capture "->"

Dear Duncan,

On Fri, Mar 1, 2024 at 11:30?AM Duncan Murdoch <murdoch.duncan at gmail.com>
wrote:

...
If you parse it with srcrefs, you could look at the source.  The parser
doesn't record whether it was A -> B or B <- A anywhere else.

Thank you, this gets me closer but it still needs a little push:

foo <- function(x) {

x <- substitute(x)
  return(attr(x, "srcref")[[2]])
}

foo(A -> B)

NULL

This seems to work, however:

foo({A -> B})

A -> B

Is there a way to treat the formula as if it was enclosed between the curly
brackets?
Dmitri