在C＃中四舍五入双重价值 [英] Rounding double values in C#

问题描述

我想在C＃双值的舍入方法。它需要能够为double值四舍五入到任何舍入precision值。手头上我的code如下：

I want a rounding method on double values in C#. It needs to be able to round a double value to any rounding precision value. My code on hand looks like:

public static double RoundI(double number, double roundingInterval) {

    if (roundingInterval == 0.0)
    {
        return;
    }

    double intv = Math.Abs(roundingInterval);
    double sign = Math.Sign(number);
    double val = Math.Abs(number);

    double valIntvRatio = val / intv;
    double k = Math.Floor(valIntvRatio);
    double m = valIntvRatio - k;

    bool mGreaterThanMidPoint = ((m - 0.5) >= 1e-14) ? true : false;
    bool mInMidpoint = (Math.Abs(m - 0.5) < 1e-14) ? true : false;
    return (mGreaterThanMidPoint || mInMidpoint) ? sign * ((k + 1) * intv) : sign * (k * intv);
}

所以RoundI（100，3）应该给99和RoundI（1.2345，0.001），应给予1.235。

So RoundI(100, 3) should give 99 and RoundI(1.2345, 0.001) should give 1.235.

问题是，RoundI（1.275，0.01）返回1.27，而不是1.28。这是执行双重valIntvRatio = VAL / INTV时因为，即双valIntvRatio = 1.275 / 0.01，它给0.12749999999999。我知道这是在任何编程语言重新双击presentation一个问题。我的问题是，有没有一个标准的code到做这样的事情，而不需要担心precision双？在这里，我设置了耐1E-14，但是这个太限制对于这个问题，我不知道什么是要设置正确的耐受性。感谢您的任何帮助。

The problem is, RoundI(1.275, 0.01) returns 1.27, rather than 1.28. This is because when executing double valIntvRatio = val/intv, that is, double valIntvRatio = 1.275 / 0.01, it gives 0.12749999999999. I know this is a problem with double representation in any programming language. My question is, is there a standard code to do things like this, without the need to worry about precision on double? Here I set the tolerant to 1e-14, but this is too restrict for this problem and I don't know what is the correct tolerance to be set. Thank you for any help.

推荐答案

使用示例小数，为的基比的指出

double d = 1.275;
Math.Round(d, 2);          // 1.27
Math.Round((decimal)d, 2); // 1.28 

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