PLAI那些事06 FAE

没有了with表达，with，exp，body的id换成exp的lambda函数，从而可以没有with来进行实现。即，{with {id exp} body}换成了{{fun {id} body} exp}。

FAE : Concrete syntax

<FWAE> ::= <num>
         | {+ <FWAE> <FWAE>}
         | {- <FWAE> <FWAE>}
         | <id>
         | {fun {<id>} <FWAE>}
         | {<FWAE> <FWAE>}

FAE : Abstrac syntax

(define-type FAE
  [num (n number?)]
  [add (lhs FAE?) (rhs FAE?)]
  [sub (lhs FAE?) (rhs FAE?)]
  [id (name symbol?)]
  [fun (param symbol?) (body FAE?)]
  [app (ftn FAE?) (arg FAE?)])

parse : sexp -> FAE

(define (parse sexp)
  (match sexp
    [(? number?) (num sexp)]
    [(list '+ l r) (add (parse l) (parse r))]
    [(list '- l r) (sub (parse l) (parse r))]
    [(? symbol?) (id sexp)]
    [(list 'fun (list x) b) (fun x (parse b))]
    [(list f a) (app (parse f) (parse a))]
    [else (error 'parse "bad syntax: ~a" sexp)]))

interp : FAE -> 'FAE'

(define (num+ x y)
  (num (+ (num-n x) (num-n y))))
(define (num- x y)
  (num (- (num-n x) (num-n y))))

(define (interp fae)
  (type-case FAE fae
    [num (n) fae]
    [add (l r) (num+ (interp l) (interp r))]
    [sub (l r) (num- (interp l) (interp r))]
    [id (s) (error 'interp "free variable")]
    [fun (x b) fae]
    [app (f a) (local [(define ftn (interp f))]
                  (interp (subst (fun-body ftn)
                                 (fun-param ftn)
                                 (interp a))))]))

subst : 'FAE' 'symbol' 'FAE' -> 'FAE'

(define (subst exp sub-id val)
  (type-case FAE exp
    [num (n) exp]
    [add (l r) (add (subst l sub-id val) (subst r sub-id val))]
    [sub (l r) (sub (subst l sub-id val) (subst r sub-id val))]
    [id (name) (cond [(equal? name sub-id) val]
                     [else exp])]
    [app (f arg) (app (subst f sub-id val)
                      (subst arg sub-id val))]
    [fun (id body) (if (equal? sub-id id)
                       exp
                       (fun id (subst body sub-id val)))]))