Re: How to use-modules within macro?

[Mark H Weaver]

Hello again,

I wrote earlier:
> So, instead of using 'match' on the result of 'syntax->datum', you
> should instead use 'syntax-case' on the syntax object itself, like this
> (untested):
>
>   (let loop ((e #'exp))
>     (syntax-case e ()
>       (num
>        (number? (syntax->datum #'num))
>        #'(1+ num))
>       ((x ...)
>        (map loop #'(x ...)))
>       (y
>        #'y)))

I should mention that the use of 'map' directly on a syntax object is
only allowable in certain cases.  Here, we assume that the syntax object
produced by #'(x ...) is a normal Scheme list.  That is _not_ the case
for arbitrary syntax objects that correspond to a list.  For example, it
would _not_ be safe to pass 'e' as the list argument to 'map', although
'e' is in some sense equivalent to #(x ...) in the second clause above.

The reason is that syntax objects contain information about the
associated lexical environment which starts at the top of the expression
tree, and is *lazily* pushed down into the subexpressions as the tree is
taken apart using 'syntax-case' and put back together using 'syntax'
(a.k.a. "#'").

As a result, there are only a few cases when you can safely assume that
the top structure of a syntax object is a normal list or pair, and they
are spelled out in the documentation for 'syntax' in the R6RS Standard
Libraries specification:

  http://www.r6rs.org/final/html/r6rs-lib/r6rs-lib-Z-H-13.html#node_sec_12.4

Here's the relevant excerpt:

  The output produced by syntax is wrapped or unwrapped according to the
  following rules.

  * the copy of (<t1> . <t2>) is a pair if <t1> or <t2> contain any
    pattern variables,

  * the copy of (<t> <ellipsis>) is a list if <t> contains any pattern
    variables,

  * the copy of #(<t1> ... <tn>) is a vector if any of <t1>, ..., <tn>
    contain any pattern variables, and

  * the copy of any portion of <t> not containing any pattern variables
    is a wrapped syntax object.

A "wrapped syntax object" is one where the lexical environment
information has not yet been pushed down into the subexpressions.  It is
a special kind of object that you can only take apart using
'syntax-case'.

So, in the clause above where 'map' is used, 'e' might be a "wrapped
syntax object", but when the elements are extracted from it using the
'syntax-case' pattern (x ...) and then put back together using
#'(x ...), you can then assume that the resulting syntax object is
a normal Scheme list of syntax objects, and therefore it is safe to
use 'map' on it.

       Mark