如何将 C++ 对象（双 *）作为 SEXP 返回给 R？

Question

I have to write a linear regression program and invoke it using R. This is my code.It is not able to return back theta as SEXP.

SEXP reg(Rcpp::NumericVector  x,Rcpp::NumericVector  y){
    int i,n1,n2;
    n1=x.size();
    n2=y.size();
    if(n1!=n2)
        cout<<"Error";
    else{
        double wx[n1], wy[n1];
        for(i=0;i<n1;i++){
            wx[i]=x[i];
            wy[i]=y[i];
        }
        int iterations=1500;
        //default learning rate;
        double alpha=0.01;
        double *theta = new double[2];
        double *J = new double[iterations];
        theta = gradient_descent(wx,wy, alpha, iterations, J, n1,theta);
        return(theta);
    }
}

It gives error:Cannot convert 'double *' to 'SEXP' in return. I want some solution for this.

Answer 1

You function is improper. Try like bellow (not tested):

SEXP reg(Rcpp::NumericVector  x,Rcpp::NumericVector  y){
    int i,n1,n2;
    SEXP out = PROTECT(allocVector(REALSXP, 2));
    n1=x.size();
    n2=y.size();
    if(n1!=n2)
        cout<<"Error";
    else{
        double wx[n1], wy[n1];
        for(i=0;i<n1;i++){
            wx[i]=x[i];
            wy[i]=y[i];
        }
        int iterations=1500;
        //default learning rate;
        double alpha=0.01;
        double *theta = REAL(out);
        double *J = new double[iterations];
        theta = gradient_descent(wx,wy, alpha, iterations, J, n1,theta);
    }
    UNPROTECT(1);
    return(out);
}

I do not know your function gradient_descent and who should care about freeing resource pointed by J . Do not forget to free it.

Very good useful manual is here http://adv-r.had.co.nz/C-interface.html