> NumericVector x = as<NumericVector>(x_); ( Btw, this is exactly the same as > NumericVector x(x_), both resulting in a shallow copy, isn't it? ) produces > a shallow copy, while > vector<double> x = as< vector<double> >(x_); produces a deep copy. > > Is it because for NumvericVector the as() function returns the address of > x_? But why does it work differently for vector<double>?
Hello Zhongyi, I think if you put the namespaces on, explicitly, it becomes clearer: Rcpp::NumericVector x = as<Rcpp::NumericVector>(x_); std::vector<double> x = as< std::vector<double> >(x_); (I'm assuming x_ is the parameter being passed in from R.) x_ is effectively a pointer to a block of memory owned by R. Rcpp classes are just wrappers around that same pointer. Hence a shallow copy is possible. std::vector is a block of memory allocated and owned by your C++ code, so the bytes need to be physically copied over. Darren P.S. You'll see a lot of C++ code that keeps the std:: namespace definition in, rather than a using namespace std; at the top of the file. It adds 5 characters each time, but on the other hand it *only* adds 5 characters each time. I don't know who invented this convention, or all the reasons it is good, but I personally do it that way, without regret. -- Darren Cook, Software Researcher/Developer http://dcook.org/work/ (About me and my work) http://dcook.org/blogs.html (My blogs and articles) _______________________________________________ Rcpp-devel mailing list Rcpp-devel@lists.r-forge.r-project.org https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/rcpp-devel
