Duncan Murdoch <[EMAIL PROTECTED]> writes:
> >> > substitute(function(a=a) 1, list(a=quote(foo)))
> >> function(a = a) 1
> >
> >a bug for sure
>
> Yes.
Hmm... The issue is that the argument to `function` is a pairlist (not
LANGSXP), and we don't recurse into those. Any R object can turn up
On 9/14/2006 3:49 PM, Robert Gentleman wrote:
>
> Duncan Murdoch wrote:
>> On 9/14/2006 3:01 PM, Seth Falcon wrote:
>>> Hi,
>>>
>>> Can someone help me understand why
>>>
>>> substitute(function(a) a + 1, list(a=quote(foo)))
>>>
>>> gives
>>>
>>> function(a) foo + 1
>>>
>>> and not
>>>
>>> f
On Thu, 14 Sep 2006, Robert Gentleman wrote:
> > > substitute(c( a = 1, b = a), list(a = quote(foo)))
> > c(a = 1, b = foo)
> >
> > The "a" in function(a) is the name of the arg, it's not the arg itself
>
> yes, but the logic seems to be broken. In Seth's case there seems to be
> no way to use su
Duncan Murdoch wrote:
> On 9/14/2006 3:01 PM, Seth Falcon wrote:
>> Hi,
>>
>> Can someone help me understand why
>>
>> substitute(function(a) a + 1, list(a=quote(foo)))
>>
>> gives
>>
>> function(a) foo + 1
>>
>> and not
>>
>> function(foo) foo + 1
>>
>> The man page leads me to believe thi
On Thu, 14 Sep 2006, Duncan Murdoch wrote:
> I think it's the same reason that this happens:
>
> > substitute(c( a = 1, b = a), list(a = quote(foo)))
> c(a = 1, b = foo)
>
> The "a" in function(a) is the name of the arg, it's not the arg itself
> (which is missing). Now a harder question to answ
On 9/14/2006 3:01 PM, Seth Falcon wrote:
> Hi,
>
> Can someone help me understand why
>
> substitute(function(a) a + 1, list(a=quote(foo)))
>
> gives
>
> function(a) foo + 1
>
> and not
>
> function(foo) foo + 1
>
> The man page leads me to believe this is related to lazy evaluation of
Hi,
Can someone help me understand why
substitute(function(a) a + 1, list(a=quote(foo)))
gives
function(a) foo + 1
and not
function(foo) foo + 1
The man page leads me to believe this is related to lazy evaluation of
function arguments, but I'm not getting the big picture.
Thanks,
+ s
