[racket] Delimited continuations and parameters

On May 15, 2012, at 3:11 PM, Asumu Takikawa wrote:

> On 2012-05-15 10:49:20 -0400, Matthias Felleisen wrote:
>>       (parameterize ([p 1])
>>         (λ (zzz) ;; <=================   INCLUDING THE parameterize r while not including parameterize p is an arbitrary choice
> 
> I only included `r` because it is the only parameterization in the
> source code up to the delimiter, which is relevant for Kiselyov et al's
> argument. Otherwise it's arbitrary.



NO, their argument is arbitrary. As I tried to explain, one can view delimiters as opaque and absolute or as transparent and pieces of a large stack. (And probably more.) 

Your particular reduction represents one of their choices. But you can make up different reductions to represent Racket's choice. And both are equally consistent, so pragmatics must decide. 



> Since Racket can support either option, I suppose it's not an issue though.

Perhaps the answer is that both may be of equal use.