当dlopen（）时怎么知道没有.so

Question

我使用dlopen为我的项目制作了插件。

我希望让我的主程序在没有插件时再次尝试加载插件。

这有什么好的解决方案吗？ 我需要比较dlerror消息吗？

请帮帮我

Answer 1

来自manpage：

  

如果dlopen()因任何原因失败，则返回NULL。

还有一个非常广泛的例子。

  

实施例

     

下面的程序加载（glibc）数学库，查找   cos(3)函数的地址，并打印2.0的余弦值。该   以下是构建和运行程序的示例：

     

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

     

节目来源

#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);
}