| Date: Wed, 07 Mar 2007 19:32:20 -0500
| From: David Van Horn <dvanhorn_at_cs.brandeis.edu>
|
| Aubrey Jaffer wrote:
| > | From: Pascal Costanza <pc_at_p-cos.net>
| > | Date: Wed, 7 Mar 2007 16:05:53 +0100
| > |
| > | ...
| > | 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?
If go is defined with defmacro, SCM behaves as advertised:
> (defmacro (go) '((go) (go)))
#<unspecified>
> (lambda () (go))
#<CLOSURE <anon> () ((go) (go))>
But your example does loop:
> (define-syntax go (syntax-rules () ((go) (go))))
#<unspecified>
> (lambda () (go))
ERROR: user interrupt ...
R5RS-macro magic was added to SCM by Radey Shouman. It seems that the
first expression in a LAMBDA expression gets expanded early if it is
an R5RS-macro; I don't know why. If (go) is not first, then its
expansion is delayed until it is evaluated:
> (lambda () (list) (go))
#<CLOSURE <anon> () (list) (go)>
So SCM must settle for being a 99.44% pure interpreter.
Received on Wed Mar 07 2007 - 20:13:11 UTC