[Rd] anonymous function parsing bug?

William Dunlap
Fri, 21 Oct 2016 17:54:29 -0000

Am 21.10.2016 um 18:10 schrieb William Dunlap:
>
> Are you saying that

    f1 <- function(x) log(x)

    f2 <- function(x) { log } (x)

should act differently?

yes.


But that would mean that {log} would act differently than log.
I suppose it is a matter of taste, but I say yuck.

As for 'return', don't use it if you want readable code.  It is
like a goto but worse.  It is never necessary.


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Fri, Oct 21, 2016 at 10:17 AM, Wilm Schumacher <wilm.schumacher at
gmail.com
> wrote:

> Hi,
>
>
> Am 21.10.2016 um 18:10 schrieb William Dunlap:
>
>> Are you saying that
>>     f1 <- function(x) log(x)
>>     f2 <- function(x) { log } (x)
>> should act differently?
>>
> yes. Or more precisely: I would expect that. "Should" implies,
that I want
> to change something. I just want to understand the behavior (or file a bug,
> if this would have been one).
>
> As I wrote, in e.g. node.js the pendents to the lines that you wrote are
> treated differently (the first is a function, the latter is a parsing
> error).
>
> Let's use this example instead:
> x <- 20
> f1 <- function(x) { x<-x+1; log(x) }
> f2 <- function(x) { x<-x+1; log } (x)
> which act equally.
>
> But as the latter is a legal statement, I would read it as
> f2 <- (function(x) { x<-x+1; log }) (x)
>
> thus, I would expect the first to be a function, the latter to be a
> numeric ( log(20) in this case ).
>
>
> Using 'return' complicates the matter, because it affects
evaluation, not
>> parsing.
>>
>
> But perhaps it illustrates my problem a little better:
> x <- 20
> f1 <- function(x) return(log(x))
> f2 <- function(x) { return(log) } (x)
>
> f1(10) is a numeric, f2(10) is the log function. Again: as the latter is a
> legal statement, I would expect:
> f2 <- (function(x) { x<-x+1; log }) (x)
>
> However, regarding the answers I will try to construct the AST regarding
> the grammar defined in gramm.y of that statement
> f2 <- function(x) { x<-x+1; log } (x)
> to understand what the R interpreter really does.
>
> Best wishes,
>
> Wilm
>
>
>
>
>

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