| /* Copyright (C) 1992, 1997 Free Software Foundation, Inc. | 
 |    This file is part of the GNU C Library. | 
 |  | 
 |  * SPDX-License-Identifier:	LGPL-2.0+ | 
 |  */ | 
 |  | 
 | typedef struct { | 
 | 	long    quot; | 
 | 	long    rem; | 
 | } ldiv_t; | 
 | /* Return the `ldiv_t' representation of NUMER over DENOM.  */ | 
 | ldiv_t | 
 | ldiv (long int numer, long int denom) | 
 | { | 
 |   ldiv_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; | 
 | } |