RcppArmadillo and C++ division issue - rcpp

Question

A very simple question regarding RcppArmadillo. Trying to multiply a vector by a scalar, and getting different results depending on small changes in syntax.
Any ideas?
// [[Rcpp::depends("RcppArmadillo")]]
// [[Rcpp::export]]
arma::vec funtemp(arma::vec x)
{
// return(x/10); // this works
// return((1/10)*x); // this does not work
return(x*(1/10)); // this does not work
}

Answer 1

Ahh, the good ol' integer vs. double division problem in C++. Before we begin, note that: arma::vec is by default a double and 1, 10, 1/10 are all ints...
Let's take a look at your functions separately:
#include <RcppArmadillo.h>
// [[Rcpp::depends("RcppArmadillo")]]
// [[Rcpp::export]]
arma::vec funtemp_one(arma::vec x)
{
return(x/10); // this works
}
// [[Rcpp::export]]
arma::vec funtemp_two(arma::vec x)
{
return((1/10)*x); // this does not work
}
// [[Rcpp::export]]
arma::vec funtemp_three(arma::vec x)
{
return(x*(1/10)); // this does not work
}
Therefore, when we run through your problem we get:
> funtemp_one(1)
[,1]
[1,] 0.1
> funtemp_two(1)
[,1]
[1,] 0
> funtemp_three(1)
[,1]
[1,] 0
In the later functions (e.g. the 1/10), the operator/ that is being used is int based division. As a result, 2 ints enter and 1 int is returned. If the result is not divisible, then you end up with a zero being returned as it is outside of the integer scope.
In order to use the double version, which returns a double, at least one of the ints must be explicitly cast to a double. This happens by default in the first case as you have a double/int due to arma::vec's structure. The second and third cases have an int/int structure that can be dealt with in two ways: 1. use a .0 after the int or 2. explicitly cast the value as a double with double(int)
e.g.
// [[Rcpp::export]]
arma::vec funtemp_four(arma::vec x)
{
return(x*(1/10.0)); // this works
}
// [[Rcpp::export]]
arma::vec funtemp_five(arma::vec x)
{
return(x*(1/double(10))); // this works
}
Will give what you expect:
> funtemp_four(1)
[,1]
[1,] 0.1
> funtemp_five(1)
[,1]
[1,] 0.1