/* Copyright (C) 1991, 1994, 1996, 2004 Free Software Foundation, Inc. This file is part of the GNU C Library. The GNU C Library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU C Library; if not, see <http://www.gnu.org/licenses/>. */ /* Written by Paul Eggert <eggert@cs.ucla.edu>. */ #include <time.h> #include <limits.h> #include <float.h> #include <stdint.h> #define TYPE_BITS(type) (sizeof (type) * CHAR_BIT) #define TYPE_FLOATING(type) ((type) 0.5 == 0.5) #define TYPE_SIGNED(type) ((type) -1 < 0) /* Return the difference between TIME1 and TIME0, where TIME0 <= TIME1. time_t is known to be an integer type. */ static double subtract (time_t time1, time_t time0) { if (! TYPE_SIGNED (time_t)) return time1 - time0; else { /* Optimize the common special cases where time_t can be converted to uintmax_t without losing information. */ uintmax_t dt = (uintmax_t) time1 - (uintmax_t) time0; double delta = dt; if (UINTMAX_MAX / 2 < INTMAX_MAX) { /* This is a rare host where uintmax_t has padding bits, and possibly information was lost when converting time_t to uintmax_t. Check for overflow by comparing dt/2 to (time1/2 - time0/2). Overflow occurred if they differ by more than a small slop. Thanks to Clive D.W. Feather for detailed technical advice about hosts with padding bits. In the following code the "h" prefix means half. By range analysis, we have: -0.5 <= ht1 - 0.5*time1 <= 0.5 -0.5 <= ht0 - 0.5*time0 <= 0.5 -1.0 <= dht - 0.5*(time1 - time0) <= 1.0 If overflow has not occurred, we also have: -0.5 <= hdt - 0.5*(time1 - time0) <= 0 -1.0 <= dht - hdt <= 1.5 and since dht - hdt is an integer, we also have: -1 <= dht - hdt <= 1 or equivalently: 0 <= dht - hdt + 1 <= 2 In the above analysis, all the operators have their exact mathematical semantics, not C semantics. However, dht - hdt + 1 is unsigned in C, so it need not be compared to zero. */ uintmax_t hdt = dt / 2; time_t ht1 = time1 / 2; time_t ht0 = time0 / 2; time_t dht = ht1 - ht0; if (2 < dht - hdt + 1) { /* Repair delta overflow. The following expression contains a second rounding, so the result may not be the closest to the true answer. This problem occurs only with very large differences. It's too painful to fix this portably. */ delta = dt + 2.0L * (UINTMAX_MAX - UINTMAX_MAX / 2); } } return delta; } } /* Return the difference between TIME1 and TIME0. */ double __difftime (time_t time1, time_t time0) { /* Convert to double and then subtract if no double-rounding error could result. */ if (TYPE_BITS (time_t) <= DBL_MANT_DIG || (TYPE_FLOATING (time_t) && sizeof (time_t) < sizeof (long double))) return (double) time1 - (double) time0; /* Likewise for long double. */ if (TYPE_BITS (time_t) <= LDBL_MANT_DIG || TYPE_FLOATING (time_t)) return (long double) time1 - (long double) time0; /* Subtract the smaller integer from the larger, convert the difference to double, and then negate if needed. */ return time1 < time0 ? - subtract (time0, time1) : subtract (time1, time0); } strong_alias (__difftime, difftime)