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Is Math.IEEERemainder(x,y) equivalent to x%y?

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Is there any differences between Math.IEEERemainder(x,y) and x%y ?

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Ahmad Farid Avatar asked Dec 28 '09 21:12

Ahmad Farid


2 Answers

No, they are not equivalent. MSDN shows the different formulas used for modulo and for IEEERemainder and has a short sample program exhibiting the differences:

IEEERemainder = dividend - (divisor * Math.Round(dividend / divisor))

Modulus = (Math.Abs(dividend) - (Math.Abs(divisor) * 
      (Math.Floor(Math.Abs(dividend) / Math.Abs(divisor))))) * 
      Math.Sign(dividend)

Some examples where they have different/identical output (taken from MSDN):

                         IEEERemainder              Modulus
   3 / 2 =                          -1                    1
   4 / 2 =                           0                    0
   10 / 3 =                          1                    1
   11 / 3 =                         -1                    2
   27 / 4 =                         -1                    3
   28 / 5 =                         -2                    3
   17.8 / 4 =                      1.8                  1.8
   17.8 / 4.1 =                    1.4                  1.4
   -16.3 / 4.1 =    0.0999999999999979                   -4
   17.8 / -4.1 =                   1.4                  1.4
   -17.8 / -4.1 =                 -1.4                 -1.4

See also this good answer by sixlettervariables on a similar question.

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Dirk Vollmar Avatar answered Oct 23 '22 08:10

Dirk Vollmar


No, they're not the same; see the documentation.

Here's the source:

    public static double IEEERemainder(double x, double y) { 
        double regularMod = x % y;
        if (Double.IsNaN(regularMod)) { 
            return Double.NaN;
        }
        if (regularMod == 0) {
            if (Double.IsNegative(x)) { 
                return Double.NegativeZero;
            } 
        } 
        double alternativeResult;
        alternativeResult = regularMod - (Math.Abs(y) * Math.Sign(x)); 
        if (Math.Abs(alternativeResult) == Math.Abs(regularMod)) {
            double divisionResult = x/y;
            double roundedResult = Math.Round(divisionResult);
            if (Math.Abs(roundedResult) > Math.Abs(divisionResult)) { 
                return alternativeResult;
            } 
            else { 
                return regularMod;
            } 
        }
        if (Math.Abs(alternativeResult) < Math.Abs(regularMod)) {
            return alternativeResult;
        } 
        else {
            return regularMod; 
        } 
    }
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SLaks Avatar answered Oct 23 '22 09:10

SLaks