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

c gcc shared-libraries static-libraries position-independent-code
Luke Smith picture Luke Smith · Dec 29, 2015 · Viewed 16.4k times · Source

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

Employed Russian picture Employed Russian · Dec 30, 2015

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.

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