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d4697bc9 1/* Copyright (C) 1991-2014 Free Software Foundation, Inc.
6bc31da0 2 This file is part of the GNU C Library.
28f540f4 3
6bc31da0 4 The GNU C Library is free software; you can redistribute it and/or
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5 modify it under the terms of the GNU Lesser General Public
6 License as published by the Free Software Foundation; either
7 version 2.1 of the License, or (at your option) any later version.
28f540f4 8
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9 The GNU C Library is distributed in the hope that it will be useful,
10 but WITHOUT ANY WARRANTY; without even the implied warranty of
11 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
41bdb6e2 12 Lesser General Public License for more details.
28f540f4 13
41bdb6e2 14 You should have received a copy of the GNU Lesser General Public
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15 License along with the GNU C Library; if not, see
16 <http://www.gnu.org/licenses/>. */
28f540f4 17
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18/* Written by Paul Eggert <eggert@cs.ucla.edu>. */
19
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20#include <time.h>
21
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22#include <limits.h>
23#include <float.h>
24#include <stdint.h>
25
26#define TYPE_BITS(type) (sizeof (type) * CHAR_BIT)
27#define TYPE_FLOATING(type) ((type) 0.5 == 0.5)
28#define TYPE_SIGNED(type) ((type) -1 < 0)
29
30/* Return the difference between TIME1 and TIME0, where TIME0 <= TIME1.
31 time_t is known to be an integer type. */
32
33static double
34subtract (time_t time1, time_t time0)
35{
36 if (! TYPE_SIGNED (time_t))
37 return time1 - time0;
38 else
39 {
40 /* Optimize the common special cases where time_t
41 can be converted to uintmax_t without losing information. */
42 uintmax_t dt = (uintmax_t) time1 - (uintmax_t) time0;
43 double delta = dt;
44
45 if (UINTMAX_MAX / 2 < INTMAX_MAX)
46 {
47 /* This is a rare host where uintmax_t has padding bits, and possibly
48 information was lost when converting time_t to uintmax_t.
49 Check for overflow by comparing dt/2 to (time1/2 - time0/2).
50 Overflow occurred if they differ by more than a small slop.
51 Thanks to Clive D.W. Feather for detailed technical advice about
52 hosts with padding bits.
53
54 In the following code the "h" prefix means half. By range
55 analysis, we have:
56
57 -0.5 <= ht1 - 0.5*time1 <= 0.5
58 -0.5 <= ht0 - 0.5*time0 <= 0.5
59 -1.0 <= dht - 0.5*(time1 - time0) <= 1.0
60
61 If overflow has not occurred, we also have:
62
63 -0.5 <= hdt - 0.5*(time1 - time0) <= 0
64 -1.0 <= dht - hdt <= 1.5
65
66 and since dht - hdt is an integer, we also have:
67
68 -1 <= dht - hdt <= 1
69
70 or equivalently:
71
72 0 <= dht - hdt + 1 <= 2
73
74 In the above analysis, all the operators have their exact
75 mathematical semantics, not C semantics. However, dht - hdt +
76 1 is unsigned in C, so it need not be compared to zero. */
77
78 uintmax_t hdt = dt / 2;
79 time_t ht1 = time1 / 2;
80 time_t ht0 = time0 / 2;
81 time_t dht = ht1 - ht0;
82
83 if (2 < dht - hdt + 1)
84 {
85 /* Repair delta overflow.
86
87 The following expression contains a second rounding,
88 so the result may not be the closest to the true answer.
89 This problem occurs only with very large differences.
90 It's too painful to fix this portably. */
91
92 delta = dt + 2.0L * (UINTMAX_MAX - UINTMAX_MAX / 2);
93 }
94 }
95
96 return delta;
97 }
98}
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99
100/* Return the difference between TIME1 and TIME0. */
101double
29311370 102__difftime (time_t time1, time_t time0)
28f540f4 103{
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104 /* Convert to double and then subtract if no double-rounding error could
105 result. */
28f540f4 106
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107 if (TYPE_BITS (time_t) <= DBL_MANT_DIG
108 || (TYPE_FLOATING (time_t) && sizeof (time_t) < sizeof (long double)))
28f540f4 109 return (double) time1 - (double) time0;
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110
111 /* Likewise for long double. */
112
113 if (TYPE_BITS (time_t) <= LDBL_MANT_DIG || TYPE_FLOATING (time_t))
6bc31da0 114 return (long double) time1 - (long double) time0;
28f540f4 115
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116 /* Subtract the smaller integer from the larger, convert the difference to
117 double, and then negate if needed. */
118
119 return time1 < time0 ? - subtract (time0, time1) : subtract (time1, time0);
28f540f4 120}
47eb6e76 121strong_alias (__difftime, difftime)