1/*-
2 * SPDX-License-Identifier: BSD-2-Clause-FreeBSD
3 *
4 * Copyright (c) 2002,2005 Marcel Moolenaar
5 * Copyright (c) 2002 Hiten Mahesh Pandya
6 * All rights reserved.
7 *
8 * Redistribution and use in source and binary forms, with or without
9 * modification, are permitted provided that the following conditions
10 * are met:
11 * 1. Redistributions of source code must retain the above copyright
12 *    notice, this list of conditions and the following disclaimer.
13 * 2. Redistributions in binary form must reproduce the above copyright
14 *    notice, this list of conditions and the following disclaimer in the
15 *    documentation and/or other materials provided with the distribution.
16 *
17 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
18 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
19 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
20 * ARE DISCLAIMED.  IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
21 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
22 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
23 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
24 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
25 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
26 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
27 * SUCH DAMAGE.
28 *
29 * $FreeBSD$
30 */
31
32#include <uuid.h>
33
34/*
35 * uuid_is_nil() - return whether the UUID is a nil UUID.
36 * See also:
37 *	http://www.opengroup.org/onlinepubs/009629399/uuid_is_nil.htm
38 */
39int32_t
40uuid_is_nil(const uuid_t *u, uint32_t *status)
41{
42	const uint32_t *p;
43
44	if (status)
45		*status = uuid_s_ok;
46
47	if (!u)
48		return (1);
49
50	/*
51	 * Pick the largest type that has equivalent alignment constraints
52	 * as an UUID and use it to test if the UUID consists of all zeroes.
53	 */
54	p = (const uint32_t*)u;
55	return ((p[0] == 0 && p[1] == 0 && p[2] == 0 && p[3] == 0) ? 1 : 0);
56}
57