[plt-scheme] Why isn't the car of a list of symbols a symbol?

Joe (cc'ing plt-scheme)-

On Jul 9, 2009, at 7:10 PM, Joe Marshall wrote:

> On Thu, Jul 9, 2009 at 1:21 PM, Robby
> Findler<robby at eecs.northwestern.edu> wrote:
>> On Thu, Jul 9, 2009 at 3:18 PM, Carl  
>> Eastlund<carl.eastlund at gmail.com> wrote:
>>> Following those rules, '('yes 'no) is (quote ((quote yes) (quote
>>> no))), and it produces the value ((quote yes) (quote no)).  Apply
>>> those two rules, and quote shouldn't have any more surprises.
>>
>> ((quote yes) (quote no))
>>
>> is not a value. It is an application expression.
>
> It most certainly is a value.  It is a list of two elements.
> There are many ways to construct such a value.  Carl
> suggested evaluating '('yes 'no).
>
>> The way to keep your sanity here is to think of the printer as
>> stripping off the outer quote when it prints out a value.
>
> What are you smoking?!  The printer isn't involved here and
> it certainly isn't stripping anything.

Oh please, please, please, lets *not* have another debate between the  
"reductionists" and "non-reductionists"!

http://list.cs.brown.edu/pipermail/plt-scheme/2008-October/027923.html

-Felix Klock