[r6rs-discuss] Interpreters need not apply?

From: David Van Horn <dvanhorn>
Date: Wed Mar 7 19:32:36 2007

Aubrey Jaffer wrote:
> | From: Pascal Costanza <pc_at_p-cos.net>
> | Date: Wed, 7 Mar 2007 16:05:53 +0100
> |
> | On 7 Mar 2007, at 15:51, AndrevanTonder wrote:
> |
> | > On Wed, 7 Mar 2007, Pascal Costanza wrote:
> | >
> | >> The terms "compiler" and "interpreter" are not well-defined. But
> | >> AFAICT, a compiler typically works in two phases: A translation
> | >> from one representation to another one, where the latter is
> | >> typically a representation that can be executed by some
> | >> interpreter (for example, a CPU).
> | >
> | > No r5rs-conformant Scheme interpreter will start evaluating a
> | > form without macro-expanding it first. R6RS requires nothing
> | > more than this.
> |
> | Here is an example:
> |
> | (if expression form1 form2)
> |
> | Assume that both form1 and form2 are macro invocations. Will form1
> | and form2 both be macroexpanded before the if statement is
> | evaluated, or will first the expression be evaluated and depending
> | on its outcome only either form1 or form2 be expanded and then
> | evaluated?
>
> In SCM, only form1 or form2 will be expanded and evaluated.

Given this, the following behavior is curious to me:

    $ scm -r r5rs
> (define-syntax go (syntax-rules () ((go) (go))))
    #<unspecified>
> (if #t #t (go))
    #t
> (lambda () (go)) ;; loops

 From my understanding of this discussion, this implies SCM is not a
"pure interpreter", correct?

David
Received on Wed Mar 07 2007 - 19:32:20 UTC