# [racket] Evaluation problem

You are missing a pair of parens around '(func first)' and 'second'. The current code in your message will return second, no matter what function you pass in as 'func'.
The correct definition for make-pair should be:
>* define make-pair
*>* (lambda (first)
*>* (lambda (second)
*>* (lambda (func)
*>* ((func first) second)))))
*
Jim
On Sep 30, 2012, at 4:43 PM, Mikko Tiihonen <mikko.tiihonen at tmtiihonen.fi> wrote:
>* Hi!
*>*
*>* I have been going through some lambda-calculus exercises with Racket. I have stumbled on a puzzling problem where beta-reduction by hand leads to different result from that generated by Racket.
*>*
*>*
*>* (define select-first
*>* (lambda (first)
*>* (lambda (second)
*>* first)))
*>*
*>* (define select-second
*>* (lambda (first)
*>* (lambda (second)
*>* second)))
*>*
*>* (define make-pair
*>* (lambda (first)
*>* (lambda (second)
*>* (lambda (func)
*>* (func first) second))))
*>*
*>* -----
*>>* ((select-first 'a) 'b)
*>* 'a
*>>* ((select-second 'a) 'b)
*>* 'b
*>>* (((make-pair 'a) 'b) select-first)
*>* 'b
*>>* (((make-pair 'a) 'b) select-second)
*>* 'b
*>* -------
*>*
*>* So, select-first returns the correct result when applied directly but returns the second argument when applied via make-pair. Also, I have checked that the implementation of make-pair gives the correct result if I do the beta-reduction by hand.
*>*
*>* I suspect that I'm missing something obvious, but I really cannot find the reason why make-pair misbehaves - you gurus on this mailing-list probably can pinpoint the problem instantly :)
*>*
*>* Best regards,
*>*
*>* -Mikko Tiihonen
*>*
*>* P.S. Having used Racket for some months now, I find both Racket and the community around it really great!
*>*
*>*
*>* ____________________
*>* Racket Users list:
*>* http://lists.racket-lang.org/users
*