gcc 使用Debian libpam 0 g-dev获取对misc_conv的未定义引用

tpxzln5u  于 2022-11-24  发布在  其他
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编码:

#include <security/pam_appl.h>
#include <security/pam_misc.h>
#include <stdio.h>

static struct pam_conv conv = {
    misc_conv,
    NULL
};

int main(int argc, char *argv[])
{
    pam_handle_t *pamh=NULL;
    int retval;
    const char *user="nobody";

    if(argc == 2) {
        user = argv[1];
    }

    if(argc > 2) {
        fprintf(stderr, "Usage: check_user [username]\n");
        exit(1);
    }

    retval = pam_start("check_user", user, &conv, &pamh);

    if (retval == PAM_SUCCESS)
        retval = pam_authenticate(pamh, 0);    /* is user really user? */

    if (retval == PAM_SUCCESS)
        retval = pam_acct_mgmt(pamh, 0);       /* permitted access? */

    /* This is where we have been authorized or not. */

    if (retval == PAM_SUCCESS) {
        fprintf(stdout, "Authenticated\n");
    } else {
        fprintf(stdout, "Not Authenticated\n");
    }

    if (pam_end(pamh,retval) != PAM_SUCCESS) {     /* close Linux-PAM */
        pamh = NULL;
        fprintf(stderr, "check_user: failed to release authenticator\n");
        exit(1);
    }

    return ( retval == PAM_SUCCESS ? 0:1 );       /* indicate success */
}

二库我又道:

  • libpam 0 g-开发
  • libpam0g

在/usr/include/security/中我有:

_pam_compat.h
_pam_macros.h
_pam_types.h
pam_appl.h
pam_client.h
pam_ext.h
pam_filter.h
pam_misc.h
pam_modules.h
pam_modutil.h

我编译:gcc -o check_user -lpam check_user.c
我得到:

/usr/bin/ld: /tmp/cc9ldItB.o:(.data.rel+0x0): undefined reference to `misc_conv'
collect2: error: ld returned 1 exit status

那么,为什么添加了libpam0g-dev后会得到undefined reference to "misc_conv'呢?

gdx19jrr

gdx19jrr1#

您应该在命令中添加-lpam_misc选项:

gcc -o check_user -lpam -lpam_misc check_user.c

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