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Diffstat (limited to 'newlib/libc/stdlib/div.c')
| -rw-r--r-- | newlib/libc/stdlib/div.c | 132 |
1 files changed, 0 insertions, 132 deletions
diff --git a/newlib/libc/stdlib/div.c b/newlib/libc/stdlib/div.c deleted file mode 100644 index 2c070dec7..000000000 --- a/newlib/libc/stdlib/div.c +++ /dev/null @@ -1,132 +0,0 @@ -/* -FUNCTION -<<div>>---divide two integers - -INDEX - div - -ANSI_SYNOPSIS - #include <stdlib.h> - div_t div(int <[n]>, int <[d]>); - -TRAD_SYNOPSIS - #include <stdlib.h> - div_t div(<[n]>, <[d]>) - int <[n]>, <[d]>; - -DESCRIPTION -Divide -@tex -$n/d$, -@end tex -@ifinfo -<[n]>/<[d]>, -@end ifinfo -returning quotient and remainder as two integers in a structure <<div_t>>. - -RETURNS -The result is represented with the structure - -. typedef struct -. { -. int quot; -. int rem; -. } div_t; - -where the <<quot>> field represents the quotient, and <<rem>> the -remainder. For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then -<[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'. - -To divide <<long>> rather than <<int>> values, use the similar -function <<ldiv>>. - -PORTABILITY -<<div>> is ANSI. - -No supporting OS subroutines are required. -*/ - -/* - * Copyright (c) 1990 Regents of the University of California. - * All rights reserved. - * - * This code is derived from software contributed to Berkeley by - * Chris Torek. - * - * Redistribution and use in source and binary forms, with or without - * modification, are permitted provided that the following conditions - * are met: - * 1. Redistributions of source code must retain the above copyright - * notice, this list of conditions and the following disclaimer. - * 2. Redistributions in binary form must reproduce the above copyright - * notice, this list of conditions and the following disclaimer in the - * documentation and/or other materials provided with the distribution. - * 3. All advertising materials mentioning features or use of this software - * must display the following acknowledgement: - * This product includes software developed by the University of - * California, Berkeley and its contributors. - * 4. Neither the name of the University nor the names of its contributors - * may be used to endorse or promote products derived from this software - * without specific prior written permission. - * - * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND - * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE - * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE - * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE - * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL - * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS - * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) - * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT - * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY - * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF - * SUCH DAMAGE. - */ - -#include <_ansi.h> -#include <stdlib.h> /* div_t */ - -div_t -_DEFUN (div, (num, denom), - int num _AND - int denom) -{ - div_t r; - - r.quot = num / denom; - r.rem = num % denom; - /* - * The ANSI standard says that |r.quot| <= |n/d|, where - * n/d is to be computed in infinite precision. In other - * words, we should always truncate the quotient towards - * 0, never -infinity or +infinity. - * - * Machine division and remainer may work either way when - * one or both of n or d is negative. If only one is - * negative and r.quot has been truncated towards -inf, - * r.rem will have the same sign as denom and the opposite - * sign of num; if both are negative and r.quot has been - * truncated towards -inf, r.rem will be positive (will - * have the opposite sign of num). These are considered - * `wrong'. - * - * If both are num and denom are positive, r will always - * be positive. - * - * This all boils down to: - * if num >= 0, but r.rem < 0, we got the wrong answer. - * In that case, to get the right answer, add 1 to r.quot and - * subtract denom from r.rem. - * if num < 0, but r.rem > 0, we also have the wrong answer. - * In this case, to get the right answer, subtract 1 from r.quot and - * add denom to r.rem. - */ - if (num >= 0 && r.rem < 0) { - ++r.quot; - r.rem -= denom; - } - else if (num < 0 && r.rem > 0) { - --r.quot; - r.rem += denom; - } - return (r); -} |