On 20 March 2018 at 11:42, Iñaki Úcar wrote:
| I may have my own reasons for that, but that's not the point here. The
| point is that I expected a homogeneous behaviour across Rcpp classes
| when any object is passed to operator<< (i.e., print *something*).
|
| By grepping the source, I discovere
2018-03-20 4:33 GMT+01:00 Tim Keitt :
> Why not something like:
>
> Rcpp::sourceCpp(code='
> #include
>using Rcpp::Rcout;
>
>// [[Rcpp::export]]
>void print_addr(SEXP x){
> Rcout << static_cast(x) << std::endl;
> }')
>
> I'm not sure why one would expect Rcpp types t
http://www.keittlab.org/
On Wed, Mar 14, 2018 at 12:51 PM, Iñaki Úcar wrote:
> Hi all,
>
> I'm not sure whether this is a bug or not, so I think this is the
> right place to start with. Consider the following code:
>
> Rcpp::sourceCpp(code='
> #include
> using namespace Rcpp;
>
> // [[Rcp
On 19 March 2018 at 08:24, Iñaki Úcar wrote:
| Hi -- I hope my last email didn't hit the spam folder. :-)
It didn't but it is a little hard to say anything here. Sometimes the
compiler needs help with disambiguation as you said.
Dirk
| Iñaki
|
| 2018-03-14 18:51 GMT+01:00 Iñaki Úcar :
| > Hi
Hi -- I hope my last email didn't hit the spam folder. :-)
Iñaki
2018-03-14 18:51 GMT+01:00 Iñaki Úcar :
> Hi all,
>
> I'm not sure whether this is a bug or not, so I think this is the
> right place to start with. Consider the following code:
>
> Rcpp::sourceCpp(code='
> #include
> using nam
Hi all,
I'm not sure whether this is a bug or not, so I think this is the
right place to start with. Consider the following code:
Rcpp::sourceCpp(code='
#include
using namespace Rcpp;
// [[Rcpp::export]]
void print_fun(Function x) {
Rcout << x << std::endl;
}
// [[Rcpp::export]
