如何正确地将dlsym返回的指针分配给函数指针类型的变量?

问题描述

我正在尝试在代码中使用dlopen()和dlsym()并使用gcc进行编译.

I am trying to use dlopen() and dlsym() in my code and compile it with gcc.

这是第一个文件.

/* main.c */

#include <dlfcn.h>

int main()
{
    void *handle = dlopen("./foo.so", RTLD_NOW);

    if (handle) {
        void (*func)() = dlsym(handle, "func");
        func();
    }

    return 0;
}

这是第二个文件.

/* foo.c */

#include <stdio.h>

void func()
{
    printf("hello, world\n");
}

这是我编译和运行代码的方式.

Here is how I compile and run the code.

$ gcc -std=c99 -pedantic -Wall -Wextra -shared -fPIC -o foo.so foo.c
$ gcc -std=c99 -pedantic -Wall -Wextra -ldl -o main main.c
main.c: In function ‘main’:
main.c:10:26: warning: ISO C forbids initialization between function pointer and ‘void *’ [-Wpedantic]
         void (*func)() = dlsym(handle, "func");
                          ^
$ ./main
hello, world

如何摆脱警告?

类型转换无济于事.如果我尝试将dlsym()的返回值类型转换为函数指针，则会收到此警告.

Type casting doesn't help. If I try to type cast the return value of dlsym() into a function pointer, I get this warning instead.

main.c:10:26: warning: ISO C forbids conversion of object pointer to function pointer type [-Wpedantic]
         void (*func)() = (void (*)()) dlsym(handle, "func");
                          ^

什么会使编译器相信此代码是正确的?

What would convince the compiler that this code is fine?

推荐答案

如果您想获得正确的信息，请不要尝试解析函数的地址.而是从动态库中导出某种结构:

If you want to be pedantically correct, don't try to resolve the address of a function. Instead, export some kind of structure from the dynamic library:

struct export_vtable {
   void (*helloworld)(void);
};
struct export_vtable exports = { func };

在呼叫者中

struct export_vtable {
   void (*helloworld)(void);
};

int main() {
   struct export_vtable* imports;
   void *handle = dlopen("./foo.so", RTLD_NOW);

   if (handle) {
        imports = dlsym(handle, "exports");
        if (imports) imports->helloworld();
    }

    return 0;
}

这种技术实际上很常见，不是为了可移植性-POSIX保证可以将函数指针与void *进行相互转换，但是因为它具有更大的灵活性.

This technique is actually quite common, not for portability -- POSIX guarantees that function pointers can be converted to and from void* -- but because it allows more flexibility.

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