Passing a gcc flag through makefile

Question

Hi,

I am trying to build a pass using llvm and I have finished building llvm and its associated components. However, when I run make after following all the steps to build a pass including the makefile, I get the following error:

*relocation R_X86_64_32 against `a local symbol' can not be used when making a shared object; recompile with -fPIC*

After tyring to find a fix by googling the error message, I came to know that this is not specific to llvm. A few solutions suggested that I should use "--enable-shared" while running configure but that didn't help my case. Now I want to re-build llvm using fPIC, as the error says. But how do I do this using the makefile?

Answer 1

Looks like you could add the -fPIC (for position-independent code, something you want for a shared library that could be loaded at any address) by setting shell variables:

export CFLAGS="$CFLAGS -fPIC"
export CXXFLAGS="$CXXFLAGS -fPIC"

Looking at Makefile.rules, these will be picked up and used. Seems strange that it wasn't there to begin with.

EDIT:

Actually, reading more in the makefiles, I found this link to the LLVM Makefile Guide. From Makefile.rules, setting either SHARED_LIBRARY=1 or LOADABLE_MODULE=1 (which implies SHARED_LIBRARY) in Makefile will put -fPIC in the compiler flags.

Answer 2

If you are moderately convinced that you should use '-fPIC' everywhere (or '-m32' or '-m64', which I need more frequently), then you can use the 'trick':

CC="gcc -fPIC" ./configure ...

This assumes a Bourne/Korn/POSIX/Bash shell and sets the environment variable CC to 'gcc -fPIC' before running the configure script. This (usually) ensures that all compilations are done with the specified flags. For setting the correct 'bittiness' of the compilation, this sometimes works better than the various other mechanisms you find - it is hard for a compilation to wriggle around it except by completely ignoring the fact you specified the C compiler to use.