_ref; get_ref(i); // This will return me a reference to the integer Let me know if I got that right. But since the template arguments are type names is there any reason why we need to use identity in template arguments? For example, in the code snippet given in the documentation, (my question embedded in the code fragment below as comments), typedef multi_index_container< employee, indexed_by< ordered_unique<identity<employee> >, // Why is identity required here? // Why is ordered_unique<employee> not sufficient? ordered_non_unique<member<employee,std::string,&employee::name> >, ordered_unique<member<employee,int,&employee::ssnumber> >
employee_set;
Since the std::set doesn't require anything other than the type name of the elements it would hold, why does boost::multi_index::ordered_unique need this identity<>? What are the additional benefits of this added complexity? Rgds, anna -- Abusive Language on Internet http://missingrainbow.blogspot.com/2008/08/abusive-language-on-internet.html