xref: /netbsd-src/lib/libc/time/difftime.c (revision 4439cfd0acf9c7dc90625e5cd83b2317a9ab8967)
1 /*	$NetBSD: difftime.c,v 1.24 2024/01/20 14:52:49 christos Exp $	*/
2 
3 /* Return the difference between two timestamps.  */
4 
5 /*
6 ** This file is in the public domain, so clarified as of
7 ** 1996-06-05 by Arthur David Olson.
8 */
9 
10 #include <sys/cdefs.h>
11 #if defined(LIBC_SCCS) && !defined(lint)
12 #if 0
13 static char	elsieid[] = "@(#)difftime.c	8.1";
14 #else
15 __RCSID("$NetBSD: difftime.c,v 1.24 2024/01/20 14:52:49 christos Exp $");
16 #endif
17 #endif /* LIBC_SCCS and not lint */
18 
19 /*LINTLIBRARY*/
20 
21 #include "private.h"	/* for time_t and TYPE_SIGNED */
22 
23 /* Return -X as a double.  Using this avoids casting to 'double'.  */
24 static double
25 dminus(double x)
26 {
27 	return -x;
28 }
29 
30 double
31 difftime(time_t time1, time_t time0)
32 {
33 	/*
34 	** If double is large enough, simply convert and subtract
35 	** (assuming that the larger type has more precision).
36 	*/
37 	/*CONSTCOND*/
38 	if (sizeof(time_t) < sizeof(double)) {
39 		double t1 = time1, t0 = time0;
40 		return t1 - t0;
41  	}
42 
43 	/*
44 	** The difference of two unsigned values can't overflow
45 	** if the minuend is greater than or equal to the subtrahend.
46 	*/
47 	if (!TYPE_SIGNED(time_t))
48 		/*NOTREACHED*/
49 		return time0 <= time1 ? time1 - time0 :
50 		    dminus((double)(time0 - time1));
51 
52 	/* Use uintmax_t if wide enough.  */
53 	/*CONSTCOND*/
54 	if (sizeof(time_t) <= sizeof(uintmax_t)) {
55 		uintmax_t t1 = time1, t0 = time0;
56 		return time0 <= time1 ? t1 - t0 : dminus((double)(t0 - t1));
57 	}
58 
59 	/*
60 	** Handle cases where both time1 and time0 have the same sign
61 	** (meaning that their difference cannot overflow).
62 	*/
63 	if ((time1 < 0) == (time0 < 0))
64 		return time1 - time0;
65 
66 	/*
67 	** The values have opposite signs and uintmax_t is too narrow.
68 	** This suffers from double rounding; attempt to lessen that
69 	** by using long double temporaries.
70 	*/
71 	{
72 		long double t1 = time1, t0 = time0;
73 		return t1 - t0;
74 	}
75 }
76