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Cc: linux-kernel@vger.kernel.org, Adrian Bunk <bunk@stusta.de>
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From: Kyle Moffett <mrmacman_g4@mac.com>
Subject: Re: [2.6 patch] make UNIX a bool
Date: Sat, 4 Mar 2006 17:57:08 -0500
To: "James C. Georgas" <jgeorgas@rogers.com>
Sender: linux-kernel-owner@vger.kernel.org
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On Mar 3, 2006, at 16:31:51, James C. Georgas wrote:
> On Thu, 2006-02-03 at 23:32 -0500, Kyle Moffett wrote:
>> af_unix (IE: CONFIG_UNIX) currently uses the symbol  
>> get_max_files.  It is the only module that uses that symbol, and  
>> that symbol probably should not be exported as it's kind of an  
>> internal API.  Therefore if we mandate that CONFIG_UNIX != m, then  
>> that symbol may be properly unexported and made private, because  
>> nothing modular would use it.  Does that clear things up?
>
> Yes, I think I understand.
>
> However, even if you don't export the symbol, I don't see how you  
> can make it private (i.e. static declaration) to file_table.c,  
> since it has to remain extern, in order to be visible to af_unix.c.

You're missing some crucial information about how Linux operates.   
EXPORT_SYMBOL != extern.  Basically, Linux maintains a list of  
symbols that dynamically loaded modules are allowed to use, along  
with some technical usage restrictions on each  (Symbols exported  
with EXPORT_SYMBOL_GPL may only be used by modules that declare  
'MODULE_LICENSE("GPL");'.)  Exporting a symbol increases the  
likliehood that some module author will use it inappropriately, and  
bloats the kernel.  In this case, removing the EXPORT_SYMBOL() would  
be a good thing.

Cheers,
Kyle Moffett

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