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 <[email protected] > <mailto:[email protected]>> 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 [email protected]. For more options, visit https://groups.google.com/d/optout.
