R function closure environments

Question

I recently ran into some unexpected R behavior, which can be reproduced in its simplest form with the following code:

make.adder <- function(a) {function(x) {x + a}}
s <- list(1, 2)
adders <- lapply(s, make.adder)

We might now expect adders[[1]] to be a function which adds 1, and adders[[2]] to be a function which adds 2. However,

adders[[1]](1)

returns 3 (when we might've expected 2). Taking a look at the environment with

environment(adders[[1]])$a

returns 2, which is again surprising (but consistent). What's going on here?

We see similar behavior if we try to instead use a for loop:

adders <- list()
for (i in seq(1, 2)) {adders[[i]] <- make.adder(i)}

Again, adders[[1]] (1) returns 3. How can we create a list of 100 functions, such that the i'th function is make.adder(i)?

and try this question http://stackoverflow.com/questions/16129902/explain-a-lazy-evaluation-quirk — rawr, Nov 14 '15 at 22:29
And read through [here](http://adv-r.had.co.nz/Functions.html#function-arguments) which I might add contains nearly exactly that example. — joran, Nov 14 '15 at 22:31
@BenBolker I think the dupe rawr found might have the better answers. — joran, Nov 14 '15 at 22:34
OK, I'll vote to re-open so we can re-close it with the right dupe ... — Ben Bolker, Nov 14 '15 at 22:35

score 2 · Answer 1 · answered Nov 14 '15 at 22:27

I think you need to use force() to make sure the parameter is evaluated when you expect it to be. See ?force for more information ...

make.adder <- function(a) { force(a); function(x) {x + a}}
s <- list(1, 2)
adders <- lapply(s, make.adder)

adders[[1]](1) ## 2

1 Answers1