On Dec 28, 2005, at 3:53 PM, Peter O'Gorman wrote:
GNU libtool will behave differently should libtool-2.0 ever be
released (there will be libfoo.dylib and libfoo.X.dylib, removing a
redundant symlink, and libraries will, by default, link -
single_module).
Hooray!!
-- Dave
Le 27 déc. 2005 à 14:52, Martin Costabel a écrit :
Michèle Garoche wrote:
I try to compile a package which constructs three plugins during
the compilation.
[]
$(PLUGIN): $(OBJECTS)
$(LIBTOOL) --mode=link $(CC) -module $(LIBS) -shared -o $@ $
(LOBJECTS)
What happens if you remove the
Michèle Garoche wrote:
Le 27 déc. 2005 à 14:52, Martin Costabel a écrit :
[]
What happens if you remove the -shared from this line?
It does not change anything, apart that it does not tell that -shared is
undefined of course.
Libtool should recognize the -module flag and translate it to
Le 28 déc. 2005 à 10:27, Martin Costabel a écrit :
Michèle Garoche wrote:
Le 27 déc. 2005 à 14:52, Martin Costabel a écrit :
[]
What happens if you remove the -shared from this line?
It does not change anything, apart that it does not tell that -
shared is undefined of course.
Libtool
On Dec 27, 2005, at 1:52 PM, Martin Costabel wrote:
-Wl,-bind_at_load -flat_namespace -undefined suppress -bundle
If you want to have a two-level image (and don't get an answer from
pogma explaining how to talk to libtool in its native language),
you can replace the -module -shared simply
I try to compile a package which constructs three plugins during the
compilation.
An example of a typical Makefile.in for the plugin is:
PLUGIN=about.so
PLUGINDIR=${pkglibdir}
OBJECTS= about.o
LOBJECTS = $(OBJECTS:.o=.lo)
$(PLUGIN): $(OBJECTS)
$(LIBTOOL) --mode=link $(CC) -module
Michèle Garoche wrote:
I try to compile a package which constructs three plugins during the
compilation.
[]
$(PLUGIN): $(OBJECTS)
$(LIBTOOL) --mode=link $(CC) -module $(LIBS) -shared -o $@ $(LOBJECTS)
What happens if you remove the -shared from this line?
Libtool should recognize the
