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c++ - dylib 与另一个 dylib 有何不同?

转载 作者:塔克拉玛干 更新时间:2023-11-03 07:27:24 25 4
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我正在多次加载 dylib 并尝试理解是什么让这些符号不同。

这是我的步骤:

  • 使用以下入口点构建 lib_a.dylib:

    FactoryA : IFActory () {}
    extern "C" IFactory* GetFactory () { return new FactoryA(); }
  • 复制 lib_a.dylib 到 lib_b.dylib
  • 从 lib_a.dylib 和 lib_b.dylib 加载 GetFactory

    void * module=dlopen(fileName,RTLD_LAZY);
    void * proc = (void *)dlsym(module, "GetFactory");

当加载第二个动态库(lib_b.dylib)时,GetFactory 被认为已经被 lib_a.dylib 定义。
实际上,nm 输出具有相同的结果。

但我认为编译标记 -two_level_namespace 保证 2 个 dylib 位于一种不同的命名空间中,我错了吗?

我可以更改什么来加载我的两个 dylib?


下面是我的测试。

我的类.h :

#include <stdio.h>  
class IFactory {
public:
virtual int GetCount() = 0;
};
extern "C"
{
extern IFactory* GetFactory ();
}

我的类.cpp

#include <stdio.h>
#include "myclass.h"

class MyFactory : public IFactory {
public:
virtual int GetCount() { mCount++; return mCount; }
static int mCount;
};
int MyFactory::mCount = 0;


IFactory* GetFactory () {
return new MyFactory;
}

我的测试.cpp

#include <stdio.h>
#include <dlfcn.h>
#include <mach-o/dyld.h>
#include "myclass.h"

typedef IFactory* (*factoryPtr)();

int main()
{
void* handleA = dlopen("libmylib.dylib", RTLD_LAZY);
factoryPtr functionA = (IFactory*(*)())dlsym(handleA, "GetFactory");
IFactory* factoryA = (*functionA)();
fprintf(stderr, "Handle A: %p\tFunction A: %p\t Count: %d\n", handleA, functionA, factoryA->GetCount());

// Reload same library
void* handleB = dlopen("libmylib.dylib", RTLD_LAZY);
factoryPtr functionB = (IFactory*(*)())dlsym(handleB, "GetFactory");
IFactory* factoryB = (*functionB)();
fprintf(stderr, "Handle B: %p\tFunction B: %p\t Count: %d\n", handleB, functionB, factoryB->GetCount());

// Load copy of first library (just rename)
void* handleC = dlopen("libmylib_copy.dylib", RTLD_LAZY);
factoryPtr functionC = (IFactory*(*)())dlsym(handleC, "GetFactory");
IFactory* factoryC = (*functionC)();
fprintf(stderr, "Handle C: %p\tFunction C: %p\t Count: %d\n", handleC, functionC, factoryC->GetCount());

return 0;
}

命令:

clang++ -dynamiclib myclass.cpp -o libmylib.dylib  
cp libmylib.dylib libmylib_copy.dylib
clang++ mytest.cpp -o mytest
./mytest

输出:

Handle A: 0x7fe5dac039b0    Function A: 0x106d49d30  Count: 1    
Handle B: 0x7fe5dac039b0 Function B: 0x106d49d30 Count: 2
Handle C: 0x7fe5dac03e00 Function C: 0x106d7cd30 Count: 3

为什么我们最后有 count = 3?


属性 -fvisibility=hidden -fvisibility-inlines-hidden 允许做同样的事情。

修改后的 myclass.h :

#include <stdio.h>  
#define EXPORT_FACTORY __attribute__ ((visibility ("default")))
class IFactory {
public:
virtual int GetCount() = 0;
};
extern "C"
{
extern EXPORT_FACTORY IFactory* GetFactory ();
}

构建:

clang++ -dynamiclib myclass.cpp  -o libmylib.dylib  -fvisibility=hidden -fvisibility-inlines-hidden
cp libmylib.dylib libmylib_copy.dylib
clang++ mytest.cpp -o mytest
./mytest

输出:

Handle A: 0x7fe078c039b0    Function A: 0x1076e1c00  Count: 1
Handle B: 0x7fe078c039b0 Function B: 0x1076e1c00 Count: 2
Handle C: 0x7fe078c03e20 Function C: 0x107714c00 Count: 1

最佳答案

这里的根本问题是库中符号的可见性。通过nm -m可以看到,除了要导出的GetFactory函数外,还有几个外部符号:

$ nm -m libmylib.dylib
0000000000000f30 (__TEXT,__text) external _GetFactory
0000000000001068 (__DATA,__common) external __ZN9MyFactory6mCountE
0000000000000f50 (__TEXT,__text) weak external __ZN9MyFactory8GetCountEv
0000000000001038 (__DATA,__data) weak external __ZTI8IFactory
0000000000001050 (__DATA,__data) weak external __ZTI9MyFactory
0000000000000f91 (__TEXT,__const) weak external __ZTS8IFactory
0000000000000f86 (__TEXT,__const) weak external __ZTS9MyFactory
0000000000001020 (__DATA,__data) weak external __ZTV9MyFactory
(undefined) external __ZTVN10__cxxabiv117__class_type_infoE (from libc++)
(undefined) external __ZTVN10__cxxabiv120__si_class_type_infoE (from libc++)
(undefined) weak external __Znwm (from libc++)
(undefined) external dyld_stub_binder (from libSystem)

标记为 weak external 的符号是导致问题的原因。

使用 DYLD_PRINT_BINDINGS=YES 运行您的测试应用程序显示:

$ DYLD_PRINT_BINDINGS=YES ./mytest
[ … output showing initialization of libstdc++.dylib and libmylib.dylib omitted …]
Handle A: 0x7fc729c03810 Function A: 0x102a51ee0 Count: 1
Handle B: 0x7fc729c03810 Function B: 0x102a51ee0 Count: 2
dyld: bind: libmylib_copy.dylib:0x102A85038 = libc++abi.dylib:__ZTVN10__cxxabiv117__class_type_infoE, *0x102A85038 = 0x7FFF7CA67B50 + 16
dyld: bind: libmylib_copy.dylib:0x102A85050 = libc++abi.dylib:__ZTVN10__cxxabiv120__si_class_type_infoE, *0x102A85050 = 0x7FFF7CA67BD0 + 16
dyld: bind: libmylib_copy.dylib:0x102A85018 = libstdc++.6.dylib:__Znwm, *0x102A85018 = 0x7FFF938F0325
dyld: bind: libmylib_copy.dylib:0x102A85000 = libdyld.dylib:dyld_stub_binder, *0x102A85000 = 0x7FFF9084E878
dyld: weak bind: libmylib_copy.dylib:0x102A85030 = libmylib.dylib:__ZN9MyFactory8GetCountEv, *0x102A85030 = 0x102A51F00
dyld: weak bind: libmylib_copy.dylib:0x102A85060 = libmylib.dylib:__ZTI8IFactory, *0x102A85060 = 0x102A52038
dyld: weak bind: libmylib_copy.dylib:0x102A85028 = libmylib.dylib:__ZTI9MyFactory, *0x102A85028 = 0x102A52050
dyld: weak bind: libmylib_copy.dylib:0x102A85040 = libmylib.dylib:__ZTS8IFactory, *0x102A85040 = 0x102A51F41
dyld: weak bind: libmylib_copy.dylib:0x102A85058 = libmylib.dylib:__ZTS9MyFactory, *0x102A85058 = 0x102A51F36
dyld: weak bind: libmylib_copy.dylib:0x102A85010 = libmylib.dylib:__ZTV9MyFactory, *0x102A85010 = 0x102A52020
dyld: weak bind: libmylib_copy.dylib:0x102A85018 = libstdc++.6.dylib:__Znwm, *0x102A85018 = 0x7FFF938F0325
Handle C: 0x7fc729c03c20 Function C: 0x102a84ee0 Count: 3

如果您查看 weak bind 行,您会看到我们之前看到的标记为 weak external 的符号正在解析为相同的符号libmylib.dylib 中的名称。我相信这种行为与 C++ 的单一定义规则有关(“每个程序都应包含该程序中使用的每个非内联函数或对象的一个​​定义”)。由于您在程序中对同一个非内联函数有多个定义,因此链接器试图在加载时合并这些符号,因此只使用了一个函数。

这里最好的解决方案是让每个库将其符号放在不同的命名空间中。由于您打算仅导出 C 工厂函数,因此匿名命名空间就足够了。这也具有将其余符号标记为非外部符号的效果,这对库加载时间有益。

或者,您可以通过仅导出库的客户需要的符号来稍微欺骗一下单一定义规则:

$ clang++ -Wl,-exported_symbol -Wl,_GetFactory -dynamiclib myclass.cpp -o libmylib.dylib
$ cp libmylib.dylib libmylib_copy.dylib
$ ./mytest
Handle A: 0x7fc593403910 Function A: 0x1009e4e90 Count: 1
Handle B: 0x7fc593403910 Function B: 0x1009e4e90 Count: 2
Handle C: 0x7fc593403b10 Function C: 0x1009e7e90 Count: 1

这会导致我们之前看到的 weak external 符号被标记为私有(private),因此它们只会解析为同一图像中的符号。

关于c++ - dylib 与另一个 dylib 有何不同?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17218320/

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