The details of your answer are in Reference section 1.1.10 http://docs.racket-lang.org/reference/eval-model.html#(part._module-eval-model)
Basically, the runtime and compile time of every module is distinct and not shared between compiles of different modules. Racket guarantees that they are never shared, no matter how many modules are already (or not already) compiled when you start. You may think there is sharing, but there is not. Instead, when d runs it runs the syntax toplevel of b and c, despite having compiled them separately. (You could observe the difference if you changed your begin-for-syntax code into code that happens to run during the expansion of b and c.) The justification for this behavior is elaborated in Matthew Flatt's "Metaprogramming Time!" [1] and the paper "Composable and Compilable Macros" [2] Jay 1. http://www.infoq.com/presentations/racket 2. http://www.cs.utah.edu/plt/publications/macromod.pdf Jay On Tue, Sep 24, 2013 at 1:38 PM, Dmitry Pavlov <dpav...@ipa.nw.ru> wrote: > Hello, > > I have a problem that originates from the fact that I do not > really understand the innards of the mechanism of module loading. > > Basically, I want to share a hashtable stored in some module > between a number of other modules. Moreover, I want to do > that on the syntax level. In this example, I will do that > in parallel on run level and syntax level, showing problems > in the second case. > > ~~~~~ a.rkt: > > #lang racket > (provide h > (for-syntax s-h)) > > (define h (begin (printf "making h\n") (make-hash))) > > (define-for-syntax s-h (begin (printf "making s-h\n") (make-hash))) > > > ~~~~~ b.rkt: > > #lang racket > > (require "a.rkt") > (hash-set! h 1 "x") > (begin-for-syntax > (hash-set! s-h 1 "xx")) > > > ~~~~~ c.rkt: > > #lang racket > (require "a.rkt") > (hash-set! h 2 "y") > (begin-for-syntax > (hash-set! s-h 2 "yy")) > > ~~~~~ d.rkt: > > #lang racket > (require "a.rkt") > (require "b.rkt") > (require "c.rkt") > > (define-syntax (print-s-h stx) > (syntax-case stx () > ((_) > (for (((key val) s-h)) > (printf "~a : ~a\n" key val)) > #'(void)))) > > (for (((key val) h)) > (printf "~a : ~a\n" key val)) > > (print-s-h) > > > > What I get: > > making s-h > making s-h > making s-h > making s-h > making s-h > making s-h > making s-h > 1 : xx > 2 : yy > making s-h > making s-h > making h > 1 : x > 2 : y > > > What I see: on the execution level everything works as I want. > (Is it guaranteed to work that way?) > > On the syntax level, the hash table is created eight times instead > of (as I would expect) one, but still only one of the instances > is in action, and is shared across modules. Is it guaranteed to > work that way? In my real (more complicated) code a similar > hash table is not shared, and in effect, the "top" procedure gets > a fresh and empty instance of it, therefore I am in trouble. > I am wondering what are Racket's rules for sharing module data on the > syntax level. > > > Regards, > > Dmitry > ____________________ > Racket Users list: > http://lists.racket-lang.org/users -- Jay McCarthy <j...@cs.byu.edu> Assistant Professor / Brigham Young University http://faculty.cs.byu.edu/~jay "The glory of God is Intelligence" - D&C 93 ____________________ Racket Users list: http://lists.racket-lang.org/users
