我应该如何比较这些双打以获得所需的结果？

我这里有一个简单的示例应用程序，我在其中进行乘法和加法double变量，然后将它们与预期结果进行比较。在这两种情况下，结果都等于预期结果，但当我进行比较时，它失败了。

static void Main(string[] args)
{
    double a = 98.1;
    double b = 107.7;
    double c = 92.5;
    double d = 96.5;

    double expectedResult = 88.5;
    double result1 = (1*2*a) + (-1*1*b);
    double result2 = (1*2*c) + (-1*1*d);            

    Console.WriteLine(String.Format("2x{0} - {1} = {2}\nEqual to 88.5? {3}\n", a, b, result1, expectedResult == result1));
    Console.WriteLine(String.Format("2x{0} - {1} = {2}\nEqual to 88.5? {3}\n", c, d, result2, expectedResult == result2));

    Console.Read();
}

这是输出：

2x98.1 - 107.7 = 88.5
Equal to 88.5? False

2x92.5 - 96.5 = 88.5
Equal to 88.5? True

我需要能够捕捉到事实确实如此True在这两种情况下。我该怎么做呢？

由于浮点数存储数字的方式，浮点数通常不包含数学告诉我们的精确值。

为了仍然进行可靠的比较，您需要允许一些差异：

private const double DoubleEpsilon = 2.22044604925031E-16;

/// <summary>Determines whether <paramref name="value1"/> is very close to <paramref name="value2"/>.</summary>
/// <param name="value1">The value1.</param>
/// <param name="value2">The value2.</param>
/// <returns><c>true</c> if <paramref name="value1"/> is very close to value2; otherwise, <c>false</c>.</returns>
public static bool IsVeryCloseTo(this double value1, double value2)
{
    if (value1 == value2)
        return true;

    var tolerance = (Math.Abs(value1) + Math.Abs(value2)) * DoubleEpsilon;
    var difference = value1 - value2;

    return -tolerance < difference && tolerance > difference;
}

还请务必阅读这篇博文 http://realtimecollisiondetection.net/blog/?p=89.