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64 lines
2.3 KiB
C
64 lines
2.3 KiB
C
/* imaxdiv() function: division of 'intmax_t'.
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Copyright (C) 2006, 2009-2025 Free Software Foundation, Inc.
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This file is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as
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published by the Free Software Foundation, either version 3 of the
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License, or (at your option) any later version.
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This file 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
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>. */
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#include <config.h>
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/* Specification. */
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#include <inttypes.h>
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#include <stdlib.h>
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imaxdiv_t
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imaxdiv (intmax_t numer, intmax_t denom)
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{
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imaxdiv_t result;
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result.quot = numer / denom;
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result.rem = numer % denom;
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/* Verify the requirements of ISO C 99 section 6.5.5 paragraph 6:
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"When integers are divided, the result of the / operator is the
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algebraic quotient with any fractional part discarded. (This is
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often called "truncation toward zero".) If the quotient a/b is
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representable, the expression (a/b)*b + a%b shall equal a." */
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if (!(denom == 0
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|| (INTMAX_MIN + INTMAX_MAX < 0
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&& denom == -1
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&& numer < - INTMAX_MAX)))
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{
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if (!(result.quot * denom + result.rem == numer))
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/* The compiler's implementation of / and % is broken. */
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abort ();
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if (!(numer >= 0
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? result.rem >= 0
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&& (denom >= 0
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? result.rem < denom
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: /* Don't write result.rem < - denom,
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as it gives integer overflow if denom == INTMAX_MIN. */
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- result.rem > denom)
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: result.rem <= 0
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&& (denom >= 0
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? result.rem > - denom
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: result.rem > denom)))
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/* The compiler's implementation of / and % may be ok according to
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C89, but not to C99. Please report this to <bug-gnulib@ngu.org>.
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This might be a big portability problem. */
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abort ();
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}
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return result;
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}
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