What is the idiomatic way in CMAKE to add the -fPIC compiler option?

Question

I've come across at least 3 ways to do this and I'm wondering which is the idiomatic way. This needs to be done almost universally to any static library. I'm surprised that the Makefile generator in CMake doesn't automatically add this to static libraries. (unless I'm missing something?)

target_compile_options(myLib PRIVATE -fPIC)

add_compile_options(-fPIC)

set(CMAKE_C_FLAGS "${CMAKE_C_FLAGS} -fpic")
set(CMAKE_CXX_FLAGS "${CMAKE_CXX_FLAGS} -fpic")

I believe there might also be other variations. (please edit my question if you find one)

If you happen to know the answer to this question, do you also know if there is a way to cause a 3rd party CMake project to be compiled with this flag without modifying its CMakeLists.txt file? I have run across static libraries missing that flag. It causes problems when compiling a static library into a dynamic library.

You get:

relocation R_X86_64_32 against `.rodata' can not be used when making a shared object; recompile with -fPIC

this set(CMAKE_CXX_FLAGS "${CMAKE_CXX_FLAGS} -fpic") worked for me. — moki
^ Do not do that. Use the POSITION_INDEPENDENT_CODE property, as suggested in the top answer. — Alex Reinking

Amadeus Amadeus · Accepted Answer · 2016-07-10T23:45:23

You can set the position independent code property on all targets:

set(CMAKE_POSITION_INDEPENDENT_CODE ON)

or in a specific library:

add_library(lib1 lib1.cpp)
set_property(TARGET lib1 PROPERTY POSITION_INDEPENDENT_CODE ON)

Reference: CMAKE_POSITION_INDEPENDENT_CODE cmake build system

What is the idiomatic way in CMAKE to add the -fPIC compiler option?

2 Answers