On Thu, Jun 3, 2010 at 2:34 AM, David Aldrich <[email protected]> wrote: > Our software application consists of a kernel executable and several shared > libraries. Our makefiles build the libraries and executable correctly. > However, there is one minor inconvenience, when a source file changes, which > is a component of a shared library, the compilation command for the cpp file > is displayed on the console, but the libtool command is not. So it is not > clear that the library has been updated. > > So I just see something like: > > $ make > g++ -c myfile.cpp > $
Do any of your Makefiles (or the files they include) use the .SILENT target? > # Rule for building release library > $(SOLIBDIR)/$(OBJDIR_R)/lib$(STARLIBNAME).so : $(patsubst > %,$(OBJDIR_R)/%,$(OBJFILES)) > $(TRUNKDIR)/$(SVI) $(CURDIR) $(STARLIBNAME) > $(CXX) -c $(CXXFLAGS_R) SourceFileInfo.cpp -o > $(OBJDIR_R)/SourceFileInfo.o > @rm -f SourceFileInfo.cpp > $(LIBTOOLCMD) $(CXX) -shared -o > $(SOLIBDIR)/$(OBJDIR_R)/lib$(STARLIBNAME).so $(patsubst > %,$(OBJDIR_R)/%,$(OBJFILES)) $(OBJDIR_R)/SourceFileInfo.o $(LIBDEPEND) Is the LIBTOOLCMD defined to start with an '@' ? Philip Guenther _______________________________________________ Help-make mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-make
