From hb at stat.berkeley.edu  Thu Jul  8 09:14:35 2010
From: hb at stat.berkeley.edu (Henrik Bengtsson)
Date: Thu, 8 Jul 2010 09:14:35 +0200
Subject: [Rd] pass-by-reference
In-Reply-To: <1278564749228-2281802.post@n4.nabble.com>
References: <1278564749228-2281802.post@n4.nabble.com>
Message-ID: <AANLkTil0RDbUb5LydnddH2OG5YLsZyxGYF8-wm1FxMnq@mail.gmail.com>

See packages R.oo and proto.

If you wish to do it yourself, you want to utilize environments for this.

/Henrik

On Thu, Jul 8, 2010 at 6:52 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)
>
> 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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>



[Rd] pass-by-reference
Henrik Bengtsson