Hi All, The Writing R-extensions document says this on whether I should prefer .C or .Call: "Before you decide to use .Call or .External, you should look at other alternatives. First, consider working in interpreted R code; if this is fast enough, this is normally the best option. You should also see if using .C is enough. If the task to be performed in C is simple enough requiring no call to R, .C suffices. The new interfaces are relatively recent additions to S and R, and a great deal of useful code has been written using just .C before they were available. The .Call and .External interfaces allow much more control, but they also impose much greater responsibilities so need to be used with care. Neither .Call nor .External copy their arguments. You should treat arguments you receive through these interfaces as read-only." So I am using .C; I am not referencing R objects in my code and .C appears simpler for me to use. First, am I correct about using .C? Next, my C function has a char* argument and I cannot figure out how to pass that from my R code. The argument specifies the mode in which computations will be performed by the function; it is a single character actually. (The code was written as a standalone program and I want to make it into an R package.) For the R character type the doc lists char** as equivalent. I am not following that at all... That is used for an array of pointers (to chars) not the chars themselves. I am confused. Finally, I read that .C does a lot of copying of the arguments on call and return. How can I avoid that? Is DUP=FALSE to be used for that? I am hesitant because I read that DUP=FALSE is not to be used with characters and further that I should treat the arguments read-only in the C-code. (and I am told several times that it is dangerous...) In my code a matrix is passed to the C code (an initial value) and it is modified (updated) by the C code to be returned. So I have something like this presently: tmp <- .C("foo", DUP = FALSE, NAOK = TRUE, as.character('d'), X = as.double(X), as.double(Y)) where Y will not be modified by foo() and X is changed and treated as a return value. Your insight would be appreciated. -Matyas [[alternative HTML version deleted]]