Re: [racket-users] Macro help

[Michael Murdock MacLeod]

Does this work? It uses a helper function, `prune`, to parse the var-val 
clauses.

#lang racket

(define-for-syntax (prune stx)
  (syntax-case stx ()
[()
 #'()]
[((var val) others ...)
 (cons #'(var val)
   (prune #'(others ...)))]
[(var others ...)
 (cons #'(var #f)
   (prune #'(others ...)))]))

(define-syntax (aux stx)
  (syntax-case stx ()
[(_ terms ...)
 (with-syntax ([((var val) ...) (prune #'(terms ...))])
   #'(define-values (var ...) (values val ...)))]))

(aux a (b (* 2 pi)) c (d pi))
a
b
c
d

;; output shown below

#f
6.283185307179586
#f
3.141592653589793

On Thursday, May 23, 2019 9:41:17 PM PDT Kevin Forchione wrote:
> Hi guys,
> I’ve been wracking my brains all day trying to come up with a macro that
> would convert this syntax:
> 
> ;; (aux a (b (* 2 pi)) c (d pi))
> ;; => (define-values (a b c d) (values #f 6.28318530717958 #f
> 3.141592653589793)
> 
> 
> I’m missing some part of the picture. The closest I’ve come is to create a
> list of the pairs:
> 
> #lang racket
> 
> (define-syntax (aux stx)
>   (syntax-case stx ()
> [(_ (var val)) #'`((var ,val))]
> [(_ var) #''((var #f))]
> [(_ (var0 val0) var1 ...) #'(append `((var0 ,val0)) (aux var1 ...))]
> [(_ var0 var1 ...) #'(append '((var0 #f)) (aux var1 ...))]))
> 
> (aux a (b (* 2 pi)) c (d pi)) ;=> '((a #f) (b 6.283185307179586) (c #f) (d
> 3.141592653589793))
> 
> 
> Any help is greatly appreciated!
> 
> Kevin


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Re: [racket-users] question about matching with "(? ...)"

[Michael Murdock MacLeod]

Match treats `empty` as an identifier to bind, not as a pattern for the empty 
list. This means that the first case will always be matched. You can use 
`(list)` or `'()` to match an empty list.

Also, in the last catch-all case, you pass the function `rest` which returns 
the tail of a list to `filter-odds`. You probably wanted to shadow that binding 
by using `(cons first rest)` instead of `_`.

Here's the version with the fixes above:

(define (filter-odds v)
  (match v
[(list) (list)]
[(cons (? odd? first) rest)
 (cons first (filter-odds rest))]
[(cons first rest) (filter-odds rest)]))

(filter-odds '(1 2 3 4 5)) ; ==> '(1 3 5)

Best,
Michael MacLeod

On Sunday, April 21, 2019 1:19:03 PM PDT Tim Meehan wrote:
> Forgive this naive question, I am having some trouble understanding some
> matching forms. If I wanted to filter out odd numbers like this:
> 
> (filter odd? '(1 2 3 4 5)) => '(1 3 5)
> 
> but I wanted to do this with a match:
> (define (filter-odds v)
>   (match v
> [empty empty]
> [(cons (? odd? first) rest) (cons first (filter-odds rest))]
> [_ (filter-odds rest)]))
> 
> (filter-odds '(1 2 3 4 5)) => '(1 2 3 4 5)


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Re: [racket-users] catch and bind an unbound id in a macro

[Michael Murdock MacLeod]

I'm in no ways a macro expert, but will this work? It uses identifier-binding 
to check if the identifier for the hash table is bound.

(define-syntax (set/define stx)
  (syntax-case stx ()
[(_ ht key value)
 (cond [(identifier-binding #'ht)
#'(hash-set! ht key value)]
   [else
#'(begin
(define ht (make-hash))
(hash-set! ht key value))])]))

On Friday, April 19, 2019 2:33:54 PM PDT zeRusski wrote:
> Not quite what I had in mind. The following examples must all work without
> errors:
> 
> #lang racket
> 
> ;; set/define defined here
> 
> (set/define h a 1)
> (set/define h a 2)
> (hash-set! h 'b 42)
> 
> (define bar (make-hash))
> (set/define bar a 1)
> 
> (define (foo)
>   (set/define local-h a 1)
>   (set/define local-h a 2)
>   (hash-set! local-h 'b 42)
>   local-h)
> 
> (foo)
> 
> Basically, if the identifier h is unbound at runtime
>   (set/define h key val)
> is the same as
>  (define h (make-hash))
>  (hash-set! h 'key val)
> 
> If h is bound
>  (hash-set! h 'key val)


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Re: [racket-users] Help understanding cond expression

[Michael Murdock MacLeod]

On Saturday, January 12, 2019 8:34:35 PM PST Hassan Shahin wrote:
> I have this definition for a procedure:
> 
> (define type-of (lambda (item)
>  (cond
>[(pair? item) 'pair]
>[(null? item) 'empty-list]
>[(number? item) 'number]
>[(symbol? item) 'symbol]
>[else 'some-other-type])))
> 
> My understanding is that if the first 4 conditions fail (=> #f
> ), then the last expression (the else
> expression) is evaluated.
> When I apply this procedure to John, as in (type-of John) I get an error
> (; john: undefined; ; cannot reference an identifier before its definition)
> .
> 
> What is going on?
> Thanks

The issue is that John is evaluated before it is passed to type-of. For 
example, the expression (type-of (+ 2 3)) is equivalent to (type-of 5), 
because the expression (+ 2 3) is evaluated to 5 before it is supplied to the 
type-of function.

In the Google+ post you write "cond behaves as if (not evaluating its 
arguments)". It is true that the arguments to cond, i.e. [(pair? item) 'pair], 
[(null? item) 'empty-list], and so on are not immediately evaluated. However, 
item is evaluated before it is passed to type-of, hence the "cannot reference" 
error.

See https://docs.racket-lang.org/reference/if.html and 
https://docs.racket-lang.org/reference/eval-model.html?q=evaluation%20model for 
more information 
about how Racket evaluates expressions.


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