Using substitute in multiple function calls

Question

foo <- function(arg) {
  substitute(arg)
}

foo1 <- function(parm) {
  foo(param)
}
foo1(2 + 2)

output is:

param

How can I use substitute inside foo such that the output will be the expression 2 + 2?

Saurabh · Answer 1 · 2021-01-12T17:32:30.093

1

Here's the rlang approach:

foo <- function(...) {
  x <- rlang::enexprs(...)
  x[[1]]
}
foo(2 + 2)

Result is:

2 + 2

edited Jan 12 '21 at 17:32

answered Jan 11 '21 at 07:42

Saurabh

The input to `foo1` could be any expression, not limited to a string. And it is not supposed to evaluate either. So outputting 4 is not desired. – Bear Bile Farming is Torture Jan 11 '21 at 07:58
You are missing the point completely. Your answer doesn't address non-standard evaluation at all. – Roland Jan 11 '21 at 08:47

score 1 · Answer 2 · answered Jan 11 '21 at 08:01

1

foo1 <- function(parm) {
  p <- substitute(parm)
  do.call(foo, list(p))
}
foo1(2 + 2)
#2 + 2

answered Jan 11 '21 at 08:01

Roland

I am aware of that solution. However, is there a way to not use `substitute` inside `foo1`? – Bear Bile Farming is Torture Jan 11 '21 at 08:02
@code_life Not easily. Can we make any assumptions regarding the call stack? – Roland Jan 11 '21 at 08:42

2 Answers2