Compile shared object library, which call function from so too

Question

I have got a f2.cpp file

// f2.cpp #include <iostream> void f2() { std::cout << "It's a call of f2 function" << std::endl; }

I use cygwin with crosstool compiler gcc.

g++ -fPIC -c f2.cpp g++ -shared -o libf2.so f2.o

I have got a libf2.so file. Now I want to call f2 function in f1 library (shared object too) libf1.so.

It's a f1.cpp and i want take f1.so

// f1.cpp #include <iostream> void f1() { std::cout << "f1 function is calling f2()..." << std::endl; f2(); }

How i must compile f1.cpp? I don't want to use dlclose, dlerror, dlopen, dlsym... Аt last i want to use f1.so in main.cpp as a shared object library too... without using use dlclose, dlerror, dlopen, dlsym. How I must compile main.cpp, when i will have a f1.so ?

// main.cpp #include <iostream> int main() { f1(); return 0; }

balki · Accepted Answer · 2011-11-17 13:03:07Z

declare f2() in a header file. and compile libf1.so similar to libf2.

Now compile main linking against f1 and f2. It should look something like this g++ -lf2 -lf1 -L /path/to/libs main.o

Eli Bendersky · Accepted Answer · 2011-11-17 13:01:38Z

You can simply link them together (if f2 is compiled into libf2.so, you pass -lf2 to the linker). The linker will take care of connecting calls from f1 to f2. Naturally, at runtime f1 will expect to find f2 in the SO load path and the dynamic loader will load it.

Here's a more complete sample, taken from a portion of a Makefile I found lying around. Here, mylib stands for your f2, and main_linked is f1:

mylib: mylib.c mylib.h gcc $(CFLAGS) -fpic -c mylib.c gcc -shared -o libmylib.so mylib.o main_linked: main_linked.c mylib.h mylib.c gcc $(CFLAGS) -L. -lmylib main_linked.c -o main_linked

Note:

mylib is compiled into a shared library with -shared
main_linked is then built with a single gcc call passing -lmylib to specify the library to link and -L. to say where to find it (in this case - current dir)

Some programmer dude · Accepted Answer · 2011-11-17 13:06:22Z

1

Check the -L and -l flags to g++.

answered Nov 17, 2011 at 13:06

Some programmer dude

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