Yes, lambda expression have an implicit begin in the body.
> (begin . (1 2 3))
3
> (begin (1 2 3))
application: not a procedure;
expected a procedure that can be applied to arguments
given: 1
arguments...:
Here (begin . (1 2 3)) is the same as (begin 1 2 3).
The docs for lambda are here:
http://docs.racket-lang.org/reference/lambda.html?q=lambda#%28form._%28%28lib._racket%2Fprivate%2Fbase..rkt%29._lambda%29%29
Note the body ...+ which means 1 or more bodies are allowed.
/Jens Axel
2016-03-14 18:37 GMT+01:00 Pedro Caldeira <pedro.calde...@bluewin.ch>:
> 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> 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
>
>
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