[Rd] How to test if an object/argument is "parse tree" - without evaluating it?

[Hadley Wickham]

On Thu, May 1, 2014 at 4:08 PM, Kevin Ushey <kevinushey at gmail.com> wrote:
> Henrik,
>
> If I understand correctly, you want something along the lines of
> (following your example):
>
>     foo <- function(expr) {
>       if (!is.language(expr)) substitute(expr)
>       else expr
>     }
>
>     ## first example
>     expr0 <- foo({ x <- 1 })
>     expr1 <- foo(expr0)
>     stopifnot(identical(expr1, expr0))
>
>     ## second
>     expr2 <- foo(foo(foo(foo(expr0))))
>     stopifnot(identical(expr2, expr0))

Unfortunately this won't work in general because is.language evaluates expr:

foo(stop("Uh oh!"))


In general, I'm with Bill Dunlap - you're better off being explicit
with formulas.

Hadley

-- 
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