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How do I modify an argument being passed to a function in R? In C++ this would be pass by reference.

abc <- function(x) {x<-5}

I would like g to be set to 5.

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my feeling is that the R way to do is to say g = abc() where abc returns 5. why this specific syntax? –  Paul Hiemstra Dec 7 '11 at 17:52

5 Answers 5

up vote 4 down vote accepted

There are ways as @Dason showed, but really - you shouldn't!

The whole paradigm of R is to "pass by value". @Rory just posted the normal way to handle it - just return the modified value...

Environments are typically the only objects that can be passed by reference in R.

But lately new objects called reference classes have been added to R (they use environments). They can modify their values (but in a controlled way). You might want to look into using them if you really feel the need...

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There's also the proto package. –  Joshua Ulrich Dec 7 '11 at 18:00

There has got to be a better way to do this but...

abc <- function(x){eval(parse(text = paste(substitute(x), "<<- 5")))}
g <- 4

gives the output

[1] 5
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Am I missing something as to why you can't just do this?

g <- abc(g)
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With a slight modification to abc so that it actually returns a value, yes. –  joran Dec 7 '11 at 17:51
yes that would work. the point is to modify the value within teh function though without throwing around all these big objects and making copies of them. –  Alex Dec 7 '11 at 17:53

I think that @Dason's method is the only way to do it theoretically, but practically I think R's way already does it.

For example, when you do the following:

y <- c(1,2)
x <- y

x is really just a pointer to a the value c(1,2). Similarly, when you do

abc <- function(x) {x <- 5; x}
g <- abc(g)

It is not that you are spending time copying g to the function and then copying the result back into g. I think what R does with the code

g <- abc(g)


  1. The right side is looked at first. An environment for the function abc is set up.
  2. A pointer is created in that environment called x.
  3. x points to the same value that g points to.
  4. Then x points to 5
  5. The function returns the pointer x
  6. g now points to the same value that x pointed to at the time of return.

Thus, it is not that there is a whole bunch of unnecessary copying of large options.

I hope that someone can confirm/correct this.

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interesting! i would be curious to know if this is right. in my program that has a df that is approximately a gig, it seems that df gets copied several times as my RAM usage grows pretty large –  Alex Dec 7 '11 at 19:20
@Alex: In this case, yes - but only since it is trivial. If you modify a part of a large vector it is typically (but not always) copied. x <- runif(1e6); y <- x; x[4] <- 42 makes a copy so that x and y are now different. –  Tommy Dec 7 '11 at 20:56

I have a solution similar to @Dason's, and I am curious if there are good reasons not to use this or if there are important pitfalls I should be aware of:

changeMe = function(x){
assign(deparse(substitute(x)), "changed", env=.GlobalEnv)
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