/* Copyright (C) 1991-2018 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)