Hello Michel, * Michel Briand wrote on Thu, Feb 12, 2009 at 03:41:42PM CET: > plugin_LTLIBRARIES = libFreeWRLplugin.la > plugindir=$(PLUGIN_DIR) # configure puts in it /usr/lib/mozilla/plugins > libFreeWRLplugin_la_LDFLAGS = -avoid-version $(AM_LDFLAGS)
> This produces the following command line: > > /bin/bash ../../libtool --tag=CC --mode=link gcc -g -O2 -avoid-version > -o libFreeWRLplugin.la -rpath /usr/lib/mozilla/plugins plugin_main.lo > npunix.lo internal_version.lo > libtool: link: gcc -shared .libs/plugin_main.o .libs/npunix.o > .libs/internal_version.o -Wl,-soname -Wl,libFreeWRLplugin.so -o > .libs/libFreeWRLplugin.so > > Why -rpath /usr/lib/mozilla/plugins ??? Quoting '(libtool.info.gz)Link mode': | `-rpath LIBDIR' | If OUTPUT-FILE is a library, it will eventually be installed in | LIBDIR. If OUTPUT-FILE is a program, add LIBDIR to the run-time | path of the program. It seems weird, and it is, but somebody chose '-rpath' to have this rather unusual meaning for libtool. > The rpath troubles me. I think that rpath would be use to specify > library path needed by the shared object. Not the path where it is > supposed to be installed. Am I right ? Not in this case; you are thinking about the ld option -rpath, typically passed to compilers as -Wl,-rpath,... But that's not what the above means to libtool, at least not in the case of creating a library. It does not cause /usr/lib/mozilla/plugins to be added to the run path of the library. Hope that helps. I haven't read the rest, so if there were more questions hidden there, please ping. Cheers, Ralf _______________________________________________ http://lists.gnu.org/mailman/listinfo/libtool
