R: How to return the exact form of `...` in `fn(...)` without evaluating `...`?

Question

Consider this code

fn = function(...) {
  # print the content of ... without evaluation?
}

I want the output of fn(a = b) to be "a = b" and fn(a = gn(b), 3~b~a, dd, e = 2 + f, h = hn(jn(cdf))) to be list("a=gn(b)", "3~b~a", "dd", "e=2+f", "h= hn(jn(cdf)))".

I can't seem to find the right NSE function for it. I prefer Base R so I understand the process better. The closest I got was this

fn = function(...) {
   res = rlang::enexprs(...)
   paste0(names(res), ifelse(names(res)=="",names(res) , "=") , sapply(res, capture.output))
}

score 4 · Answer 1 · answered May 02 '20 at 07:33

An alternative to match.call (see here) is the undocumented ...().

fn <- function(...) {
  x <- substitute(...())
  y <- ifelse(nchar(names(x)) > 0, paste(names(x), "=", x), as.character(x))
  as.list(y)
}

fn(a = b)
fn(a = gn(b), 3~b~a, dd, e = 2 + f, h = hn(jn(cdf)))

score 3 · Accepted Answer · answered May 02 '20 at 07:20

You can use match.call :

fn = function(...) {
  temp <- as.list(match.call())[-1]
  as.list(sub('^=', '', paste0(names(temp), '=', temp)))
}

fn(a = b)
#[[1]]
#[1] "a=b"

fn(a = gn(b), 3~b~a, dd, e = 2 + f, h = hn(jn(cdf)))
#[[1]]
#[1] "a=gn(b)"

#[[2]]
#[1] "3 ~ b ~ a"

#[[3]]
#[1] "dd"

#[[4]]
#[1] "e=2 + f"

#[[5]]
#[1] "h=hn(jn(cdf))"

R: How to return the exact form of `...` in `fn(...)` without evaluating `...`?

2 Answers2