有谁用过pam，进来交流一下

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

  You need to add the following (or equivalent) to the /etc/pam.conf file.
  # check authorization
  check_user auth    required    /usr/lib/security/pam_unix_auth.so
  check_user account required    /usr/lib/security/pam_unix_acct.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)
{
      printf("ss\n"

;
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 */
}

像以上这段代码，它是利用misc_conv来作为提示用户输入密码，那么如何编写类似的鉴权呢？