当动态库使用静态库的符号时，请从静态库中打开一个动态库 [英] dlopen a dynamic library from a static library, when the dynamic library uses symbols of the static one

问题描述

此问题与 dlopen动态相关静态库linux C ++中的库，但是包含了进一步的复杂性(并且使用C ++而不是C):

This question is closely related to dlopen a dynamic library from a static library linux C++, but contains a further complication (and uses C++ instead of C):

我有一个链接到静态库(.a)的应用程序，该库使用dlopen函数加载动态库(.so).另外，动态库调用在静态库中定义的函数.

I have an application that links against a static library (.a) and that library uses the dlopen function to load dynamic libraries (.so). In addition, the dynamic libraries call functions defined in the static one.

是否有一种编译方法，而无需将动态库与静态库链接，反之亦然?

Is there a way to compile this without linking the dynamic libraries against the static one or vice versa?

这是我到目前为止所做的尝试，对相关问题中的示例做了一些修改:

Here comes what I tried so far, slightly modifying the example from the related question:

app.cpp:

#include "staticlib.hpp"
#include <iostream>
int main()
{
    std::cout << "and the magic number is: " << doSomethingDynamicish() << std::endl;
    return 0;
}

staticlib.hpp:

staticlib.hpp:

#ifndef __STATICLIB_H__
#define __STATICLIB_H__

int doSomethingDynamicish();
int doSomethingBoring();
#endif

staticlib.cpp:

staticlib.cpp:

#include "staticlib.hpp"
#include "dlfcn.h"
#include <iostream>
int doSomethingDynamicish()
{
    void* handle = dlopen("./libdynlib.so",RTLD_NOW);
    if(!handle)
    {
        std::cout << "could not dlopen: " << dlerror() << std::endl;
        return 0;
    }

    typedef int(*dynamicfnc)();
    dynamicfnc func = (dynamicfnc)dlsym(handle,"GetMeANumber");
    const char* err = dlerror();
    if(err)
    {
        std::cout << "could not dlsym: " <<err << std::endl;
        return 0;
    }
    return func();
}

staticlib2.cpp:

staticlib2.cpp:

#include "staticlib.hpp"
#include "dlfcn.h"
#include <iostream>

int doSomethingBoring()

{
    std::cout << "This function is so boring." << std::endl;
    return 0;

}

dynlib.cpp:

dynlib.cpp:

#include "staticlib.hpp"

extern "C" int GetMeANumber()
{
    doSomethingBoring();
    return 1337;
}

并构建:

g++ -c -o staticlib.o staticlib.cpp
g++ -c -o staticlib2.o staticlib2.cpp
ar rv libstaticlib.a staticlib.o staticlib2.o
ranlib libstaticlib.a
g++ -rdynamic -o app app.cpp libstaticlib.a -ldl 
g++ -fPIC -shared -o libdynlib.so dynlib.cpp

当我使用 ./app 运行它时，我得到了

When I run it with ./app I get

could not dlopen: ./libdynlib.so: undefined symbol: _Z17doSomethingBoringv
and the magic number is: 0

推荐答案

来自

如果可执行文件与标志"-rdynamic"(或同义为"--export-dynamic")链接，则可执行文件中的全局符号也将用于解析动态加载的库中的引用./p>

If the executable was linked with the flag "-rdynamic" (or, synonymously, "--export-dynamic"), then the global symbols in the executable will also be used to resolve references in a dynamically loaded library.

这意味着要使应用程序导出其符号以在动态库中使用，必须将应用程序与 -rdynamic 标志链接.

That means that for the application to export its symbols for use in the dynamic library, you have to link your application with the -rdynamic flag.

除了上述问题外，还有另一个问题与静态库有关:问题在于，由于在主程序中未调用 doSomethingBoring 函数，因此目标文件静态库中的 staticlib2.o 未链接.

Besides the problem described above, there is another problem and that has to do with the static library: The problem is namely that since the doSomethingBoring function is not called in your main program, the object file staticlib2.o from the static library is not linked.

答案可以在例如这个旧问题，它告诉您添加-whole-archive 链接器标志:

The answer can be found in e.g. this old question, which tells you to add the --whole-archive linker flag:

g++ -rdynamic -o app app.cpp -L. \
    -Wl,--whole-archive -lstaticlib \
    -Wl,--no-whole-archive -ldl

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