Issue
When compiling a shared library with gcc / g++, why is -fPIC not implied by -shared option? Or, said differently, is the option -fPIC required at link time?
For short, should I write:
gcc -c -fPIC foo.c -o foo.o
gcc -shared -fPIC foo.o -o libfoo.so // with -fPIC
or is the following sufficient:
gcc -c -fPIC foo.c -o foo.o
gcc -shared foo.o -o libfoo.so // without -fPIC
Solution
Code that is built into shared libraries should normally but not mandatory be position independent code, so that the shared library can readily be loaded at any address in memory. The -fPIC option ensures that GCC produces such code. It is not required, thus it is not implied in -shared so GCC gives a freedom of choice. Without this option the compiler can make some optimizations on that position dependent code.
Position dependent code may occur an error if one process wants to load more than one shared library at the same virtual address. Since libraries cannot predict what other libraries could be loaded, this problem is unavoidable with the traditional shared library concept. Virtual address space doesn't help here. If your application does not use a lot of other shared libraries and they are loaded before yours, you can predict the loading address of your library and you can define it as a base address of your position dependent library.
The shared library is supposed to be shared between processes, it may not always be possible to load the library at the same address in both. If the code were not position independent, then each process would require its own copy.
Answered By - 273K Answer Checked By - David Goodson (WPSolving Volunteer)