Emacs ffi (was: Re: as for Calc and the math library)

[Andrea Corallo <acorallo@gnu.org>]

Eli Zaretskii <eliz@gnu.org> writes:

>> From: Suhail Singh <suhailsingh247@gmail.com>
>> Cc: Suhail Singh <suhailsingh247@gmail.com>,  gerd.moellmann@gmail.com,
>>  nicolas@n16f.net,  arthur.miller@live.com,  emacs-devel@gnu.org
>> Date: Wed, 14 Aug 2024 11:08:02 -0400
>> 
>> Eli Zaretskii <eliz@gnu.org> writes:
>> 
>> >> On the topic of what would be acceptable for an FFI, wouldn't something
>> >> akin to what's done for modules be sufficient ?  I.e., have the users of
>> >> the interface explicitly state that they are compliant.
>> >> 
>> >> It would scale better than an allow-list.  IIUC, Arthur mentioned this
>> >> in another thread.  If this wouldn't be sufficient for an FFI, could you
>> >> please elaborate on why that's the case ?
>> >
>> > What exactly are you suggesting?  IOW, please describe what you have
>> > in mind in more detail, because I don't think I understand it.
>> 
>> Specifically, modify the `define-ffi-library' macro that emacs-ffi
>> provides.
>> 
>> Presently, it takes two arguments: a SYMBOL and a NAME.  I am proposing
>> that it be updated to take three arguments: a SYMBOL, a NAME and a
>> GPL-COMPATIBLE-P token.  A value of `t' would be necessary for creating
>> a reference to the library.
>
> And if the value is not t, then the load will fail?  If so, then this
> additional argument makes very little sense: you could instead say
> that just by loading the library, the Lisp program which uses
> emacs-ffi "declares" the loaded library to be GPL-compatible.  And we
> are back where we began.

Agree

> The way we do it when loading modules requires the _loaded_ library to
> declare itself compatible, by exporting a symbol of a certain name.
> That is an action by the library we load, not by the Lisp program
> which loads it.

But we could do the same for our hypothetical ffi machinery, that is:
`define-ffi-library' could fail if the shared library is not exporting
the predetermined symbol we expect no?

  Andrea