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To get a flavor of the way a <span class="emphasis"><em>Linux-PAM</em></span>
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 (<span class="returnvalue">0</span> for success;
<span class="returnvalue">1</span> for failure).
</p><pre class="programlisting">
/*
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 */
}
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