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/* imaxdiv() function: division of 'intmax_t'.
Copyright (C) 2006, 2009 Free Software Foundation, Inc.
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 3 of the License, or
(at your option) any later version.
This program 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 General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>. */
#include <config.h>
/* Specification. */
#include <inttypes.h>
#include <stdlib.h>
imaxdiv_t
imaxdiv (intmax_t numer, intmax_t denom)
{
imaxdiv_t result;
result.quot = numer / denom;
result.rem = numer % denom;
/* Verify the requirements of ISO C 99 section 6.5.5 paragraph 6:
"When integers are divided, the result of the / operator is the
algebraic quotient with any fractional part discarded. (This is
often called "truncation toward zero".) If the quotient a/b is
representable, the expression (a/b)*b + a%b shall equal a." */
if (!(denom == 0
|| (INTMAX_MIN + INTMAX_MAX < 0
&& denom == -1
&& numer < - INTMAX_MAX)))
{
if (!(result.quot * denom + result.rem == numer))
/* The compiler's implementation of / and % is broken. */
abort ();
if (!(numer >= 0
? result.rem >= 0
&& (denom >= 0
? result.rem < denom
: /* Don't write result.rem < - denom,
as it gives integer overflow if denom == INTMAX_MIN. */
- result.rem > denom)
: result.rem <= 0
&& (denom >= 0
? result.rem > - denom
: result.rem > denom)))
/* The compiler's implementation of / and % may be ok according to
C89, but not to C99. Please report this to <bug-gnulib@ngu.org>.
This might be a big portability problem. */
abort ();
}
return result;
}
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