Dear Dirk,
Thank you very much for this. I modified slightly your own code so that
it is very similar (for the names of the arguments) to the one on
StackOverflow:
http://stackoverflow.com/questions/26181068/rcpp-error-not-compatible-with-requested-type
This gives me this code saved in /tmp/pierre.cpp:
-----------------------------------------------------------------------------------------------
#include <Rcpp.h>
// [[Rcpp::export]]
Rcpp::List gensampleRcpp2(Rcpp::Function rlawfunc, Rcpp::IntegerVector n) {
Rcpp::RNGScope __rngScope;
return Rcpp::List::create(Rcpp::Named("sample") = rlawfunc(n),
Rcpp::Named("law.name") = " ",
Rcpp::Named("law.pars") = R_NilValue);
}
// [[Rcpp::export]]
Rcpp::List compQuantRcpp2(Rcpp::IntegerVector n, Rcpp::IntegerVector M,
Rcpp::Function Rlaw) {
int i;
GetRNGstate();
for (i=1;i<=M[0];i++) {
Rcpp::List resultsample = gensampleRcpp2(Rlaw, n);
Rcpp::NumericVector mysample = resultsample["sample"];
}
PutRNGstate();
return Rcpp::List::create(Rcpp::Named("law.pars") = "");
}
/*** R
res <- compQuantRcpp2(n=50, M=10^3, Rlaw=rnorm)
print(str(res))
*/
-----------------------------------------------------------------------------------------------
Please have a look to compare both codes. They are very similar.
I then issue this command in R:
for (i in 1:100) Rcpp::sourceCpp("tmp/pierre.cpp")
Magic! No error occur.
So now I will try to implement all this in a package (I guess I should
read more carefully the Attributes vignette that you told me about. Am I
right? Is it the good document to read for that purpose?).
By the way, since both codes are such similar, would you have an
explanation on what goes wrong with what I have written on StackOverflow?
Regards,
Pierre L.
Le 03/10/2014 12:49, Dirk Eddelbuettel a écrit :
Oh, and I missed one call to a subordinate function. So add
// [[Rcpp::export]]
SEXP newCompQuant2(int n, int M, Rcpp::Function f) {
SEXP res = compquantRcpp2(n, M, f);
return res;
}
and at the bottom call it:
res <- newCompQuant2(n=50, M=10^3, f=rnorm)
print(str(res))
and then the result is
R> res <- newCompQuant2(n=50, M=10^3, f=rnorm)
R> print(str(res))
List of 1
$ law.pars: chr ""
NULL
R>
just like you were told as well on StackOverflow.
Dirk
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