函数在RcppArmadillo中通过引用传递 [英] function pass by reference in RcppArmadillo

问题描述

我有一个用RcppArmadillo样式编写的函数，我想将其用于在调用环境中更改变量.我知道这样做是不明智的，但对我来说是有帮助的.具体来说，我正在尝试:

I have a function written in RcppArmadillo style and I want to use it for changing variables in the calling environment. I know it is not advisable to do things like this, but it's helpful in my case. Concretely, I'm trying this:

#include <RcppArmadillo.h>
#include <iostream>

//[[Rcpp::export]]
void myfun(double &x){
  arma::mat X = arma::randu<arma::mat>(5,5);
  arma::mat Y = X.t()*X;
  arma::mat R1 = chol(Y);

  x = arma::det(R1);
  std::cout << "Inside myfun: x = " << x << std::endl;
}


/*** R
x = 1.0  // initialize x 
myfun(x) // update x to a new value calculated internally
x        // return the new x; it should be different from 1
*/ 

我想念什么?为什么不起作用?

What am I missing ? Why is not working?

推荐答案

double不是本机R类型(因此总是复制 )，并且没有传递引用是可能的.

A double is not a native R type (so there is always a copy being made) and no pass-through reference is possible.

相反，请使用Rcpp::NumericVector，它是SEXP类型的代理.这有效:

Instead, use Rcpp::NumericVector which is a proxy for a SEXP type. This works:

R> sourceCpp("/tmp/so44047145.cpp")

R> x = 1.0  

R> myfun(x) 
Inside myfun: x = 0.0361444

R> x        
[1] 0.0361444
R> 

下面是完整的代码，另有一两次小修:

Below is the full code with another small repair or two:

#include <RcppArmadillo.h>

// [[Rcpp::depends(RcppArmadillo)]]

//[[Rcpp::export]]
void myfun(Rcpp::NumericVector &x){
  arma::mat X = arma::randu<arma::mat>(5,5);
  arma::mat Y = X.t()*X;
  arma::mat R1 = chol(Y);

  x[0] = arma::det(R1);
  Rcpp::Rcout << "Inside myfun: x = " << x << std::endl;
}


/*** R
x = 1.0  // initialize x 
myfun(x) // update x to a new value calculated internally
x        // return the new x; it should be different from 1
*/ 

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