Subject: Re: Rationale behind 'extern' on protypes in .h files

From: Jed Brown <[email-masked]>
Date: Mon, 23 Dec 2013 09:59:25 -0700

Ravi Shekhar Jethani <[email-masked]> writes:
> To check this I installed the libgit2-dev package which installed:
> /usr/include/git2/*.h , /usr/lib/libgit2.so
> Now, I exported all symbols using:
> $ readelf -s /usr/lib/libgit2.so
> and tried to match these with 'externed' prototypes in the Git source
> directory..no matches.
> I am confused!!!.

libgit2 is an entirely different package from Git.  If you look at the
libgit2 sources (https://github.com/libgit2/libgit2), look in
include/git2/common.h:

/** Declare a public function exported for application use. */
#if __GNUC__ >= 4
# define GIT_EXTERN(type) extern \
                         __attribute__((visibility("default"))) \
                         type
#elif defined(_MSC_VER)
# define GIT_EXTERN(type) __declspec(dllexport) type
#else
# define GIT_EXTERN(type) extern type
#endif

I have always used __attribute__((visibility("default"))), but the gcc
man page says

    extern declarations are not affected by -fvisibility, so a lot of
    code can be recompiled with -fvisibility=hidden with no
    modifications.  However, this means that calls to "extern" functions
    with no explicit visibility use the PLT, so it is more effective to
    use "__attribute ((visibility))" and/or "#pragma GCC visibility" to
    tell the compiler which "extern" declarations should be treated as
    hidden.

However, I don't understand what the first statement means
(documentation bug?) since -fvisibility=hidden causes functions declared
with 'extern' to be hidden.

symbols.c:
EXTERN int foo(void);
int foo(void) {return 1;}

$ gcc -fvisibility=hidden -DEXTERN=extern -shared -o libsymbols.so symbols.c
$ nm -D libsymbols.so | grep foo
$

meanwhile,

$ gcc -fvisibility=hidden -DEXTERN='__attribute((visibility("default")))' -shared -o libsymbols.so symbols.c
$ nm -D libsymbols.so | grep foo
000000000000055c T foo

> Also I checked this:
> $ ldd git
> There is no 'gitish' .so in the output; it seems everything is packed
> inside one executable.
> So your second point 'skipping the PLT...' also doesn't seem to apply here.

My comment applied to shared libraries in general, not specifically to
Git (which isn't a shared library).