/* imaxdiv() function: division of 'intmax_t'. Copyright (C) 2006 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 2, 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, write to the Free Software Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */ #include /* Specification. */ #include #include 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 . This might be a big portability problem. */ abort (); } return result; }