Marco Antoniotti  <marcoxa@parades.rm.cnr.it> wrote:
+---------------
| > But since in Scheme the "function" position is evaluated with the *same*
| > rules as the "arg" positions -- that is, the function position can be
| > *any* expression that evaluates to a function -- you can accomplish the
| > above in Scheme with:
| > 
| >   	((if multiprocessing-p process-run-function (lambda (x) (x)))
| >   	         (lambda ()
| >   		     ...))
| 
| If multiprocessing is on then 'process-run-function' executes the
| lambda, otherwise the lambda is sipmply returned as a value.
+---------------

You're not reading what I typed clsoely enough. What you said would have
be true if I had typed *this*:

      	((if multiprocessing-p process-run-function (lambda (x) x))
      	         (lambda ()
      		     ...))

but I didn't. In what I typed (which you quoted correctly), if
multiprocessing is on the lambda on the following lines is passed
as a parameter to the "process-run-function", otherwise the lambda
is passed as a parameter to the unnamed function "(lambda (x) (x))",
which *calls* it! (...with no args.)  Again:

	(lambda (x) (x))	; calls its arg
	(lambda (x) x)		; returns its arg

+---------------
| > p.s. A definition of "funcall" that would work with the above *and*
| > handle the more general uses of "funcall" is:
| > 	(define (funcall f . rest) (apply f rest))
| > Then you could say:
| > 
| >   	((if multiprocessing-p process-run-function funcall)
| >   	         (lambda ()
| >   		     ...))
| 
| Now, this would work.
+---------------

Both forms work as originally posted.


-Rob

-----
Rob Warnock, 8L-846		rpw3@sgi.com
Applied Networking		http://reality.sgi.com/rpw3/
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