.net - 有没有办法得到小数的 "significant figures"？

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好的，经过一番调查，这在很大程度上要感谢乔恩和汉斯提供的有用答案，这就是我能够整理出来的内容。到目前为止，我认为它似乎运作良好。当然，我不会用我的生命去赌它的完全正确性。

public static int GetSignificantDigitCount(this decimal value)
{
    /* So, the decimal type is basically represented as a fraction of two
     * integers: a numerator that can be anything, and a denominator that is 
     * some power of 10.
     * 
     * For example, the following numbers are represented by
     * the corresponding fractions:
     * 
     * VALUE    NUMERATOR   DENOMINATOR
     * 1        1           1
     * 1.0      10          10
     * 1.012    1012        1000
     * 0.04     4           100
     * 12.01    1201        100
     * 
     * So basically, if the magnitude is greater than or equal to one,
     * the number of digits is the number of digits in the numerator.
     * If it's less than one, the number of digits is the number of digits
     * in the denominator.
     */

    int[] bits = decimal.GetBits(value);

    if (value >= 1M || value <= -1M)
    {
        int highPart = bits[2];
        int middlePart = bits[1];
        int lowPart = bits[0];

        decimal num = new decimal(lowPart, middlePart, highPart, false, 0);

        int exponent = (int)Math.Ceiling(Math.Log10((double)num));

        return exponent;
    }
    else
    {
        int scalePart = bits[3];

        // Accoring to MSDN, the exponent is represented by
        // bits 16-23 (the 2nd word):
        // http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx
        int exponent = (scalePart & 0x00FF0000) >> 16;

        return exponent + 1;
    }
}

我还没有彻底测试过。不过，这里有一些示例输入/输出:

Value          Precision0              1 digit(s).0.000          4 digit(s).1.23           3 digit(s).12.324         5 digit(s).1.2300         5 digit(s).-5             1 digit(s).-5.01          3 digit(s).-0.012         4 digit(s).-0.100         4 digit(s).0.0            2 digit(s).10443.31       7 digit(s).-130.340       6 digit(s).-80.8000       6 digit(s).

Using this code, I imagine I would accomplish my goal by doing something like this:

public static decimal DivideUsingLesserPrecision(decimal x, decimal y)
{
    int xDigitCount = x.GetSignificantDigitCount();
    int yDigitCount = y.GetSignificantDigitCount();

    int lesserPrecision = System.Math.Min(xDigitCount, yDigitCount);

    return System.Math.Round(x / y, lesserPrecision);
}

不过，我还没有真正完成这个工作。任何想要分享想法的人:我们将不胜感激!

<小时/>

原始问题

假设我写了这段代码:

decimal a = 1.23M;
decimal b = 1.23000M;

Console.WriteLine(a);
Console.WriteLine(b);

以上将输出:

1.231.23000

I find that this also works if I use decimal.Parse("1.23") for a and decimal.Parse("1.23000") for b (which means this question applies to cases where the program receives user input).

So clearly a decimal value is somehow "aware" of what I'll call its precision. However, I see no members on the decimal type that provide any way of accessing this, aside from ToString itself.

Suppose I wanted to multiply two decimal values and trim the result to the precision of the less precise argument. In other words:

decimal a = 123.4M;
decimal b = 5.6789M;

decimal x = a / b;

Console.WriteLine(x);

以上输出:

21.729560302171195125816619416

我要问的是:我如何编写一个返回 21.73 的方法(因为 123.4M 有四个有效数字)？

需要明确的是:我意识到我可以对两个参数调用 ToString，计算每个字符串中的有效数字，并使用该数字对计算结果进行四舍五入。如果可能的话，我正在寻找一种不同的方式。

(我还意识到，在大多数处理有效数字的情况下，您可能不需要使用decimal类型。但是我'我问这个问题是因为，正如我在开头提到的，decimal 类型似乎包含有关精度的信息，而 double 则不包含有关精度的信息。我知道。)

Answer 1

您可以使用Decimal.GetBits获取原始数据，并从中计算出来。

不幸的是，我现在没有时间编写示例代码 - 您可能想使用 BigInteger对于一些操作，如果您使用的是 .NET 4 - 但希望这能让您继续前进。只需计算出精度，然后对原始结果调用 Math.Round 可能是一个好的开始。