Re: Is it possible for a macro to expand to nothing?

[Pascal J. Bourguignon]

"Drew Adams" <drew.adams@oracle.com> writes:

>> > (defmacro titi (fn)
>> >   `(defun ,fn ()
>> >      (setq bar 1)
>> >      ,@(ifdef baz '((setq bar 2))))))
>> >
>> > Assuming that ifdef returns nil if baz is nil, that should 
>> > give you (defun foo () (setq bar 1)). If baz is not nil,
>> > it should give you this:
>> >
>> > (defun foo ()
>> >  (setq bar 1)
>> >  (setq bar 2))
>> 
>> Yes.  Unfortunately, (ifdef baz '((setq bar 2))) doesn't produce a
>> valid form when baz is not nil.
>
> Huh? It produces the list ((setq bar 2)), assuming `ifdef' acts like `and'
> (`ifdef' is undefined AFAIK).

Yes.  And ((setq bar 2)) is not a valid form.  So it is not a good
idea, not a good style to call (ifdef baz '((setq bar 2))) and use its
result like that.


>> This is not a good property.  It would be
>> better to keep the contract of macros, that is they take code, and
>> they produce code, that is, valid forms.
>
> No idea what you are talking about.

Yes, that's the problem.


> It's simple, really: The code resulting from macroexpansion is a list. 

No.  CODE resulting from macroexpansion should be CODE.  Not just a list.
If you want to get random lists, then use functions, not macros.  


-- 
__Pascal Bourguignon__