c++ - 如何知道 dlopen() 时没有 .so

Question

我使用 dlopen 为我的项目制作了插件。

我想让我的主程序在没有这样的情况下停止尝试再次加载插件。

有什么好的解决办法吗？ 我需要比较 dlerror 消息吗？

请帮帮我

Answer 1

来自manpage :

If dlopen() fails for any reason, it returns NULL.

还有一个非常广泛的例子。

EXAMPLE

The program below loads the (glibc) math library, looks up the address of the cos(3) function, and prints the cosine of 2.0. The following is an example of building and running the program:

$ cc dlopen_demo.c -ldl
$ ./a.out
-0.416147

Program source

#include <stdio.h>
#include <stdlib.h>
#include <dlfcn.h>
#include <gnu/lib-names.h>  /* Defines LIBM_SO (which will be a
                               string such as "libm.so.6") */
int
main(void)
{
    void *handle;
    double (*cosine)(double);
    char *error;

    handle = dlopen(LIBM_SO, RTLD_LAZY);
    if (!handle) {
        fprintf(stderr, "%s\n", dlerror());
        exit(EXIT_FAILURE);
    }

    dlerror();    /* Clear any existing error */

    cosine = (double (*)(double)) dlsym(handle, "cos");

    /* According to the ISO C standard, casting between function
       pointers and 'void *', as done above, produces undefined results.
       POSIX.1-2003 and POSIX.1-2008 accepted this state of affairs and
       proposed the following workaround:

           *(void **) (&cosine) = dlsym(handle, "cos");

       This (clumsy) cast conforms with the ISO C standard and will
       avoid any compiler warnings.

       The 2013 Technical Corrigendum to POSIX.1-2008 (a.k.a.
       POSIX.1-2013) improved matters by requiring that conforming
       implementations support casting 'void *' to a function pointer.
       Nevertheless, some compilers (e.g., gcc with the '-pedantic'
       option) may complain about the cast used in this program. */

    error = dlerror();
    if (error != NULL) {
        fprintf(stderr, "%s\n", error);
        exit(EXIT_FAILURE);
    }

    printf("%f\n", (*cosine)(2.0));
    dlclose(handle);
    exit(EXIT_SUCCESS);
}