Re: GNU Libtool 1.5.8 released.

[Bob Friesenhahn]

On Fri, 13 Aug 2004, Ralf Corsepius wrote:
> > Well, the difference, in my little mind at least, is that the commercial
> > unixes can all be identified in libtool using $host,
> Right, you can identify $host and in case of Solaris even the OS version
> as part of $host (solaris2.8).
>
> However, you can not identify the multilib-variant and the multilib
> subdir being used from $host, because it is chosen depending upon the
> flags being passed to gcc:
> sparc-sun-solaris2.8-gcc .. -> . (sparcv7)
> sparc-sun-solaris2.8-gcc -m64 .. -> sparcv9

If I wrap gcc up in a script which provides secret -m options, then 
you can't see what I am doing.  Neener, neener, neener! :-)

It seems that the only valid test to determine the default output 
architecture is to compile code and then use 'file' or some other 
utility to determine the architecture.  In order to produce multilib 
output, libtool would need to know the specific options necessary to 
obtain each desired variant.  These options are compiler and linker 
specific.

Bob
======================================
Bob Friesenhahn
address@hidden
http://www.simplesystems.org/users/bfriesen