$ cat pam-sample.c
/* pam-sample.c  from 
 * 4.1. Creating a PAM-Aware Application - Linux Security Cookbook [Book]
 * https://www.oreilly.com/library/view/linux-security-cookbook/0596003919/ch04s01.html
 *  gcc pam-sample.c -lpam -lpam_misc -o pam-sample
 */

#include <security/pam_appl.h>
#include <security/pam_misc.h>
#include <pwd.h>
#include <sys/types.h>
#include <stdio.h>
#define MY_CONFIG "su"
static struct pam_conv conv = { misc_conv, NULL };

int main( )
{
  pam_handle_t *pamh;
  int result;
  struct passwd *pw;
  if ((pw = getpwuid(getuid( ))) == NULL)
    perror("getpwuid");
  else if ((result = pam_start(MY_CONFIG, pw->pw_name, &conv, &pamh)) != PAM_SUCCESS)
    fprintf(stderr, "start failed: %d\n", result);
  else if ((result = pam_authenticate(pamh, 0)) != PAM_SUCCESS)
    fprintf(stderr, "authenticate failed: %d\n", result);
  else if ((result = pam_acct_mgmt(pamh, 0)) != PAM_SUCCESS)
    fprintf(stderr, "acct_mgmt failed: %d\n", result);
  else if ((result = pam_end(pamh, result)) != PAM_SUCCESS) {
    fprintf(stderr, "end failed: %d\n", result);
    return result;
  } else {
    /* Run_My_Big_Application( ) */ ;            /* Run your application code */
    printf("authentication - PASS.\n");
  }
  return 0;
}

$ gcc pam-sample.c -lpam -lpam_misc -o pam-sample

$ ls -l pam-sample*
-rwxr-xr-x 1 user user 17176  8月 27 12:52 pam-sample
-rw-r--r-- 1 user user  1231  8月 27 12:50 pam-sample.c

$ ./pam-sample 
Password:                   # 正しいパスワード
authentication - PASS.

$ ./pam-sample 
Password:                   # 間違えると PAM_AUTH_ERR
authenticate failed: 7
$