Is div function useful (stdlib.h)?

Question

There is a function called div in C,C++ (stdlib.h)

div_t div(int numer, int denom);

typedef struct _div_t
{
  int quot;
  int rem;
} div_t;

But C,C++ have / and % operators.

My question is: "When there are / and % operators, Is div function useful?"

Answer 1

The div() function returns a structure which contains the quotient and remainder of the division of the first parameter (the numerator) by the second (the denominator). There are four variants:

1. div_t div(int, int)
2. ldiv_t ldiv(long, long)
3. lldiv_t lldiv(long long, long long)
4. imaxdiv_t imaxdiv(intmax_t, intmax_t (intmax_t represents the biggest integer type available on the system)

The div_t structure looks like this:

typedef struct
  {
    int quot;           /* Quotient.  */
    int rem;            /* Remainder.  */
  } div_t;

The implementation does simply use the / and % operators, so it's not exactly a very complicated or necessary function, but it is part of the C standard (as defined by [ISO 9899:201x][1]).

See the implementation in GNU libc:

/* Return the `div_t' representation of NUMER over DENOM.  */
div_t
div (numer, denom)
     int numer, denom;
{
  div_t result;

  result.quot = numer / denom;
  result.rem = numer % denom;

  /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
     NUMER / DENOM is to be computed in infinite precision.  In
     other words, we should always truncate the quotient towards
     zero, never -infinity.  Machine division and remainer may
     work either way when one or both of NUMER or DENOM is
     negative.  If only one is negative and QUOT has been
     truncated towards -infinity, REM will have the same sign as
     DENOM and the opposite sign of NUMER; if both are negative
     and QUOT has been truncated towards -infinity, REM will be
     positive (will have the opposite sign of NUMER).  These are
     considered `wrong'.  If both are NUM and DENOM are positive,
     RESULT will always be positive.  This all boils down to: if
     NUMER >= 0, but REM < 0, we got the wrong answer.  In that
     case, to get the right answer, add 1 to QUOT and subtract
     DENOM from REM.  */

  if (numer >= 0 && result.rem < 0)
    {
      ++result.quot;
      result.rem -= denom;
    }

  return result;
}