Right, but why cannot we forge an identifier easily? I'm happy getting an armed identifier. What are the reasons for preventing such a construction?
On Thu, May 23, 2013 at 6:04 PM, Carl Eastlund <c...@ccs.neu.edu> wrote: > Essentially yes. It doesn't do anything else, but it needs an identifier to > do it. Currently, TR starts with a module and a symbol, goes through an > expensive process to forge an identifier from them, just to call > free-identifier=? to compare based on the module and the symbol after all. > Doing the comparison directly, without ever forging the identifier, would be > quicker. > > > On Thu, May 23, 2013 at 8:43 PM, Eric Dobson <eric.n.dob...@gmail.com> > wrote: >> >> Isn't that exactly what free-indentifier=? is checking for on >> identfiers with a module level binding? Or is there something else it >> does? >> >> On Thu, May 23, 2013 at 3:13 PM, Carl Eastlund <c...@ccs.neu.edu> wrote: >> > On Thu, May 23, 2013 at 4:13 PM, Ryan Culpepper <ry...@ccs.neu.edu> >> > wrote: >> >> >> >> On 05/23/2013 01:57 AM, Eric Dobson wrote: >> >>> >> >>> Some modules have macros which expand into identifiers that are not >> >>> exported, as they want to protect those bindings. TR currently has the >> >>> following code which allows it to generate an identifier which is >> >>> free-identifier=? to what would appear in the output of the macros. >> >>> >> >>> define (make-template-identifier what where) >> >>> (let ([name (module-path-index-resolve (module-path-index-join >> >>> where >> >>> #f))]) >> >>> (parameterize ([current-namespace (make-empty-namespace)]) >> >>> (namespace-attach-module (current-namespace) ''#%kernel) >> >>> (parameterize ([current-module-declare-name name]) >> >>> (eval `(,#'module any '#%kernel >> >>> (#%provide ,what) >> >>> (define-values (,what) #f)))) >> >>> (namespace-require `(for-template ,name)) >> >>> (namespace-syntax-introduce (datum->syntax #f what))))) >> >>> >> >>> This turns out to be a slightly slow part of the initialization of TR. >> >>> Does anyone know another way to get such an identifier? >> >> >> >> >> >> There's another way around this issue, which is to avoid creating these >> >> identifiers at all. In other words, change the representation of the >> >> type >> >> environment to something that supports symbol+module pairs as keys in >> >> addition to identifiers. The easiest way to do that is to add in a hash >> >> table behind the current free-id-table, since the two tables would >> >> handle >> >> disjoint sets of identifiers. >> >> >> >> Ryan >> > >> > >> > I would not have thought that'd work, but apparently identifier-binding >> > will >> > give one that information. Nice going, Ryan! >> > >> > --Carl >> > _________________________ Racket Developers list: http://lists.racket-lang.org/dev