Re: [Rd] pass-by-reference

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Thu, 08 Jul 2010 07:02:31 -0400

On Thu, Jul 8, 2010 at 12:52 AM, Paul Bailey <pdbailey_at_umd.edu> wrote:
>
> I'm working with a large object that I want to modify slightly in a function.
> Pass-by-reference would make a lot of sense, but I don't know how to do it.
>
> I've searched this archive and thought that I can do something like
>
> f <- function(x) {
>  v1 <- list(a=x,b=3)
>  g(x)
>  v1
> }
> g <- function(x) {
>  frame <- parent.frame()
>  assign("v1",list(a=x,b=x),frame)
> }
> f(4)
> returns list(a=4,b=4)
>
> but what if I wanted to make v1[[1]] = v1[[1]] + v1[[2]] without creating a
> copy of v1?
>
> f2 <- function(x) {
>  v1 <- list(a=x,b=3)
>  g2(x)
>  v1
> }
> g2 <- function(x) {
>  frame <- parent.frame()
>  v1 <- get("v1",envir=frame)
>  v1[[1]] <- v1[[1]] + v1[[2]]
> }
> f2(4)

Try:

g2 <- function(x, env = parent.frame()) {

   with(env, v1[[1]] <- v1[[1]] + v1[[2]]) }

or

g2 <- function(x, env = parent.frame()) {

   env$v1[[1]] <- env$v1[[1]] + env$v1[[2]] }

Sometimes the reason people want pass by reference is not so much for efficiency is but that they are trying to recreate an object oriented structure without realizing it. In that case the packages referred to by Henrik would be useful.

>
> but this fails. (it returns list(a=4,b=3) because v1 was copied into g2, not
> passed by reference) Is there a way to do this?
> --
> View this message in context:
http://r.789695.n4.nabble.com/pass-by-reference-tp2281802p2281802.html
> Sent from the R devel mailing list archive at Nabble.com.
>
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>



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https://stat.ethz.ch/mailman/listinfo/r-devel Received on Thu 08 Jul 2010 - 11:06:40 GMT

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