问题

Update

OK, after some investigation, and thanks in big part to the helpful answers provided by Jon and Hans, this is what I was able to put together. So far I think it seems to work well. I wouldn't bet my life on its total correctness, of course.

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;
    }
}

I haven't tested it all that thoroughly. Here are a few sample inputs/outputs, though:

Value          Precision
0              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);
}

I haven't really finished working through this, though. Anybody who wants to share thoughts: that would be much appreciated!

Original Question

Suppose I have write this code:

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

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

The above will output:

1.23
1.23000

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);

The above outputs:

21.729560302171195125816619416

What I'm asking is: how could I write a method that would return 21.73 instead (since 123.4M has four significant figures)?

To be clear: I realize I could call ToString on both arguments, count the significant figures in each string, and use this number to round the result of the calculation. I'm looking for a different way, if possible.

(I also realize that in most scenarios where you're dealing with significant figures, you probably don't need to be using the decimal type. But I'm asking because, as I mentioned in the beginning, the decimal type appears to include information about precision, whereas double does not, as far as I know.)

回答1:

You can use Decimal.GetBits to get the raw data, and work it out from that.

Unfortunately I don't have time to write sample code at the moment - and you'll probably want to use BigInteger for some of the manipulation, if you're using .NET 4 - but hopefully this will get you going. Just working out the precision and then calling Math.Round on the original result may well be a good start.

回答2:

Yes, unlike floating point types, System.Decimal keeps track of the number of digits in the literal. This is a feature of decimal.Parse(), whether executed by your code yourself or by the compiler when it parses a literal in your program. You can recover this information, check out the code in my answer in this thread.

Recovering the number of significant digits after you do math on the value strikes me as a long shot. No idea if the operators preserve them, please let us know what you find out.

回答3:

The solution above falls out of bed quickly with values like 1200 where the number of significant digits is 2 but the function returns 4. Perhaps there is a consistent manner to deal with such a situation i.e. check to make sure return value does not include trailing zeros in the whole number portion without a fractional part.

来源：https://stackoverflow.com/questions/3683718/is-there-a-way-to-get-the-significant-figures-of-a-decimal

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