I misunderstood because of your mismatched parentheses. Without eval, you can’t do better than that. Parens rock.
> On May 28, 2017, at 9:51 PM, Vityou <zlee...@gmail.com> wrote: > > On Sunday, May 28, 2017 at 2:01:09 PM UTC-6, Matthias Felleisen wrote: >> You need to apply the function, not just compute the substitution. See >> applicable struct in previous message. This should just work out, w/o much >> ado. >> >> >> >>> On May 28, 2017, at 12:40 AM, Vityou <zlee...@gmail.com> wrote: >>> >>> On Thursday, May 25, 2017 at 5:50:29 PM UTC-6, Matthias Felleisen wrote: >>>> The client module can refer to all imported #% forms. If you don’t export >>>> it from the language module, it’s not there. [Well, mostly] Implicitly the >>>> client module already refers to #% forms already. >>>> >>>> >>>> >>>>> On May 25, 2017, at 7:09 PM, Vityou <zlee...@gmail.com> wrote: >>>>> >>>>> On Wednesday, May 24, 2017 at 2:05:23 PM UTC-6, Matthias Felleisen wrote: >>>>>> Don’t eval. This is a bit crude but it now your lam-s keep track of your >>>>>> environment, too. >>>>>> >>>>>> #lang racket ;; new-lang.rkt >>>>>> >>>>>> (provide >>>>>> #%app >>>>>> #%datum >>>>>> #%top-interaction >>>>>> (rename-out >>>>>> (new-lambda lambda) >>>>>> (new-mb #%module-begin))) >>>>>> >>>>>> (require racket/stxparam) >>>>>> >>>>>> (define-syntax (new-lambda stx) >>>>>> (syntax-case stx () >>>>>> [(_ (x ...) e ...) >>>>>> #`(letrec ([L (lam '(x ...) >>>>>> '(e ...) >>>>>> (*env) >>>>>> (lambda (x ...) >>>>>> (syntax-parameterize ([*env >>>>>> (lambda (stx) >>>>>> (syntax-case stx () >>>>>> [(_) #`(append '(x >>>>>> ...) (lam-environment L))]))]) >>>>>> e) >>>>>> ...))]) >>>>>> L)])) >>>>>> (define-syntax-parameter *env >>>>>> (syntax-rules () [(_) '()])) >>>>>> (struct lam (parameters bodies environment closure) #:property >>>>>> prop:procedure 3) >>>>>> >>>>>> (define-syntax (new-mb stx) >>>>>> (syntax-case stx () >>>>>> [(_ e ...) >>>>>> #'(#%module-begin >>>>>> (let ([v e]) >>>>>> (if (lam? v) >>>>>> `(let (,@(map (lambda (x) `(,x --some-value--)) >>>>>> (lam-environment v))) >>>>>> (lambda ,(lam-parameters v) ,@(lam-bodies v))) >>>>>> v)) >>>>>> ...)])) >>>>>> >>>>>> >>>>>> >>>>>>> On May 24, 2017, at 3:41 PM, Vityou <zlee...@gmail.com> wrote: >>>>>>> >>>>>>> On Wednesday, May 24, 2017 at 12:05:19 PM UTC-6, Vityou wrote: >>>>>>>> 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. >>>>>>>>> >>>>>>>>> -- >>>>>>>>> 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...@googlegroups.com. >>>>>>>>> For more options, visit https://groups.google.com/d/optout. >>>>>>>>> >>>>>>>>> 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))) >>>>>>> >>>>>>> I was able to get it to work by processing updating the list parts of >>>>>>> the struct "In parallel" with the normal evaluation by applying the old >>>>>>> s-exp processing functions to the correct part of the struct: >>>>>>> >>>>>>> (define-syntax (new-app stx) >>>>>>> (syntax-case stx () >>>>>>> [(_ f x) >>>>>>> #'(let ([result (#%app f x)]) >>>>>>> (if (lam? result) >>>>>>> (struct-copy lam >>>>>>> result >>>>>>> [lex (third (eval `((λ (,(lam-parameter f)) >>>>>>> ,(lam-body f)) >>>>>>> ,(if (lam? x) >>>>>>> `(λ (,(lam-parameter >>>>>>> x)) ,(lam-body x)) >>>>>>> x)) >>>>>>> (lam-lex f)))]) >>>>>>> result))])) >>>>> >>>>> Thanks that works great, I just changed (append '(x) (lam-environment L)) >>>>> to (cons `(x ,x) (lam-environment L)) to get the actual value. And this >>>>> is unrelated but is there a way to restrict the user of the >>>>> lambda-calculus language from using the exported #%... macros? >>>>> >>>>> -- >>>>> 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. >>> >>> I was looking at some different lambda calculus interpreters, and I noticed >>> that most of them just substituted the values of the parameter with the >>> actual value when the function was called. I was able to achieve this by >>> making a substitute function and a syntax parameter "body" that basically >>> does the same thing as env did accept it takes syntax and returns it quoted >>> and the syntax-parameterize part made body substitute the parameter for its >>> value: >>> >>> (define-syntax (new-lambda stx) >>> (syntax-case stx () >>> [(_ (parameter) body) >>> #`(letrec ([L (lambda-structure 'parameter >>> (*body body) >>> (lambda (parameter) >>> (syntax-parameterize ([*body >>> (lambda (stx) >>> (syntax-case >>> stx () >>> [(_ b) >>> >>> #`(substitute 'b >>> >>> 'parameter >>> >>> parameter)]))]) >>> body)))]) >>> L)])) >>> >>> (define-syntax-parameter *body >>> (syntax-rules () [(_ b) 'b])) >>> >>> This works ok, but it doesn't actually evaluate the body, ((lambda (x) >>> (lambda (y) (x y))) (lambda (z) f)) will display (lambda (y) (lambda (z) f) >>> y) instead of (lambda (y) f). I can make this work by evaluating >>> s-expressions since I can just eval the body after the substitution, but is >>> there any way to do this in racket? > > I don't seem to understand what you mean by apply the function. The quoted > parts of the struct are what is being printed, and since they are just data, > I don't see how they can be changed without evaluating them seperately. -- 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.
