xref: /openbmc/u-boot/lib/ldiv.c (revision fbe502e9aba098b5ad500d1cdb6b376f56f9ddbb)
1  // SPDX-License-Identifier: GPL-2.0+
2  /* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
3     This file is part of the GNU C Library.
4   */
5  
6  typedef struct {
7  	long    quot;
8  	long    rem;
9  } ldiv_t;
10  /* Return the `ldiv_t' representation of NUMER over DENOM.  */
11  ldiv_t
12  ldiv (long int numer, long int denom)
13  {
14    ldiv_t result;
15  
16    result.quot = numer / denom;
17    result.rem = numer % denom;
18  
19    /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
20       NUMER / DENOM is to be computed in infinite precision.  In
21       other words, we should always truncate the quotient towards
22       zero, never -infinity.  Machine division and remainer may
23       work either way when one or both of NUMER or DENOM is
24       negative.  If only one is negative and QUOT has been
25       truncated towards -infinity, REM will have the same sign as
26       DENOM and the opposite sign of NUMER; if both are negative
27       and QUOT has been truncated towards -infinity, REM will be
28       positive (will have the opposite sign of NUMER).  These are
29       considered `wrong'.  If both are NUM and DENOM are positive,
30       RESULT will always be positive.  This all boils down to: if
31       NUMER >= 0, but REM < 0, we got the wrong answer.  In that
32       case, to get the right answer, add 1 to QUOT and subtract
33       DENOM from REM.  */
34  
35    if (numer >= 0 && result.rem < 0)
36      {
37        ++result.quot;
38        result.rem -= denom;
39      }
40  
41    return result;
42  }
43