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Re: two instances of global from shared lib linked with -Bsymbolic

From: Rafal Dabrowa
Subject: Re: two instances of global from shared lib linked with -Bsymbolic
Date: Tue, 6 Apr 2004 20:03:58 +0200
User-agent: KMail/1.6.1

On Tuesday 06 of April 2004 18:22, Ian Lance Taylor wrote:
> Using -Bsymbolic breaks language rules.  That is why it is not the
> default.  The option exists because it can make shared library code
> more efficient, for people who do not require strict adherence to
> normal variable usage.
> Why are you using -Bsymbolic?

I want to have the following behavior. I have created a shared library with a 
lot of functions and global variables exposed to others. But I don't want up 
to anybody would override my functions. Suppose I have two functions in my 

        void fun1() { ... }
        void fun2() { fun1(); ... }

Function fun2 invokes fun1. I want up to fun2 would always invoke fun1 from my 
library. It is surprising for me, that if I create fun1 in main program, then 
fun2 invokes fun1 from main program instead of library. From this reason I 
have added -Bsymbolic option.

But, this option caused that I can't access any global variable from library. 
If a library function changes a global from library (e.g. MyErrno), I can't 
see the varibale change in main program.

Do you know any solution ?


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