* Copyright (c) 1990 Regents of the University of California.
* This code is derived from software contributed to Berkeley by
* %sccs.include.redist.c%
#if defined(LIBC_SCCS) && !defined(lint)
static char sccsid
[] = "@(#)div.c 5.2 (Berkeley) %G%";
#endif /* LIBC_SCCS and not lint */
#include <stdlib.h> /* div_t */
* 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
* 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
* If both are num and denom are positive, r will always
* 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 && r
.rem
< 0) {