C++ Linux: lib1 shared library uses shared library lib2, main program uses lib1, why shall it be linked to lib2?

问题

I'm using g++ (Ubuntu 5.4.0-6ubuntu1~16.04.12) 5.4.0 20160609.

I have a shared library lib2:

lib2.h:

#pragma once
class Lib2
{
public:
    Lib2();
};

lib2.cpp:

#include "lib2.h"

Lib2::Lib2()
{
}

I have a shared library lib1 that uses and so needs lib2:

lib1.h:

#pragma once
class Lib1
{
public:
    Lib1();
};

lib1.cpp:

#include "lib1.h"
#include "lib2.h"

Lib1::Lib1()
{
    Lib2 lib2;
}

Now I have a program that uses lib1, and so does not "directly" use lib2 but would need it at runtime.

main.cpp:

#include "lib1.h"
int main(int argc, char *argv[])
{
    Lib1 lib;
    return 0;
}

Let's compile this:

I compile lib2:

g++ -fPIC -O -g lib2.cpp -c -o lib2.o
g++ -shared lib2.o -L. -o liblib2.so

I compile lib1, linking with lib2:

g++ -fPIC -O -g lib1.cpp -c -o lib1.o
g++ -shared lib1.o -L. -llib2 -o liblib1.so

Now I compile main:

g++ -fPIC -O -g main.cpp -c -o main.o

And when I try to create the executable:

g++ main.o -L. -llib1 -o main

I get the error:

./liblib1.so : référence indéfinie vers « Lib2::Lib2() »
collect2: error: ld returned 1 exit status

If I run:

g++ main.o -L. -llib1 -llib2 -o main

it works. But why should my program explcitely link to lib2 while it does not directly use it? Am I doing something wrong? Under Windows, main only links with lib1.lib, lib2.lib is not needed when creating the executable.

Please do not mark as duplicate of Linking a shared library with another shared lib in linux. I checked this post, it mentions that -fPIC has to be used, and I'm using it.

来源：https://stackoverflow.com/questions/65603457/c-linux-lib1-shared-library-uses-shared-library-lib2-main-program-uses-lib1

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