| Line 96... |
Line 96... |
r.rem = num % denom;
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r.rem = num % denom;
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/*
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/*
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* The ANSI standard says that |r.quot| <= |n/d|, where
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* The ANSI standard says that |r.quot| <= |n/d|, where
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* n/d is to be computed in infinite precision. In other
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* n/d is to be computed in infinite precision. In other
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* words, we should always truncate the quotient towards
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* words, we should always truncate the quotient towards
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* 0, never -infinity.
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* 0, never -infinity or +infinity.
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*
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*
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* Machine division and remainer may work either way when
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* Machine division and remainer may work either way when
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* one or both of n or d is negative. If only one is
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* one or both of n or d is negative. If only one is
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* negative and r.quot has been truncated towards -inf,
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* negative and r.quot has been truncated towards -inf,
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* r.rem will have the same sign as denom and the opposite
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* r.rem will have the same sign as denom and the opposite
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| Line 114... |
Line 114... |
*
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*
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* This all boils down to:
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* This all boils down to:
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* if num >= 0, but r.rem < 0, we got the wrong answer.
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* if num >= 0, but r.rem < 0, we got the wrong answer.
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* In that case, to get the right answer, add 1 to r.quot and
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* In that case, to get the right answer, add 1 to r.quot and
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* subtract denom from r.rem.
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* subtract denom from r.rem.
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* if num < 0, but r.rem > 0, we also have the wrong answer.
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* In this case, to get the right answer, subtract 1 from r.quot and
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* add denom to r.rem.
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*/
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*/
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if (num >= 0 && r.rem < 0) {
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if (num >= 0 && r.rem < 0) {
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r.quot++;
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++r.quot;
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r.rem -= denom;
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r.rem -= denom;
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}
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}
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else if (num < 0 && r.rem > 0) {
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--r.quot;
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r.rem += denom;
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}
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return (r);
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return (r);
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}
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}
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No newline at end of file
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No newline at end of file
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