Re: [racket-users] namespace-variable-value thinks functions are bound to syntax

2017-10-31 Thread Junsong Li
That works! Thanks.

On Mon, Oct 30, 2017 at 9:35 AM Shu-Hung You <
shu-hung@eecs.northwestern.edu> wrote:

> Yes, because keyword functions expand to macro definitions that try to
> optimize keyword applications.
>
> namespace-require and namespace-variable-value are relatively
> low-level operations on modules and namespaces. Using dynamic-require
> could be a bit easier:
>
> #lang racket
>
> (module p racket
>   (provide (all-defined-out))
>
>   (define (foo0 a b)
> (+ a b))
>   (define (foo1 a b [c 3])
> (+ a b c))
>   (define (foo2 a b [c 3] #:kw1 kw1)
> (+ a b c))
>   (define (foo3 a b [c 3] #:kw1 kw1 #:kw2 [kw2 1])
> (+ a b c))
>   )
>
> (require syntax/location)
>
> (define ns (make-base-empty-namespace))
>
> (parameterize ([current-namespace ns])
>   (dynamic-require (quote-module-path p) 'foo0))
> ;=> #
>
> (parameterize ([current-namespace ns])
>   (dynamic-require (quote-module-path p) 'foo1))
> ;=> #
>
> (parameterize ([current-namespace ns])
>   (dynamic-require (quote-module-path p) 'foo2))
> ;=> #
>
> (parameterize ([current-namespace ns])
>   (dynamic-require (quote-module-path p) 'foo3))
> ;=> #
>
> Please note that dynamic-require handles macro exports specially, as
> stated in the doc:
>
>
> https://docs.racket-lang.org/reference/Module_Names_and_Loading.html#%28def._%28%28quote._~23~25kernel%29._dynamic-require%29%29
> > If the module exports _provided_ as syntax,
> > then a use of the binding is expanded and
> > evaluated in a fresh namespace to which
> > the module is attached, which means that
> > the module is visited in the fresh namespace.
> > The expanded syntax must return a single value.
>
> Best,
> Shu-Hung
>
> On Sun, Oct 29, 2017 at 7:40 PM, Junsong Li <ljs.darkf...@gmail.com>
> wrote:
> > Hello list,
> >
> > In the following program, racket thinks foo2, and foo3 are bound to
> > syntax. Is this expected behavior?
> >
> > -- File: go.rkt
> >
> > #lang racket
> >
> > (module p racket
> >   (provide (all-defined-out))
> >
> >   (define (foo0 a b)
> > (+ a b))
> >   (define (foo1 a b [c 3])
> > (+ a b c))
> >   (define (foo2 a b [c 3] #:kw1 kw1)
> > (+ a b c))
> >   (define (foo3 a b [c 3] #:kw1 kw1 #:kw2 [kw2 1])
> > (+ a b c))
> >   )
> >
> > (require syntax/location)
> >
> > (define ns (make-base-empty-namespace))
> > (parameterize ([current-namespace ns])
> >   (namespace-require (quote-module-path p)))
> >
> > (namespace-variable-value 'foo0 #t #f ns) ; OK
> > (namespace-variable-value 'foo1 #t #f ns) ; OK
> > (namespace-variable-value 'foo2 #t #f ns) ; namespace-variable-value:
> > bound to syntax in: foo2
> > (namespace-variable-value 'foo3 #t #f ns) ; namespace-variable-value:
> > bound to syntax in: foo3
> >
> > --
> > 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.
>

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[racket-users] namespace-variable-value thinks functions are bound to syntax

2017-10-29 Thread Junsong Li
Hello list,

In the following program, racket thinks foo2, and foo3 are bound to
syntax. Is this expected behavior?

-- File: go.rkt

#lang racket

(module p racket
  (provide (all-defined-out))

  (define (foo0 a b)
(+ a b))
  (define (foo1 a b [c 3])
(+ a b c))
  (define (foo2 a b [c 3] #:kw1 kw1)
(+ a b c))
  (define (foo3 a b [c 3] #:kw1 kw1 #:kw2 [kw2 1])
(+ a b c))
  )

(require syntax/location)

(define ns (make-base-empty-namespace))
(parameterize ([current-namespace ns])
  (namespace-require (quote-module-path p)))

(namespace-variable-value 'foo0 #t #f ns) ; OK
(namespace-variable-value 'foo1 #t #f ns) ; OK
(namespace-variable-value 'foo2 #t #f ns) ; namespace-variable-value:
bound to syntax in: foo2
(namespace-variable-value 'foo3 #t #f ns) ; namespace-variable-value:
bound to syntax in: foo3

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Re: [racket-users] namespace-require a submodule

2017-10-29 Thread Junsong Li
That explains. Thanks!

On Sun, Oct 29, 2017 at 11:20 AM, Sam Tobin-Hochstadt
<sa...@cs.indiana.edu> wrote:
> If you look at what `(quote-module-path food)` produces, it's just
> `(submod "/path/to/go.rkt" food)`. The issue is just that
> `namespace-require` needs a reference that it can resolve without
> knowing what file it's in -- `namespace-require`, unlike `require`, is
> a just a function and doesn't know anything about where it is in the
> source code. So you need to use something like `(quote-module-path)`
> that knows where it is.
>
> Sam
>
> On Sun, Oct 29, 2017 at 2:12 PM, Junsong Li <ljs.darkf...@gmail.com> wrote:
>> Thanks! That works.
>>
>> I am still not sure what is the thing that I don't know. Why do we need a
>> compile-time reference here?
>>
>> On Sun, Oct 29, 2017 at 9:59 AM, Sam Tobin-Hochstadt <sa...@cs.indiana.edu>
>> wrote:
>>>
>>> I think you want to use `quote-module-path` here:
>>>
>>> #lang racket
>>>
>>> (require syntax/location)
>>>
>>> (module food racket
>>>   (provide apple)
>>> (define apple "pie"))
>>>
>>> (parameterize ([current-namespace (make-base-namespace)])
>>>   (namespace-require (quote-module-path food))
>>> (eval 'apple))
>>>
>>> On Sun, Oct 29, 2017 at 12:19 PM, Junsong Li <ljs.darkf...@gmail.com>
>>> wrote:
>>> > Hello list,
>>> >
>>> > I am trying to do namespace-require to a submodule, racket complains
>>> > about
>>> > unknown module, as below:
>>> >
>>> > $ cat go.rkt
>>> > #lang racket
>>> >
>>> > (module food racket
>>> >   (provide apple)
>>> > (define apple "pie"))
>>> >
>>> > (parameterize ([current-namespace (make-base-namespace)])
>>> >   (namespace-require ''food)
>>> > (eval 'apple))
>>> >
>>> > $ racket go.rkt
>>> > require: unknown module
>>> >   module name: #<resolved-module-path:'food>
>>> >   context...:
>>> >
>>> >
>>> > Simply namespace-require food in toplevel has the same issue:
>>> >
>>> > #lang racket
>>> >
>>> > (module food racket
>>> >   (provide apple)
>>> > (define apple "pie"))
>>> >
>>> > (namespace-require ''food)
>>> >
>>> > Finally, the above code works if I type them in the interactive mode.
>>> >
>>> > I tried namespace-attach-module, and I understand why that failed: it
>>> > didn't
>>> > work because (module food racket ...) wouldn't instantiate the module,
>>> > and a
>>> > instantiated module is what namespace-attach-module wants.
>>> >
>>> > What is it that I am missing in above code to require a submodule in a
>>> > namespace? Why is submodule food invisible here?
>>> >
>>> >
>>> > - Junsong
>>> >
>>> > --
>>> > 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.
>>
>>
>> --
>> You received this message because you are subscribed to the Google Groups
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Re: [racket-users] namespace-require a submodule

2017-10-29 Thread Junsong Li
Thanks! That works.

I am still not sure what is the thing that I don't know. Why do we need a
compile-time reference here?

On Sun, Oct 29, 2017 at 9:59 AM, Sam Tobin-Hochstadt <sa...@cs.indiana.edu>
wrote:

> I think you want to use `quote-module-path` here:
>
> #lang racket
>
> (require syntax/location)
>
> (module food racket
>   (provide apple)
> (define apple "pie"))
>
> (parameterize ([current-namespace (make-base-namespace)])
>   (namespace-require (quote-module-path food))
>     (eval 'apple))
>
> On Sun, Oct 29, 2017 at 12:19 PM, Junsong Li <ljs.darkf...@gmail.com>
> wrote:
> > Hello list,
> >
> > I am trying to do namespace-require to a submodule, racket complains
> about
> > unknown module, as below:
> >
> > $ cat go.rkt
> > #lang racket
> >
> > (module food racket
> >   (provide apple)
> > (define apple "pie"))
> >
> > (parameterize ([current-namespace (make-base-namespace)])
> >   (namespace-require ''food)
> > (eval 'apple))
> >
> > $ racket go.rkt
> > require: unknown module
> >   module name: #<resolved-module-path:'food>
> >   context...:
> >
> >
> > Simply namespace-require food in toplevel has the same issue:
> >
> > #lang racket
> >
> > (module food racket
> >   (provide apple)
> > (define apple "pie"))
> >
> > (namespace-require ''food)
> >
> > Finally, the above code works if I type them in the interactive mode.
> >
> > I tried namespace-attach-module, and I understand why that failed: it
> didn't
> > work because (module food racket ...) wouldn't instantiate the module,
> and a
> > instantiated module is what namespace-attach-module wants.
> >
> > What is it that I am missing in above code to require a submodule in a
> > namespace? Why is submodule food invisible here?
> >
> >
> > - Junsong
> >
> > --
> > 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.
>

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[racket-users] namespace-require a submodule

2017-10-29 Thread Junsong Li
Hello list,

I am trying to do namespace-require to a submodule, racket complains about
unknown module, as below:

$ cat go.rkt
#lang racket

(module food racket
  (provide apple)
(define apple "pie"))

(parameterize ([current-namespace (make-base-namespace)])
  (namespace-require ''food)
(eval 'apple))

$ racket go.rkt
require: unknown module
  module name: #
  context...:


Simply namespace-require food in toplevel has the same issue:

#lang racket

(module food racket
  (provide apple)
(define apple "pie"))

(namespace-require ''food)

Finally, the above code works if I type them in the interactive mode.

I tried namespace-attach-module, and I understand why that failed: it
didn't work because (module food racket ...) wouldn't instantiate the
module, and a instantiated module is what namespace-attach-module wants.

What is it that I am missing in above code to require a submodule in a
namespace? Why is submodule food invisible here?


- Junsong

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[racket-users] BSL expansion

2015-05-22 Thread Junsong Li
Hello list,

I am wondering how to expand a BSL program into the Racket _core language_?
Let's say a simple program:

(define (id x) x)

Thanks.

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