If you start with an empty namespace, then the handful of primitive modules needed to bootstrap `racket/base` are not there. Attaching `racket/base` doesn't make "m.rkt" available, but it makes it possible to load "m.rkt", after which "m.rkt" will be declared in the namespace.
It's not a matter of initializing the module name resolver, which is not attached to a namespace. It's just a question of having the necessary primitive modules declared in a namespace. The `'#%builtin` module imports all the primitive modules needed to bootstrap `racket/base` --- that's its job --- while neither `'#%kernel` nor `'#%boot` reach all of the needed primitive modules. Ideally, you wouldn't be able to access any of those modules whose names start "'#%", because you shouldn't use them. I don't know how to hide the modules from you without also hiding them from `racket/base`. When I try (parameterize ([current-namespace (make-base-empty-namespace)]) (namespace-require "m.rkt") (module-declared? "m.rkt")) then I get #t back, so I'm not sure why you were getting #f... unless you tried (parameterize ([current-namespace (make-base-empty-namespace)]) (namespace-require "m.rkt")) (module-declared? "m.rkt") which would produce #f, because `module-declared?` consults the current namespace. At Mon, 28 Apr 2014 16:48:04 -0400, Vincent St-Amour wrote: > Extra bit of information, this works too: > > #lang racket > (define ns (make-empty-namespace)) > (namespace-attach-module (current-namespace) ''#%builtin ns) > (parameterize ([current-namespace ns]) > (namespace-require "m.rkt")) > > But replacing ''#%builtin with ''#%kernel or ''#%boot doesn't. Replacing > it with 'racket/base (to end up with the equivalent of > `make-base-empty-namespace') does work. > > Vincent > > > > At Mon, 28 Apr 2014 16:38:24 -0400, > Stephen Chang wrote: > > > > Motivated by some recent email threads, I decided to better understand > > how Racket namespaces work by re-reading some docs but I got confused. > > > > Paraphrasing the examples from this part of the guide: > > http://docs.racket-lang.org/guide/mk-namespace.html > > > > the following example fails because the module registry in the > > namespace is empty: > > > > (parameterize ([current-namespace (make-empty-namespace)]) > > (namespace-require "m.rkt")) ; error > > > > But the example works if make-base-empty-namespace is used instead: > > > > (parameterize ([current-namespace (make-base-empty-namespace)]) > > (namespace-require "m.rkt")) ; works > > > > (Assume the contents of m.rkt is "#lang racket/base" with nothing else.) > > > > According to the docs, make-base-empty-namespace "attaches" > > racket/base but why does this make "m.rkt" suddenly "available"? There > > seems to be an unexplained gap between to the two examples. (Adding to > > my confusion is that (module-declared? "m.rkt") evaluates to false, > > even in the 2nd case.) > > > > With help from Vincent, we investigated and speculate that perhaps > > attaching racket/base also runs "boot" from '#%boot, which populates > > current-module-name-resolver, but could not conclude anything > > definitively. Can someone explain what's happening? > > _________________________ > > Racket Developers list: > > http://lists.racket-lang.org/dev > _________________________ > Racket Developers list: > http://lists.racket-lang.org/dev _________________________ Racket Developers list: http://lists.racket-lang.org/dev
