Dynamic Linking of Shared Libraries with Dependencies

问题

Is there a way to dynamically link shared libraries that have dependencies?

For example, I have two libraries, libA.so and libB.so. libB.so calls functions defined in libA.so.

In my main program, I wish to load the two libraries with dlopen. However, if I try:

dlopen(libA.so);
dlopen(libB.so);

Then the second dlopen will fail as libB has unrecognized symbols.

I can think of a few workarounds such as building all the object files into a single shared library, or to have libB.so call dlopen on libA.so, but that's extra work.

I guess the way I'm imagining this to work is like in the case of kernel modules where you can use "EXPORT_SYMBOL()" to allow other modules to call functions defined in a previously loaded module.

Can something similar be done with shared libraries? Or will I have to use my workarounds?

回答1:

I experienced a similar situation, this is what worked for me (using a gcc toolchain):

When you create the shared object libB.so, you unconditionally link the library libA.so, the command should look like this:

gcc -shared -Wl,--no-as-needed -lA -o libB.so b.o

You can then check that libA.so indeed became a dependency for the dynamic linker:

$ ldd libB.so
    linux-gate.so.1 =>  (0xb77ba000)
    libA.so => not found
    libc.so.6 => /lib/i386-linux-gnu/libc.so.6 (0xb75f7000)
    /lib/ld-linux.so.2 (0xb77bb000)

In your main program it should be sufficient to only dlopen() the library libB.so, and the other library should get linked automatically by the dynamic linker.

回答2:

Have you tried using RTLD_GLOBAL?

RTLD_GLOBAL The symbols defined by this library will be made available for symbol resolution of subsequently loaded libraries.

This should work fine even if B depends on A:

void * const handleA = dlopen("libA.so", RTLD_NOW | RTLD_GLOBAL);
void * const handleB = dlopen("libB.so", RTLD_NOW);

来源：https://stackoverflow.com/questions/26619897/dynamic-linking-of-shared-libraries-with-dependencies

标签

c++

shared-libraries