Re: conditional based on binding of a variable

[Damien Mattei]

this could not works because the defined-symbol? used in the macro
if-defined return false because at the expansion of the macro the symbol is
not defined.

Instead at run-time, this works:

#|kawa:21|# (define (foo2) (define x 7) (defined-symbol? x))
#|kawa:22|# (foo2)
#t

On Sun, Apr 14, 2024 at 9:24 AM Damien Mattei <damien.mattei@gmail.com>
wrote:

> i tried this, that works in some case:
>
> ;; scheme@(guile-user)> (defined-symbol? d)
> ;; ;;; <stdin>:20:17: warning: possibly unbound variable `d'
> ;; $3 = #f
> (define-syntax defined-symbol?
>   (syntax-rules ()
>     ((_ x) (call-with-current-continuation
>    (lambda (exit)
>      (with-exception-handler
>       (lambda (e)
> (display "defined-symbol? : undefined") (newline)
> (exit #f)) ; eval failed => not defined
>       (lambda ()
> (eval x (interaction-environment))
> #t))))))) ; eval suceeded => defined
>
>
> ;; scheme@(guile-user)> (define r 2)
> ;; scheme@(guile-user)> (if-defined r 'defined (define r 7))
> ;; if-defined : where=#t
> ;; $16 = defined
> ;; scheme@(guile-user)> (if-defined t3 'defined (define t3 7))
> ;; if-defined : where=#f
> ;; scheme@(guile-user)> t3
> ;; $17 = 7
> ;; scheme@(guile-user)> (if-defined z (list z) 'not-defined)
> ;; if-defined : where=#f
> ;; $18 = not-defined
> (define-syntax if-defined
>   (lambda (stx)
>     (syntax-case stx ()
>       ((_ id iftrue iffalse)
>        (let ((where (defined-symbol? #'id))) ;;(quote id))))
> (display "if-defined : where=") (display where) (newline)
> (display "id=") (display #'id) (newline)
> (if where #'iftrue #'iffalse))))))
>
> but fais with those examples:
>
> scheme@(guile-user)> (define (foo) (define x 1) (if-defined x 'def
> 'not_def))
> defined-symbol? : undefined
> if-defined : where=#f
> id=#<syntax:unknown file:37:39 x>
> scheme@(guile-user)> (define (bar) (let ((x 1)) (if-defined x 'def
> 'not_def)))
> defined-symbol? : undefined
> if-defined : where=#f
> id=#<syntax:unknown file:38:39 x>
>
>
> On Fri, Apr 12, 2024 at 1:40 PM Damien Mattei <damien.mattei@gmail.com>
> wrote:
>
>> indeed ,seems not a thing feasible ,unless in Racket where it works great:
>>
>>
>> https://stackoverflow.com/questions/3267351/how-do-i-test-whether-a-variable-is-defined-before-referencing-it?noredirect=1&lq=1
>>
>> On Fri, Apr 12, 2024 at 12:34 AM Damien Mattei <damien.mattei@gmail.com>
>> wrote:
>>
>>>
>>> hello,
>>>
>>> i'm searching the equivalent in scheme , guile or r6rs of this code for
>>> Racket:
>>>
>>> ; Tests
>>> ;; (if-defined z (list z) 'not-defined) ; -> not-defined
>>>
>>> ;; (if-defined t (void) (define t 5))
>>> ;; t ; -> 5
>>>
>>> ;; (define x 3)
>>> ;; (if-defined x (void) (define x 6))
>>> ;; x ; -> 3
>>> (define-syntax (if-defined stx)
>>>   (syntax-case stx ()
>>>     [(_ id iftrue iffalse)
>>>      (let ([where (identifier-binding #'id)])
>>>        (display "if-defined : where=") (display where) (newline)
>>>        (display "id=") (display #'id) (newline)(newline)
>>>        (if where #'iftrue #'iffalse))]))
>>>
>>> it is based on identifier-binding
>>>
>>> i tried defined? in guile but it is not reliable other than toplevel,
>>> then i wrote this code based on exception and test the evaluation of the
>>> variable in environment:
>>>
>>> ;; scheme@(guile-user)> (defined-symbol? d)
>>> ;; ;;; <stdin>:20:17: warning: possibly unbound variable `d'
>>> ;; $3 = #f
>>> (define-syntax defined-symbol?
>>>   (syntax-rules ()
>>>     ((_ x) (call-with-current-continuation
>>>    (lambda (exit)
>>>      (with-exception-handler
>>>       (lambda (e)
>>> (display "defined-symbol? : undefined") (newline)
>>> (exit #f)) ; eval failed => not defined
>>>       (lambda ()
>>> (eval x (interaction-environment))
>>> #t))))))) ; eval suceeded => defined
>>>
>>> used in conjunction with:
>>>
>>> ;; scheme@(guile-user)> (define r 2)
>>> ;; scheme@(guile-user)> (if-defined r 'defined (define r 7))
>>> ;; if-defined : where=#t
>>> ;; $16 = defined
>>> ;; scheme@(guile-user)> (if-defined t3 'defined (define t3 7))
>>> ;; if-defined : where=#f
>>> ;; scheme@(guile-user)> t3
>>> ;; $17 = 7
>>> ;; scheme@(guile-user)> (if-defined z (list z) 'not-defined)
>>> ;; if-defined : where=#f
>>> ;; $18 = not-defined
>>> (define-syntax if-defined
>>>   (lambda (stx)
>>>     (syntax-case stx ()
>>>       ((_ id iftrue iffalse)
>>>        (let ((where (defined-symbol? #'id))) ;;(quote id))))
>>> (display "if-defined : where=") (display where) (newline)
>>> (display "id=") (display #'id) (newline)
>>> (if where #'iftrue #'iffalse))))))
>>>
>>> but in some case it fails again
>>>
>>> the goal is to be able to use this macro:
>>>
>>> (define-syntax <-
>>>
>>>   (lambda (stx)
>>>
>>>     (syntax-case stx ()
>>>
>>>       ((_ var expr)
>>>
>>>        #`(if-defined var
>>>     (set! var expr)
>>>     (define var expr))))))
>>>
>>> that define a variable when not defined or just set! it
>>>
>>> it then fails in some case:
>>>
>>> (let () (define k 0) (let loop () (if (< k 4) (let () (display k)
>>> (newline) (<- k (+ k 1)) (loop)))))
>>> defined-symbol? : undefined
>>> if-defined : where=#f
>>> id=#<syntax:unknown file:22:80 k>
>>> #<unspecified>
>>> ice-9/boot-9.scm:1685:16: In procedure raise-exception:
>>> In procedure +: Wrong type argument in position 1: #<unspecified>
>>>
>>> Entering a new prompt.  Type `,bt' for a backtrace or `,q' to continue.
>>>
>>> but works in this case:
>>>
>>> scheme@(guile-user)> (define k 0)
>>> scheme@(guile-user)> (let loop () (if (< k 4) (let () (display k)
>>> (newline) (<- k (+ k 1)) (loop))))
>>> if-defined : where=#t
>>> id=#<syntax:unknown file:25:59 k>
>>> 0
>>> 1
>>> 2
>>> 3
>>>
>>> i know it is hard and perheaps not possible but if someone has any idea?
>>>
>>> Damien
>>>
>>>