Chapter 8. An example application

To get a flavor of the way a Linux-PAM application is written we include the following example. It prompts the user for their password and indicates whether their account is valid on the standard output, its return code also indicates the success (0 for success; 1 for failure).

/*
  This program was contributed by Shane Watts
  [modifications by AGM and kukuk]

  You need to add the following (or equivalent) to the
  /etc/pam.d/check_user file:
  # check authorization
  auth       required     pam_unix.so
  account    required     pam_unix.so
 */

#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 */
}