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Re: GNU Libtool 1.5.8 released.

From: Bob Friesenhahn
Subject: Re: GNU Libtool 1.5.8 released.
Date: Thu, 12 Aug 2004 21:15:45 -0500 (CDT)

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 Friesenhahn

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