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122 lines
3.9 KiB
C
122 lines
3.9 KiB
C
/* Copyright (C) 1991-2018 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, see
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<http://www.gnu.org/licenses/>. */
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/* Written by Paul Eggert <eggert@cs.ucla.edu>. */
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#include <time.h>
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#include <limits.h>
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#include <float.h>
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#include <stdint.h>
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#define TYPE_BITS(type) (sizeof (type) * CHAR_BIT)
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#define TYPE_FLOATING(type) ((type) 0.5 == 0.5)
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#define TYPE_SIGNED(type) ((type) -1 < 0)
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/* Return the difference between TIME1 and TIME0, where TIME0 <= TIME1.
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time_t is known to be an integer type. */
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static double
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subtract (time_t time1, time_t time0)
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{
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if (! TYPE_SIGNED (time_t))
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return time1 - time0;
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else
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{
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/* Optimize the common special cases where time_t
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can be converted to uintmax_t without losing information. */
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uintmax_t dt = (uintmax_t) time1 - (uintmax_t) time0;
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double delta = dt;
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if (UINTMAX_MAX / 2 < INTMAX_MAX)
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{
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/* This is a rare host where uintmax_t has padding bits, and possibly
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information was lost when converting time_t to uintmax_t.
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Check for overflow by comparing dt/2 to (time1/2 - time0/2).
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Overflow occurred if they differ by more than a small slop.
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Thanks to Clive D.W. Feather for detailed technical advice about
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hosts with padding bits.
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In the following code the "h" prefix means half. By range
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analysis, we have:
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-0.5 <= ht1 - 0.5*time1 <= 0.5
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-0.5 <= ht0 - 0.5*time0 <= 0.5
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-1.0 <= dht - 0.5*(time1 - time0) <= 1.0
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If overflow has not occurred, we also have:
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-0.5 <= hdt - 0.5*(time1 - time0) <= 0
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-1.0 <= dht - hdt <= 1.5
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and since dht - hdt is an integer, we also have:
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-1 <= dht - hdt <= 1
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or equivalently:
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0 <= dht - hdt + 1 <= 2
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In the above analysis, all the operators have their exact
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mathematical semantics, not C semantics. However, dht - hdt +
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1 is unsigned in C, so it need not be compared to zero. */
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uintmax_t hdt = dt / 2;
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time_t ht1 = time1 / 2;
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time_t ht0 = time0 / 2;
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time_t dht = ht1 - ht0;
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if (2 < dht - hdt + 1)
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{
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/* Repair delta overflow.
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The following expression contains a second rounding,
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so the result may not be the closest to the true answer.
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This problem occurs only with very large differences.
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It's too painful to fix this portably. */
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delta = dt + 2.0L * (UINTMAX_MAX - UINTMAX_MAX / 2);
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}
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}
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return delta;
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}
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}
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/* Return the difference between TIME1 and TIME0. */
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double
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__difftime (time_t time1, time_t time0)
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{
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/* Convert to double and then subtract if no double-rounding error could
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result. */
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if (TYPE_BITS (time_t) <= DBL_MANT_DIG
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|| (TYPE_FLOATING (time_t) && sizeof (time_t) < sizeof (long double)))
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return (double) time1 - (double) time0;
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/* Likewise for long double. */
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if (TYPE_BITS (time_t) <= LDBL_MANT_DIG || TYPE_FLOATING (time_t))
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return (long double) time1 - (long double) time0;
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/* Subtract the smaller integer from the larger, convert the difference to
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double, and then negate if needed. */
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return time1 < time0 ? - subtract (time0, time1) : subtract (time1, time0);
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}
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strong_alias (__difftime, difftime)
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