David Kastrup <d...@gnu.org> writes:
> within local-eval, what is the return value of calling
> (current-module)? I would expect that it is the same as outside of
> local-eval
Actually, this is not true. Within `local-eval', (current-module) is
temporarily restored (using dynamic-wind) to the module saved in the
lexical environment.
This is probably not optimal. Among other things, it breaks the nice
equivalence <expr> == (local-eval '<expr> (the-environment)), but
`local-eval' and `local-compile' are based on `eval' and `compile', and
this is what `eval' and `compile' do (for some reason unknown to me).
> so that (define x 5) inside of local-eval
As Andy pointed out, I deliberately restrict the form passed to
`local-eval' (or `local-compile') to be an _expression_. (define x 5)
is not an expression; it is a definition. I explained the reason for
this restriction a few days ago[1] on this list.
> would _not_ be equivalent to (module-define! (current-module) 'x 5) as
> the first one would take the current module at the-environment time,
> and the second one would take it at local-eval time.
Actually, because `local-eval' restores (current-module) to the one
saved in the lexical environment, (module-define! (current-module) 'x 5)
within `local-eval' actually defines the variable within the saved
module, i.e. the module used for lookup if you replace (the-environment)
with a top-level variable reference.
Mark
[1] Message to guile-devel dated 08 Jan 2012 15:39:36 -0500, with
subject "Re: [PATCH] Implement local-eval, local-compile, and
the-environment"
