https://msdn.microsoft.com/en-us/library/dwhawy9k%28v=vs.110%29.aspx

此链接详细说明了如何处理问题，以及如果您想了解更多信息，可以做什么。为了简单起见，您想做的是

double whateverYouWantToChange = whateverYouWantToChange.ToString("F2");

如果你想要一种货币，你可以通过键入“C2”而不是“F2”来简化

下面是一个Linqpad小程序，可以显示不同的格式：

void Main()
{
    FormatDecimal(2345.94742M);
    FormatDecimal(43M);
    FormatDecimal(0M);
    FormatDecimal(0.007M);
}

public void FormatDecimal(decimal val)
{
    Console.WriteLine("ToString: {0}", val);
    Console.WriteLine("c: {0:c}", val);
    Console.WriteLine("0.00: {0:0.00}", val);
    Console.WriteLine("0.##: {0:0.##}", val);
    Console.WriteLine("===================");
}

以下是结果：

ToString: 2345.94742
c: $2,345.95
0.00: 2345.95
0.##: 2345.95
===================
ToString: 43
c: $43.00
0.00: 43.00
0.##: 43
===================
ToString: 0
c: $0.00
0.00: 0.00
0.##: 0
===================
ToString: 0.007
c: $0.01
0.00: 0.01
0.##: 0.01
===================

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"));
}

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

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。

最适用的解决方案是

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