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。

In production code I would simply write 1 <= x && x <= 100 This is easy to understand and very readable. Starting with C#9.0 we can write x is >= 1 and <= 100 Note that we must write x only once. is introduces a pattern matching expression where and is part of the pattern. && would require us to repeat x is as in x is >= 1 && x is <= 100 Here is a clever method that reduces the number of comparisons from two to one by using some math. There is not necessarily a performance advantage in doing so, but it is elegant. The idea is that one of the two factors becomes negative if the number lies outside of the range and zero if the number is equal to one of the bounds: If the bounds are inclusive: (x - 1) * (100 - x) >= 0 or (x - min) * (max - x) >= 0 If the bounds are exclusive: (x - 1) * (100 - x) > 0 or (x - min) * (max - x) > 0

如果您关心@Daap对已接受答案的注释，并且只能传递一次值，则可以尝试以下方法之一

bool TestRangeDistance (int numberToCheck, int bottom, int distance)
{
  return (numberToCheck >= bottom && numberToCheck <= bottom+distance);
}

//var t = TestRangeDistance(10, somelist.Count()-5, 10);

or

bool TestRangeMargin (int numberToCheck, int target, int margin)
{
  return (numberToCheck >= target-margin && numberToCheck <= target+margin);
}

//var t = TestRangeMargin(10, somelist.Count(), 5);

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

Using an && expression to join two comparisons is simply the most elegant way to do this. If you try using fancy extension methods and such, you run into the question of whether to include the upper bound, the lower bound, or both. Once you start adding additional variables or changing the extension names to indicate what is included, your code becomes longer and harder to read (for the vast majority of programmers). Furthermore, tools like Resharper will warn you if your comparison doesn't make sense (number > 100 && number < 1), which they won't do if you use a method ('i.IsBetween(100, 1)').

我要做的唯一另一个评论是，如果你检查输入的意图是抛出异常，你应该考虑使用代码契约:

Contract.Requires(number > 1 && number < 100)

这比if(…)抛出new Exception(…)更优雅，如果有人试图调用您的方法而没有首先确保该数字在边界内，您甚至可以得到编译时警告。

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。