static class ExtensionMethods
{
    internal static bool IsBetween(this double number,double bound1, double bound2)
    {
        return Math.Min(bound1, bound2) <= number && number <= Math.Max(bound2, bound1);
    }

    internal static bool IsBetween(this int number, double bound1, double bound2)
    {
        return Math.Min(bound1, bound2) <= number && number <= Math.Max(bound2, bound1);
    }
}

使用

double numberToBeChecked = 7;

var result = numberToBeChecked.IsBetween（100，122）;

var result = 5.IsBetween（100，120）;

var result = 8.0.IsBetween（1.2，9.6）;

static class ExtensionMethods
{
    internal static bool IsBetween(this double number,double bound1, double bound2)
    {
        return Math.Min(bound1, bound2) <= number && number <= Math.Max(bound2, bound1);
    }

    internal static bool IsBetween(this int number, double bound1, double bound2)
    {
        return Math.Min(bound1, bound2) <= number && number <= Math.Max(bound2, bound1);
    }
}

使用

double numberToBeChecked = 7;

var result = numberToBeChecked.IsBetween（100，122）;

var result = 5.IsBetween（100，120）;

var result = 8.0.IsBetween（1.2，9.6）;

在c#中，关于速度和代码原的最佳解决方案，只有一次比较，没有约束检查，并且不会因溢出而容易出错:

public static bool IsInRange(int value, int min, int max) => (uint)(value - min) <= (uint)(max - min);

最小值和最大值包括在内。

就像其他人说的，使用简单的if。

你应该考虑一下顺序。

e.g

1 <= x && x <= 100

容易读吗

x >= 1 && x <= 100

我会创建一个Range对象，就像这样:

public class Range<T> where T : IComparable
{
    public T InferiorBoundary{get;private set;}
    public T SuperiorBoundary{get;private set;}

    public Range(T inferiorBoundary, T superiorBoundary)
    {
        InferiorBoundary = inferiorBoundary;
        SuperiorBoundary = superiorBoundary;
    }

    public bool IsWithinBoundaries(T value){
        return InferiorBoundary.CompareTo(value) > 0 && SuperiorBoundary.CompareTo(value) < 0;
    }
}

那么你可以这样使用它:

Range<int> myRange = new Range<int>(1,999);
bool isWithinRange = myRange.IsWithinBoundaries(3);

这样你就可以在其他类型中重用它。

当检查一个“数字”是否在一个范围内时，你必须清楚你的意思，两个数字相等意味着什么?一般来说，你应该把所有浮点数包装在一个所谓的“epsilon球”中，这是通过选择一个小的值来完成的，如果两个值如此接近，它们就是相同的。

    private double _epsilon = 10E-9;
    /// <summary>
    /// Checks if the distance between two doubles is within an epsilon.
    /// In general this should be used for determining equality between doubles.
    /// </summary>
    /// <param name="x0">The orgin of intrest</param>
    /// <param name="x"> The point of intrest</param>
    /// <param name="epsilon">The minimum distance between the points</param>
    /// <returns>Returns true iff x  in (x0-epsilon, x0+epsilon)</returns>
    public static bool IsInNeghborhood(double x0, double x, double epsilon) => Abs(x0 - x) < epsilon;

    public static bool AreEqual(double v0, double v1) => IsInNeghborhood(v0, v1, _epsilon);

有了这两个辅助，并假设任何数字都可以转换为double而不需要所需的精度。现在需要的是一个枚举和另一个方法

    public enum BoundType
    {
        Open,
        Closed,
        OpenClosed,
        ClosedOpen
    }

另一种方法如下:

    public static bool InRange(double value, double upperBound, double lowerBound, BoundType bound = BoundType.Open)
    {
        bool inside = value < upperBound && value > lowerBound;
        switch (bound)
        {
            case BoundType.Open:
                return inside;
            case BoundType.Closed:
                return inside || AreEqual(value, upperBound) || AreEqual(value, lowerBound); 
            case BoundType.OpenClosed:
                return inside || AreEqual(value, upperBound);
            case BoundType.ClosedOpen:
                return inside || AreEqual(value, lowerBound);
            default:
                throw new System.NotImplementedException("You forgot to do something");
        }
    }

现在，这可能远远超过了您想要的，但它使您不必一直处理舍入问题，并试图记住一个值是否被舍入到哪个位置。如果你需要，你可以很容易地将它扩展到任意的情况并允许变化。