I cant remember now the syntax, but maybe the gcc hidden visibility attribute is what you are looking for. El 04/03/2013 08:06, "Satz Klauer" <satzkla...@googlemail.com> escribió:
> I'm creating a shared library using libtool: > > libtool --mode=compile g++ -Wall -fPIC -shared -Wno-unused > -Wno-unused-result $(CFLAGS) $(LFLAGS) > libtool --mode=link g++ -shared -export-symbols-regex mylib_ -rpath > /usr/lib $(LFLAGS) > > (That's a snipped only, there of course is the part that causes > compilation of sources). > > As far as I understood it, the statement "-export-symbols-regex > mylib_" should ensure only functions starting with "mylib_" should be > exported by the resulting library. But when I check the resulting > .so-file it contains all symbols and names that are used within the > sources. Independent if the symbols are private or internal-only, if > they are whole class-names or simple functions, I can see all of them > within the library. The same happens when I use "-export-symbols" > together with a symbol file instead, it seems like the symbol > definitions are ignored completely. > > So my question: how can I avoid that? How can I strip the library > completely to have only these functions exported and visible that are > really allowed to be called by programs that make use of my library? > That's important for two reasons: I want to save binary code size and > (even more important) avoid users are calling functions/using objects > that are not intended for external usage. > > Thanks! > > _______________________________________________ > https://lists.gnu.org/mailman/listinfo/libtool >
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