[Rd] quote() vs quote(expr=)

[Hadley Wickham]

On Mon, Feb 18, 2013 at 10:35 AM, peter dalgaard <pdalgd at gmail.com> wrote:
>
> On Feb 18, 2013, at 17:04 , Hadley Wickham wrote:
>
>> quote(expr =) returning missing seems like the right thing to me,
>> quote() throwing an error does not, because it violates the usual
>> semantics where f(x = ) is equivalent to f().
>
> Except that it isn't:
>
>> (function(...)nargs())()
> [1] 0
>> (function(...)nargs())(x=)
> [1] 1

But hardly any functions use nargs:

> find_funs("package:base", fun_calls, fixed("nargs"))
Using environment package:base
 [1] "-.Date"               "-.POSIXt"             "[.data.frame"
 [4] "[[.data.frame"        "[[<-.data.frame"      "[[<-.numeric_version"
 [7] "[<-.data.frame"       "+.Date"               "+.POSIXt"
[10] "diag"                 "message"              "Ops.data.frame"
[13] "Ops.Date"             "Ops.difftime"         "Ops.numeric_version"
[16] "Ops.POSIXt"           "seq.default"          "stop"
[19] "system.file"          "trace"                "warning"
> find_funs("package:stats", fun_calls, fixed("nargs"))
Using environment package:stats
[1] "model.frame.default"

The majority is used to switch between cases like -x vs x - y or to
detect extra arguments if something weird is being done with ...

So I think it's reasonable to expect that f() and f(x = ) do the same thing.

Hadley

-- 
Chief Scientist, RStudio
http://had.co.nz/