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Date: Fri, 1 Mar 1996 15:03:24 +0800 (GMT+0800)
From: "Humprey C. Sy" <humprey@linux1.dlsu.edu.ph>
To: Michael Smith <msmith@atrad.adelaide.edu.au>
cc: questions@freebsd.org
Subject: Re: Compiling with shared libraries
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On Fri, 1 Mar 1996, Michael Smith wrote:

> > I'm confused about how shared libraries actually work.  Here's my 
> > problem:
> > 
> > I was able to make a shared library file - libmac.a.  What I'm confused 
> 
> If that's really a shared library, it should be libmac.so.1.0.  How
> are you building it?  If you were using the standard /usr/share/mk 
> templates it'd be named correctly.

I used the bsd.lib.mk template.  It produced libmac.a, that's all.  
What's the difference between <lib>.a and <lib>.so.X.X files anyway?  I 
thought they're both the same, save for compatibility reasons only.

> > about is how come if I have to compile it together with my c program, I 
> > have to place the option "-lmac" at the end of the command. i.e.
> > 
> > 	cc -O2 zo.c -lmac
> > 
> > If I try to include "-lmac" before zo.c, the procedures compiled in 
> > libmac.a won't be seen at all, and the compiler returns errors saying 
> > these procedures are unreferenced from text segment.  Why is this so?   
> 
> The linker commandline is read left to right

I still don't get the difference.  Either way both files will be linked 
together, so why can't the linker find those procedures in libmac.a if I 
typed -lmac before zo.c?

- Humprey -