链接到可执行文件时如何强制将目标文件包含在静态库中? [英] How to force inclusion of an object file in a static library when linking into executable?

问题描述

我有一个 C++ 项目，由于其目录结构被设置为静态库 A，它被链接到共享库 B，它被链接到可执行文件C.(这是一个使用 CMake 的跨平台项目，所以在 Windows 上我们得到 A.lib、B.dll 和 C.exe，在 Linux 上，我们得到 libA.a、libB.so 和 C.)库 A 有一个 init函数(A_init，在A/initA.cpp中定义)，从库B的init函数(B_init，定义在 B/initB.cpp)，从 C 的 main 中调用.因此，当链接 B 时，A_init(以及在 initA.cpp 中定义的所有符号)被链接到 B(这是我们想要的行为).

I have a C++ project that due to its directory structure is set up as a static library A, which is linked into shared library B, which is linked into executable C. (This is a cross-platform project using CMake, so on Windows we get A.lib, B.dll, and C.exe, and on Linux we get libA.a, libB.so, and C.) Library A has an init function (A_init, defined in A/initA.cpp), that is called from library B's init function (B_init, defined in B/initB.cpp), which is called from C's main. Thus, when linking B, A_init (and all symbols defined in initA.cpp) is linked into B (which is our desired behavior).

问题在于 A 库还定义了一个预期的函数(Af，在 A/Afort.f 中定义)通过动态加载(即 Windows 上的 LoadLibrary/GetProcAddress 和 Linux 上的 dlopen/dlsym).由于库 B 中没有对 Af 的引用，因此 A/Afort.o 中的符号不​​包含在 B.在 Windows 上，我们可以使用 pragma 人为地创建引用:

The problem comes in that the A library also defines a function (Af, defined in A/Afort.f) that is intended to by dynamically loaded (i.e. LoadLibrary/GetProcAddress on Windows and dlopen/dlsym on Linux). Since there are no references to Af from library B, symbols from A/Afort.o are not included into B. On Windows, we can artifically create a reference by using the pragma:

#pragma comment (linker, "/export:_Af")

由于这是一个编译指示，它仅适用于 Windows(使用 Visual Studio 2008).为了让它在 Linux 上运行，我们尝试将以下内容添加到 A/initA.cpp:

Since this is a pragma, it only works on Windows (using Visual Studio 2008). To get it working on Linux, we've tried adding the following to A/initA.cpp:

extern void Af(void);
static void (*Af_fp)(void) = &Af;

这不会导致符号Af 包含在B 的最终链接中.我们如何强制将符号 Af 链接到 B 中?

This does not cause the symbol Af to be included in the final link of B. How can we force the symbol Af to be linked into B?

推荐答案

原来我最初的尝试大部分都在那里.以下作品:

It turns out my original attempt was mostly there. The following works:

extern "C" void Af(void);
void (*Af_fp)(void) = &Af;

对于那些想要一个独立的预处理器宏来封装它的人:

For those that want a self-contained preprocessor macro to encapsulate this:

#if defined(_WIN32)
# if defined(_WIN64)
#  define FORCE_UNDEFINED_SYMBOL(x) __pragma(comment (linker, "/export:" #x))
# else
#  define FORCE_UNDEFINED_SYMBOL(x) __pragma(comment (linker, "/export:_" #x))
# endif
#else
# define FORCE_UNDEFINED_SYMBOL(x) extern "C" void x(void); void (*__ ## x ## _fp)(void)=&x;
#endif

这样用的:

FORCE_UNDEFINED_SYMBOL(Af)

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