How is -fPIE passed along to llc? - clang

Question

I'm working on an LLVM backend for a new architecture and we need to have position independent executables. I can pass '-fPIE' on the clang command line, but I don't see any indication of this show up in the resulting LLVM IR. For example, if I run:
clang -v -emit-llvm -fPIC -O0 -S global_dat.c -o global_dat_x86_pic.ll
And then take a look at the resulting global_dat_x86_pic.ll file, I see the following near the bottom:
!0 = !{i32 1, !"PIC Level", i32 2}
Ok, makes sense.
However if I run:
clang -v -emit-llvm -fPIE -O0 -S global_dat.c -o global_dat_x86_pie.ll
I see that the two .ll files are identical. Near the bottom of global_cat_x86_pie.ll I see:
!0 = !{i32 1, !"PIC Level", i32 2}
Which is identical to the case where I ran with -fPIE. There's no indication of "PIE Level" in the .ll file. If this .ll file were passed on to llc how would llc know that -fPIE had been set on the clang command line?
I have run in gdb and see that in fact in the second case with -fPIE on the clang commandline there is an Opts.PIELevel (in $LLVM_HOME/tools/clang/lib/Frontend/CompilerInvocation.cpp) that gets set to 2 (in fact, both Opts.PIELevel and Opts.PICLevel are set to 2 in that case whereas in -fPIC is passed to clang only Opts.PICLevel is set to 2)

Answer 1

This depends on your default target triple, which I can't tell from your question. You can see what happens if you cut off the default architecture (or run a native clang on an arch that does support PIE)
For example, bare X86-64 shows this,
$ clang -c hello.c -target x86_64 -fPIE -emit-llvm -###
if you run that command, you'll find "-pie-level" "2" in the output, which is how llc (well, it's internal equivalent) knows about it.
The key here is that you'll have to arrange for your backend to do something with this flag. Certain platforms (like Darwin) just ignore it. If you happen to be experiment on an OSX host, you won't see -pie-level in the bogbrush output.