Why does GCC create a shared object instead of an executable binary according to file?

Question

I have a library I am building. All of my objects compile and link successively when I run either one of: ar rcs lib/libryftts.a $^

gcc -shared $^ -o lib/libryftts.so

in my Makefile. I also am able to successfully install them into /usr/local/lib When I test the file with nm, all the functions are there. My problem is that when I run gcc testing/test.c -lryftts -o test && file ./test or gcc testing/test.c lib/libryftts.a -o test && file ./test it says:

test: ELF 64-bit LSB shared object instead of test: ELF 64-bit LSB executable as I would expect. What am I doing wrong?

Answer 1

What am I doing wrong?

Nothing.

It sounds like your GCC is configured to build -pie binaries by default. These binaries really are shared libraries (of type ET_DYN), except they run just like a normal executable would.

So your should just run your binary, and (if it works) not worry about it.

Or you could link your binary with gcc -no-pie ... and that should produce a non-PIE executable of type ET_EXEC, for which file will say ELF 64-bit LSB executable.