3 <<div>>---divide two integers
10 div_t div(int <[n]>, int <[d]>);
14 div_t div(<[n]>, <[d]>)
25 returning quotient and remainder as two integers in a structure <<div_t>>.
28 The result is represented with the structure
36 where the <<quot>> field represents the quotient, and <<rem>> the
37 remainder. For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then
38 <[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'.
40 To divide <<long>> rather than <<int>> values, use the similar
46 No supporting OS subroutines are required.
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86 #include <stdlib.h> /* div_t */
89 _DEFUN (div, (num, denom),
98 * The ANSI standard says that |r.quot| <= |n/d|, where
99 * n/d is to be computed in infinite precision. In other
100 * words, we should always truncate the quotient towards
101 * 0, never -infinity or +infinity.
103 * Machine division and remainer may work either way when
104 * one or both of n or d is negative. If only one is
105 * negative and r.quot has been truncated towards -inf,
106 * r.rem will have the same sign as denom and the opposite
107 * sign of num; if both are negative and r.quot has been
108 * truncated towards -inf, r.rem will be positive (will
109 * have the opposite sign of num). These are considered
112 * If both are num and denom are positive, r will always
115 * This all boils down to:
116 * if num >= 0, but r.rem < 0, we got the wrong answer.
117 * In that case, to get the right answer, add 1 to r.quot and
118 * subtract denom from r.rem.
119 * if num < 0, but r.rem > 0, we also have the wrong answer.
120 * In this case, to get the right answer, subtract 1 from r.quot and
121 * add denom to r.rem.
123 if (num >= 0 && r.rem < 0) {
127 else if (num < 0 && r.rem > 0) {