Using shared library within shared library - c++

Question

I want to use a shared library within a shared library but g++ is complaining...
That is how my test sample looks:
.
├── extA
│   ├── a.cpp
│   ├── a.h
│   └── libA.so
├── extB
│   ├── b.cpp
│   ├── b.h
│   └── libB.so
└── main.cpp
a.h:
#pragma once
int f(void);
a.cpp:
#include "a.h"
int f(void) {
return 42;
}
b.h:
#pragma once
#include "a.h"
int g(void);
b.cpp:
#include "b.h"
int g(void) {
return f();
}
main.cpp:
#include <cstdlib>
#include <iostream>
#include "b.h"
int main(int, char **) {
std::cout << g() << std::endl;
return EXIT_SUCCESS;
}
I pretty sure that my way of compiling is wrong. That is what I do:
g++ -c -fpic a.cpp
g++ -shared -o libA.so a.o
to create libA.so,
g++ -c -fpic -I../extA b.cpp -L../extA -lA
g++ -shared -o libB.so b.o
to create libB.so (I think this way is incorrect) and finally:
g++ -IexB main.cpp -LextB -lB -o main
The last g++-call leads to the error msg:
In file included from main.cpp:4:0:
extB/b.h:3:15: fatal error: a.h: No such file or directory
#include "a.h"
^
compilation terminated.
Can someone tell me, what is the right way to compile/create libB.so?

Answer 1

main.cpp:4:15: fatal error: b.h: No such file or directory
#include "b.h"
^
The error is nothing to do with libB.so it means main.cpp cannot find b.h, which is a problem finding headers, nothing to do with the shared library file.
If b.h is not in the same directory as main.cpp then you need to tell the compiler how to find the header, e.g. by using -IextB when compiling main.cpp
The file main.cpp includes the header b.h and uses the function g() defined in libB.so so why are you compiling main.cpp telling it how to find a.h and linking to libA.so? It doesn't need that header or that library!
You probably just want to change A to B since that's what main.cpp actually depends on:
g++ -IextB main.cpp -LextB -lB -o main

Answer 2

I found the solution.
The correct way to compile libB.so is:
g++ -c -fpic -I../extA b.cpp
g++ -shared -o libB.so b.o -L../extA -lA