Does that mean that lambda expressions have an implicit (begin …) block in them?
(begin ((displayln 1) (displayln 2) (displayln 3))) leads to an error (begin . ((displayln 1) (displayln 2) (displayln 3))) displays to 1 2 3 Thank you for the detailed explanation I think I get it now. > On 13 Mar 2016, at 19:48, Jens Axel Søgaard <jensa...@soegaard.net > <mailto:jensa...@soegaard.net>> wrote: > > If we use > > (define-syntax define/memoized > (syntax-rules () > ((_ (name . args) . body) > (define name (memoize (lambda args body)))))) > > and body is bound to ((displayln x) (displayln y) (displayz)) > then > > (lambda args body) > > will become > > (lambda <something> ((displayln x) (displayln y) (displayz))) > > And when this function is called you will get an error since > > ((displayln x) ...) > > means apply the result of (displayln x) to (displayln y) (displayln x). > > /Jens Axel -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to racket-users+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
