On Thursday, July 3, 2003, at 01:44  PM, Marcus Comstedt wrote:

> Well, this works, but then how come I don't need the -E flag when I
> link against the shared object?  That is, if I do
>
>   gcc -o prog foo.o -R. bar.so
>
> then bar can access the symbols in foo without -E, but if I instead
> dlopen() bar.so from foo, -E is needed.  I think the semantics should 
> be
> the same for these two cases, if I specify RTLD_GLOBAL to dlopen()...

I think in this case you're simply linking a PIC object directly into 
the executable, right?  It's not actually recording a dependency on 
bar.so in the binary, is it?

         -- Jason R. Thorpe <thorpej@wasabisystems.com>