Re: [Rd] foo2Args()

From: Paul Gilbert <>
Date: Fri 30 Jun 2006 - 21:12:09 GMT

Duncan Murdoch wrote:

> On 6/30/2006 4:19 PM, Paul Gilbert wrote:
>> I was just considering trying to clean up the arguments to a function
>> that calls other functions,
>> and was playin with a suggestion Achim made during a conversation at
>> useR. The idea is, instead of using list(), use a small function to
>> construct and check arguments. My hope was to be able to do this
>> without making it globally visible:
>> foo <- function(x, args=foo2Args()) {
>> foo2Args <- function(a=1, b=2){list(a,b)}
>> # above would actual do more testing of args
>> #now I would call foo2 with args, but to test just
>> args
>> }
>> Now,
>> > foo(1) # should I be surprized that this works
>> [[1]]
>> [1] 1
>> [[2]]
>> [1] 2
> I don't think it really works, it's just a coincidence that the answer
> matches your expectations:
> > foo(3)
> [[1]]
> [1] 1
> [[2]]
> [1] 2
> Or maybe I am completely misunderstanding your expectations...
>> > foo(1, args=foo2Args(a=2, b=10)) # or that this does not
>> Error in foo(1, args = foo2Args(a = 2, b = 10)) :
>> could not find function "foo2Args"
> This is a somewhat subtle thing about the way args are evaluated.
> What is done makes lots of sense once you understand it:
> - When you specify an argument in the call, the expression you give
> is evaluated in the current context. In the context where you made
> this call, foo2Args is not defined, hence the error.
> - When you specify a default for an argument, it is evaluated in the
> local context of the function evaluation. So as a default, foo2Args
> is recognized, because it's a local variable within foo.
> You also need to remember "lazy evaluation": neither of the above
> actually take place until args is used.
> R is pretty flexible, so it's probably possible to do whatever you
> intended here; can you describe exactly what you want to happen?

I guess what I want to happen is that when args is used it it gets evaluated in the local context of the function, whether or not I am specifying the argument or using the default. Is there a way to do that?

What I expected to happen was that it would always get evaluated either in the local context or in the calling context, i.e., it would fail or not fail, but that would not depend on whether I am specifying the argument or using the default, so your first points explain this.


> Duncan Murdoch

La version franšaise suit le texte anglais.

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