why is an archive member linked in ?

why is an archive member linked in ?

[Philippe De Muyter]

I need to answer the following ld-related question : why is some specific
archive-member linked in ?  Is there a gld option available 
or some other more intricated way  to answer that question ?

In the specific case I encountered, removing the module from the archive
made my link command succeed, while with the module in the archive my
link command failed because of some unresolved symbols in the module
that got linked in.

Thanks in advance

Philippe De Muyter

Re: why is an archive member linked in ?

[DJ Delorie]

> In the specific case I encountered, removing the module from the archive
> made my link command succeed, while with the module in the archive my
> link command failed because of some unresolved symbols in the module
> that got linked in.

Sometimes, there might be more than one object that resolves a
specific symbol.  If you take one out, the other gets used, and that
changes everything.

For example, DJGPP has two math libraries.  One (libm.a) has a
complete set of conforming (but slightly slower) functions, and the
other (in libc.a) is smaller and faster, but less complete and less
conforming.  If you link in libm.a, you get that set, but if you leave
it out, the ones in libc.a get used.

Re: why is an archive member linked in ?

[Philippe De Muyter]

> 
> 
> > In the specific case I encountered, removing the module from the archive
> > made my link command succeed, while with the module in the archive my
> > link command failed because of some unresolved symbols in the module
> > that got linked in.
> 
> Sometimes, there might be more than one object that resolves a
> specific symbol.  If you take one out, the other gets used, and that
> changes everything.
> 
> For example, DJGPP has two math libraries.  One (libm.a) has a
> complete set of conforming (but slightly slower) functions, and the
> other (in libc.a) is smaller and faster, but less complete and less
> conforming.  If you link in libm.a, you get that set, but if you leave
> it out, the ones in libc.a get used.
> 

Thanks.  I already know that but that's the (slow and ugly) method I use
to find why an archive member causing a link error is linked in.  What I need
is a clean and fast method, e.g. a gld option or, better, a direct information
in the error message.

Philippe

Re: why is an archive member linked in ?

[Ian Lance Taylor]

   Date: Tue, 30 Dec 1997 04:09:04 +0100 (MET)
   From: "Philippe De Muyter" <phdm@macqel.be>

   I need to answer the following ld-related question : why is some specific
   archive-member linked in ?  Is there a gld option available 
   or some other more intricated way  to answer that question ?

Use the -M/-Map option, and look at the top of the mapfile.

Ian