div.c File Reference

#include <stdlib.h>

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Functions
div_t	div (int num, int denom)
	Computes both the quotient and the remainder of the division of the numerator x by the denominator y. More...

Function Documentation

div()

div_t div	(	int	x,
		int	y
	)

Computes both the quotient and the remainder of the division of the numerator x by the denominator y.

Computes both the quotient and the remainder of the division of the numerator x by the denominator y.

Parameters

x	integer values
y	integer values

Returns

If both the remainder and the quotient can be represented as objects of the corresponding type (int, long, long long, imaxdiv_t, respectively), returns both as an object of type

See also

div_t,

ldiv_t,

lldiv_t,

imaxdiv_t.

If either the remainder or the quotient cannot be represented, the behavior is undefined.

Definition at line 35 of file div.c.

{
    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.
     *
     * 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 && r.rem < 0)
    {
        r.quot++;
        r.rem -= denom;
    }
    return (r);
}

References div_t::quot, and div_t::rem.