排名靠前的答案描述了一种格式化十进制值的字符串表示的方法，它是有效的。

但是，如果您确实想将保存的精度更改为实际值，则需要编写如下内容：

public static class PrecisionHelper
{
    public static decimal TwoDecimalPlaces(this decimal value)
    {
        // These first lines eliminate all digits past two places.
        var timesHundred = (int) (value * 100);
        var removeZeroes = timesHundred / 100m;

        // In this implementation, I don't want to alter the underlying
        // value.  As such, if it needs greater precision to stay unaltered,
        // I return it.
        if (removeZeroes != value)
            return value;

        // Addition and subtraction can reliably change precision.  
        // For two decimal values A and B, (A + B) will have at least as 
        // many digits past the decimal point as A or B.
        return removeZeroes + 0.01m - 0.01m;
    }
}

单元测试示例：

[Test]
public void PrecisionExampleUnitTest()
{
    decimal a = 500m;
    decimal b = 99.99m;
    decimal c = 123.4m;
    decimal d = 10101.1000000m;
    decimal e = 908.7650m

    Assert.That(a.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
        Is.EqualTo("500.00"));

    Assert.That(b.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
        Is.EqualTo("99.99"));

    Assert.That(c.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
        Is.EqualTo("123.40"));

    Assert.That(d.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
        Is.EqualTo("10101.10"));

    // In this particular implementation, values that can't be expressed in
    // two decimal places are unaltered, so this remains as-is.
    Assert.That(e.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
        Is.EqualTo("908.7650"));
}

        var arr = new List<int>() { -4, 3, -9, 0, 4, 1 };
        decimal result1 = arr.Where(p => p > 0).Count();
        var responseResult1 = result1 / arr.Count();
        decimal result2 = arr.Where(p => p < 0).Count();
        var responseResult2 = result2 / arr.Count();
        decimal result3 = arr.Where(p => p == 0).Count();
        var responseResult3 = result3 / arr.Count();
        Console.WriteLine(String.Format("{0:#,0.000}", responseResult1));
        Console.WriteLine(String.Format("{0:#,0.0000}", responseResult2));
        Console.WriteLine(String.Format("{0:#,0.00000}", responseResult3));

你可以放任意多的0。

decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m

or

decimalVar.ToString("0.##"); // returns "0.5"  when decimalVar == 0.5m

or

decimalVar.ToString("0.00"); // returns "0.50"  when decimalVar == 0.5m

Mike M.的答案对我来说非常适合.NET，但在撰写本文时，.NET核心没有十进制舍入方法。

在.NET Core中，我必须使用：

decimal roundedValue = Math.Round(rawNumber, 2, MidpointRounding.AwayFromZero);

一种包括转换为字符串在内的黑客方法是：

public string FormatTo2Dp(decimal myNumber)
{
    // Use schoolboy rounding, not bankers.
    myNumber = Math.Round(myNumber, 2, MidpointRounding.AwayFromZero);

    return string.Format("{0:0.00}", myNumber);
}

最适用的解决方案是

decimalVar.ToString("#.##");

给定小数d=12.345；表达式d.ToString（“C”）或String.Format（“｛0:C｝”，d）产生$12.35-请注意，使用了当前区域性的货币设置，包括符号。

请注意，“C”使用当前区域性中的位数。您可以始终使用C｛precision说明符｝（如String.Format（“｛0:C2｝”，5.123d））重写默认值以强制执行必要的精度。