c++ - 如果在多个共享库中调用并且启用了libstdc++静态链接，则C++流变得很糟糕

Question

我启用了--static-libstdc++编译了2个共享库。
2个共享库具有相同的函数f，它仅向stdout输出一个字符串和一个整数。
主程序将使用dlopen加载两个共享库，并使用f在其中调用dlsym。
但是，第二个加载的共享库无法输出整数，并且C++流cout变为bad & fail。

ADD: After discussion, I know this is normal... However, I want to change my question to: what implementation of libstdc++ caused this issue? Is there any shared global state? I think if there is no shared global state, it shouldn't be a problem. I wrote the similar program in Windows by static linking to VCRuntime and using LoadLibrary, and it works normally. So why libstdc++ is designed like this?

以下是2个共享库的代码。 (他们共享相同的代码)
他们只是 cout一个字符串和一个整数。

// dll.cpp

#include <iostream>

using namespace std;

extern "C" void f()
{
    cout << "hi" << 1 << endl;

    bool is_eof = cout.eof();
    bool is_fail = cout.fail();
    bool is_bad = cout.bad();

    cout.clear();

    cout << endl;
    cout << "eof: " << to_string(is_eof) << endl;
    cout << "fail: " << to_string(is_fail) << endl;
    cout << "bad: " << to_string(is_bad) << endl;
}

这是主程序，它加载共享库并调用它们的 f函数。

// main.cpp

#include <iostream>
#include <dlfcn.h>
#include <cassert>

using namespace std;

using fn_t = void(void);

void call_f_in_dll(const char *dll_path)
{
    auto dll = dlopen(dll_path, RTLD_LAZY);
    assert(dll);
    fn_t *fn = (fn_t *)dlsym(dll, "f");
    assert(fn);
    fn();
    dlclose(dll);
}

int main()
{
    call_f_in_dll("./libmydll.so");

    cout << endl;

    call_f_in_dll("./libmydll2.so");

    return 0;
}

这是CMakeLists。

# CMakeLists.txt

cmake_minimum_required(VERSION 3.16)
project (TestGCC)

add_link_options(-static-libgcc -static-libstdc++)

add_library(mydll SHARED dll.cpp)
add_library(mydll2 SHARED dll.cpp)

add_executable (main main.cpp)
target_link_libraries(main dl)

输出为:

hox@HOX-PC:~/repos/test-gcc/out$ ./main
hi1

eof: 0
fail: 0
bad: 0

hi
eof: 0
fail: 1
bad: 1

Notice the second part, there is no 1 after hi and fail & bad become 1.

您可以在此处 checkout 代码: https://github.com/xuhongxu96/dlopen-iostream-issue

Answer 1

最后，我找到了一种解决Linux(GNU扩展)问题的方法。
使用dlmopen可以提供更好的隔离btw对象。

auto dll = dlmopen(LM_ID_NEWLM, dll_path, RTLD_LAZY);

非常感谢所有评论者!

Still welcome to explain the details about what the conflict states are.