On Tuesday, May 23, 2017 at 8:21:59 PM UTC-6, Matthias Felleisen wrote:
> Try to start with this:
>
>
>
>
>
> #lang racket ;; new-lang.rkt
>
>
> (provide
> #%app
> #%datum
> #%top-interaction
> (rename-out
> (new-lambda lambda)
> (new-mb #%module-begin)))
>
>
> (define-syntax (new-lambda stx)
> (syntax-case stx ()
> [(_ (x ...) e ...)
> #'(lam '(x ...) '(e ...) (lambda (x ...) e ...))]))
>
>
> (struct lam (parameters bodies closure) #:property prop:procedure 2)
>
>
> (define-syntax (new-mb stx)
> (syntax-case stx ()
> [(_ e ...)
> #'(#%module-begin
> (let ([v e])
> (if (lam? v)
> `(lambda ,(lam-parameters v) ,@(lam-bodies v))
> v))
> ...)]))
>
>
>
>
> ;; - - -
>
>
>
> #lang s-exp "new-lang.rkt” ;; new-lang-client.rkt
>
>
> ((lambda (x) x)
> (lambda (y) y))
>
>
>
>
>
>
> On May 23, 2017, at 10:03 PM, Vityou <zle...@gmail.com> wrote:
>
>
> On Tuesday, May 23, 2017 at 7:17:18 PM UTC-6, Matthias Felleisen wrote:
> Why do you interpret S-expressions instead of re-mapping lambda and #%app?
>
>
>
>
>
> On May 23, 2017, at 9:14 PM, Vityou <zle...@gmail.com> wrote:
>
> I might be able to do something like this, but what I'm looking for is
> something that will be able to show the variables available to it in adition
> to its source. I'll probable have to do something like what you did with the
> struct accept add a field with its available variables and modify #%app to
> add to its known variables.
>
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>
> I dont know what I could map lambda to that would let it retain and print its
> known variables besides a list.
That's probably good enough for most cases, but I tried to add a struct field
to record the lexical content, I can't fin a way to mimic evaluating the body
of the function in the struct, this is the closest I got:
(define-syntax (new-app stx)
(syntax-case stx ()
[(_ f x)
#'(let ([result (#%app f x)])
(if (lam? result)
(struct-copy lam
result
[lex (cons `(,(lam-parameter f) ,(if (lam? x)
`(λ
(,(lam-parameter x)) ,(lam-body x))
x))
(lam-lex f))])
result))]))
It sort of works, but it just blindly tacks on info if the result is a struct.
((lambda (x) x) (lambda (y) y) results in (function (lambda (y) y) (x (lambda
(y) y)))
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