link a static library to a shared library and hide exported symbols

Question

I am having an annoying problem with the linker. I want to link some symbols from a shared library to a static library, but not export its symbols (ie, I cannot simply merge the libraries or link with --whole-archive). What I want is link (as in, like linking an executable, solving undefined symbols) my shared library to a static one and remove the undefined symbols.

The thing I am looking for is probably just a linker option but I can't put my finger on it.

I'll try to describe the problem the best I can (it's not that easy) and then provide a toy minimal example to play with.

EDIT: The problem has been solved, solution posted at the bottom of the question

Quick description:

I want to use the LD_PRELOAD trick to trap some function calls in an executable. This executable is linked against a third party shared library, which contains the function definition of the functions I want to trap.

This third party library also contains symbols from yet another library, which I am also using in my library, but with a different (non-compatible) version.

What I want to do is compile my shared library and link it at compile time with the definitions of the last (static) library, without exporting the symbols, so that my shared library uses a different version from the one I want to trap.

Simplified problem description

I have a third party library called libext.so, for which I don't have the source code. This defines a function bar and uses a function foo from another library, but the symbols are both defined there:

$> nm libext.so
0000000000000a16 T bar
00000000000009e8 T foo

As I mentioned, foo is an external dependency, for which I want to use a newer version. I have an updated library for it, let's call it libfoo.a:

$> nm libfoo.a
0000000000000000 T foo

Now the problem is that I want to create a dynamic library which re-defines bar, but I want my library to use the the definition of foo from libfoo.a and i want the functions from libext.so to call the function foo from libext.so. In other words, I want a compile time linkage of my library to libfoo.a.

What I am looking for is define a library which uses libfoo.a but doesn't export its symbols. If I link my library to libfoo.a, I get:

$> nm libmine.so
0000000000000a78 T bar
0000000000000b2c T foo

Which means I overload both foo and bar (i don't want to override foo). If i don't link my library to libfoo.a, I get:

$> nm libmine.so
0000000000000a78 T bar
                 U foo

So my library will use their version of foo, which I don't want either. What I want is:

$> nm libmine.so
0000000000000a78 T bar

Where foo is linked at compile time and its symbol not exported.

Minimal example

You don't need to read this, but you can use it to play around and find a solution.

bar.cpp: represents the third party app I don't have the code for:

#include <iostream>
extern "C" void foo(){ std::cerr << "old::foo" << std::endl; }
extern "C" void bar(){ std::cerr << "old::bar" << std::endl; foo(); }

foo.cpp: represents a newer version of a function used by both my lib and the third party:

#include <iostream>
extern "C" void foo(){ std::cerr << "new::foo" << std::endl; }

trap.cpp: the code from my library, it traps bar, calls the new foo and forwards:

#include <iostream>
extern "C" {
  #include <dlfcn.h>
}
extern "C" void foo();
extern "C" void bar(){
  std::cerr << "new::bar" << std::endl;
  foo(); // Should be new::foo
  void (*fwd)() = (void(*)())dlsym(RTLD_NEXT, "bar");
  fwd(); // Should use old::foo
}

exec.cpp: a dummy executable to call bar:

extern "C" void bar();

int main(){
  bar();
}

Makefile: Unix only, sorry

default:
    # The third party library
    g++ -c -o bar.o bar.cpp -fpic
    gcc -shared -Wl,-soname,libext.so -o libext.so bar.o
    # The updated library
    g++ -c -o foo.o foo.cpp -fPIC
    ar rcs libfoo.a foo.o
    # My trapping library
    g++ -c -o trap.o trap.cpp -fPIC
    gcc -shared -Wl,-soname,libmine.so -o libmine.so trap.o -ldl -L. -lfoo
    # The dummy executable
    g++ -o test exec.cpp -L. libext.so

In this case, bar calls foo; the normal execution is:

$> ./test
old::bar
old::foo

Preloading my library intercepts bar, calls my foo and forwards bar, the current execution is:

$> LD_PRELOAD=libmine.so ./test
new::bar
new::foo
old::bar
new::foo

The last line is wrong, the desired output is:

$> LD_PRELOAD=libmine.so ./test
new::bar
new::foo
old::bar
old::foo

---

Solution

1) As pointed out in the accepted answer, we can use a linker version script to change the scope of the undesired symbols from global to local:

BAR {
  global: bar;
  local: *;
};

Compiling with the linker version shows foo is local and the program now behaves as expected:

$> gcc -shared -Wl,-soname,libmine.so -Wl,--version-script=libmine.version -o libmine.so trap.o -ldl -L. -lfoo
$> nm libmine.so
0000000000000978 T bar
0000000000000000 A BAR  
0000000000000a2c t foo
$> LD_PRELOAD=libmine.so ./test
new::bar
new::foo
old::bar
old::foo

2) An alternative is to re-compile libfoo.a with the attribute -fvisibility=hidden and link against that. The visibility of the exported symbols is then local too, and the behavior is the same as above.

Solution

You want to use a linker version script, which exports the symbol(s) you want (bar here) and hides everything else.

Example here.

Comments

[Thibaut]

That worked. It's weird that there is no linker option to lower the visibility of the linked dependencies directly at link time.