[Bug c++/67195] cpp and g++ does not define __GLIBCXX__

["pinskia at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67195

--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
(In reply to Jeffrey Walton from comment #2)
> > 
> > $ uname -a
> > Linux localhost.localdomain 4.1.3-201.fc22.x86_64 #1 SMP Wed Jul 29 19:50:22
> > UTC 2015 x86_64 x86_64 x86_64 GNU/Linux
> > 
> > $ g++ --version
> > g++ (GCC) 5.1.1 20150618 (Red Hat 5.1.1-4)
> > Copyright (C) 2015 Free Software Foundation, Inc
> 
> (In reply to Jonathan Wakely from comment #1)
> > The library macro is defined by the library, not the compiler. You didn't
> > include any library headers.
> 
> Thanks Jonathon.
> 
> The documentation does not state that requirement. Perhaps this is a
> documentation bug?

I think the document assumes that since this is the library docs rather than
the compiler or preprocessor doc n



> 
> ------
> 
> Related, here's how I am trying to use it in a makefile:
> 
> USING_GLIBCXX := $(shell $(CXX) $(CXXFLAGS) -E -dM - </dev/null | $(EGREP)
> -i -c "__GLIBCXX__")
> $(info USING_GLIBCXX: $(USING_GLIBCXX))
> 
> # For debug builds
> ifneq ($(USING_GLIBCXX),0)
> CXXFLAGS += -D_GLIBCXX_DEBUG=1 --enable-concept-checks
> endif
> 
> CXX and CXXFAGS are populated, and they hold the settings that will be used
> by make to compile the source code. It kind of begs the question, how does
> CXX know to use -stdlib=libstdc++ if its not specified in CXXFLAGS?
> Shouldn't everything "just work" or everything "just fail"?