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			58 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			58 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| /* `long long int' divison with remainder.
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|    Copyright (C) 1992, 1996, 1997 Free Software Foundation, Inc.
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|    This file is part of the GNU C Library.
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| 
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|    The GNU C Library is free software; you can redistribute it and/or
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|    modify it under the terms of the GNU Lesser General Public
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|    License as published by the Free Software Foundation; either
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|    version 2.1 of the License, or (at your option) any later version.
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| 
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|    The GNU C Library is distributed in the hope that it will be useful,
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|    but WITHOUT ANY WARRANTY; without even the implied warranty of
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|    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
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|    Lesser General Public License for more details.
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| 
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|    You should have received a copy of the GNU Lesser General Public
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|    License along with the GNU C Library; if not, write to the Free
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|    Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
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|    02111-1307 USA.  */
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| 
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| #include <stdlib.h>
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| 
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| 
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| /* Return the `lldiv_t' representation of NUMER over DENOM.  */
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| lldiv_t
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| lldiv (numer, denom)
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|      long long int numer;
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|      long long int denom;
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| {
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|   lldiv_t result;
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| 
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|   result.quot = numer / denom;
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|   result.rem = numer % denom;
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| 
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|   /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
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|      NUMER / DENOM is to be computed in infinite precision.  In
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|      other words, we should always truncate the quotient towards
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|      zero, never -infinity.  Machine division and remainer may
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|      work either way when one or both of NUMER or DENOM is
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|      negative.  If only one is negative and QUOT has been
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|      truncated towards -infinity, REM will have the same sign as
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|      DENOM and the opposite sign of NUMER; if both are negative
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|      and QUOT has been truncated towards -infinity, REM will be
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|      positive (will have the opposite sign of NUMER).  These are
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|      considered `wrong'.  If both are NUM and DENOM are positive,
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|      RESULT will always be positive.  This all boils down to: if
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|      NUMER >= 0, but REM < 0, we got the wrong answer.  In that
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|      case, to get the right answer, add 1 to QUOT and subtract
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|      DENOM from REM.  */
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| 
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|   if (numer >= 0 && result.rem < 0)
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|     {
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|       ++result.quot;
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|       result.rem -= denom;
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|     }
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| 
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|   return result;
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| }
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