Question

我已经实现了语言“ ROL”，现在我正试图扩展该语言以支持fun，并称呼我正在尝试在替代模型中实现它。

这是我的eval函数代码（请注意，此代码在没有乐趣和调用的情况下也可以正常工作。因此，这两个实现一定是问题所在

(: eval : RegE -> RES)
;; evaluates RegE expressions by reducing them to bit-lists
(define (eval expr)
  (cases expr
    [(Reg right) (RegV right)]
    [(Bool b) (RES_Bool b)]
    [(Id name) (error 'eval "free identifier: ~s" name)]
    [(And left right) (reg-arith-op bit-and  (eval left ) (eval right) )]
    [(Or left right) (reg-arith-op bit-or (eval left )  (eval right) )]
    [(Shl E1) (RegV (shift-left (RegV->bit-list (eval E1))))]
    [(Maj E1) (RES_Bool (majority? (RegV->bit-list (eval E1))))]
    [(Geq E1 E2) (RES_Bool (geq-bitlists? (RegV->bit-list (eval E1)) (RegV->bit-list (eval E2))))]
    [(With bound-id named-expr bound-body)
       (eval (subst bound-body
                    bound-id
                    (Reg (RegV->bit-list(eval named-expr)))))]
    [(Fun bound-id bound-body) expr]
    [(Call fun-expr arg-expr)
       (let ([fval (eval fun-expr)])
         (cases fval
           [(Fun bound-id bound-body)
            (eval (subst bound-body
                   bound-id
                   (eval arg-expr)))]
           [else (error 'eval "`call' expects a function, got: ~s"
                              fval)]))]
    [(If E1 E2 E3) (if (RegV->boolean (eval E1)) (eval E2) (eval E3))]



))

这是我最聪明的功能：

(define (subst expr from to)
  (cases expr
    [(Reg g) expr]
    [(Bool g) expr]
    [(And left right)(And (subst left from to)(subst right from to))]
    [(Or left right)(Or (subst left from to)(subst right from to))]
    [(If bool ifBody elseBody) (If (subst bool from to) (subst ifBody from to) (subst elseBody from to))]
    [(Maj left)(Maj (subst left from to))]
    [(Geq left right)(Geq (subst left from to)(subst right from to))]
    [(Shl left)(Shl (subst left from to))]
    [(Id name) (if (eq? name from) to expr)]
    [(With bound-id named-expr bound-body)
           (if (eq? bound-id from)
               expr
               (With bound-id
                     named-expr
               (subst bound-body from to)))]
    [(Call left right) (Call (subst left from to) (subst right from to))]
    [(Fun bound-id bound-body)
         (if (eq? bound-id from)
           expr
           (Fun bound-id (subst bound-body from to)))]))

我遇到了错误：

Type Checker: type mismatch
  expected: RegE
  given: RES in: fval
. Type Checker: type mismatch
  expected: RegE
  given: RES in: (eval arg-expr)
. Type Checker: type mismatch
  expected: RES
  given: Fun in: (cases expr ((Reg right) (RegV right)) ((Bool b) (RES_Bool b)) ((Id name) (error (quote eval) "free identifier: ~s" name)) ((And left right) (reg-arith-op bit-and (eval left) (eval right))) ((Or left right) (reg-arith-op bit-or (eval left) (eval right))) ((Shl E1) (RegV (shift-left (RegV->bit-list (eval E1))))) ((Maj E1) (RES_Bool (majority? (RegV->bit-list (eval E1))))) ((Geq E1 E2) (RES_Bool (geq-bitlists? (RegV->bit-list (eval E1)) (RegV->bit-list (eval E2))))) ((With bound-id named-expr bound-body) (eval (subst bound-body bound-id (Reg (RegV->bit-list (eval named-expr)))))) ((Fun bound-id bound-body) expr) ((Call fun-expr arg-expr) (let ((fval (eval fun-expr))) (cases fval ((Fun bound-id bound-body) (eval (subst bound-body bound-id (eval arg-expr)))) (else (error (quote eval) "`call' expects a function, got: ~s" fval))))) ((If E1 E2 E3) (if (RegV->boolean (eval E1)) (eval E2) (eval E3))))
. Type Checker: Summary: 3 errors encountered in:
  fval
  (eval arg-expr)
  (cases expr ((Reg right) (RegV right)) ((Bool b) (RES_Bool b)) ((Id name) (error (quote eval) "free identifier: ~s" name)) ((And left right) (reg-arith-op bit-and (eval left) (eval right))) ((Or left right) (reg-arith-op bit-or (eval left) (eval right))) ((Shl E1) (RegV (shift-left (RegV->bit-list (eval E1))))) ((Maj E1) (RES_Bool (majority? (RegV->bit-list (eval E1))))) ((Geq E1 E2) (RES_Bool (geq-bitlists? (RegV->bit-list (eval E1)) (RegV->bit-list (eval E2))))) ((With bound-id named-expr bound-body) (eval (subst bound-body bound-id (Reg (RegV->bit-list (eval named-expr)))))) ((Fun bound-id bound-body) expr) ((Call fun-expr arg-expr) (let ((fval (eval fun-expr))) (cases fval ((Fun bound-id bound-body) (eval (subst bound-body bound-id (eval arg-expr)))) (else (error (quote eval) "`call' expects a function, got: ~s" fval))))) ((If E1 E2 E3) (if (RegV->boolean (eval E1)) (eval E2) (eval E3))))
>

我的数据类型是：

(define-type RegE
[Reg Bit-List]
[Xor RegE RegE]
[And RegE RegE]
[Or RegE RegE]
[Shl RegE]
[Id Symbol]
[With Symbol RegE RegE]
[Bool Boolean]
[Geq RegE RegE]
[Maj RegE]
[If RegE RegE RegE]
[Fun Symbol RegE]
[Call RegE RegE])

和：

(define-type RES
[RES_Bool Boolean]
[RegV Bit-List])

我可以看到，对于(let ([fval (eval fun-expr)而言，(eval fun-expr)会返回RES，我想这可能是问题所在，还有其他方法可以解决吗？任何帮助将不胜感激..

Answer 1

关键问题是如何在运行时表示函数。 eval的定义是：

(: eval : RegE -> RES)
;; evaluates RegE expressions by reducing them to bit-lists
(define (eval expr)
  (cases expr
  ...
  [(Fun bound-id bound-body) expr]
  ...

评估(Fun bound-id bound-body)的结果必须是RES。也许将RES扩展为：

(define-type RES
    [RES_Bool Boolean]
    [RegV Bit-List]
    [FunV Symbol RegE])

然后让eval返回一个(FunV bound-id bound-body)。

在“通话”评估中，您将需要切换到

     (cases fval
       [(FunV bound-id bound-body)

因为fval是RES。

使用替代模型的eval函数中的DrRacket类型不匹配错误

1 个答案: