像这样的怎么样?

if (theNumber.isBetween(low, high, IntEx.Bounds.INCLUSIVE_INCLUSIVE))
{
}

扩展方法如下(已测试):

public static class IntEx
{
    public enum Bounds 
    {
        INCLUSIVE_INCLUSIVE, 
        INCLUSIVE_EXCLUSIVE, 
        EXCLUSIVE_INCLUSIVE, 
        EXCLUSIVE_EXCLUSIVE
    }

    public static bool isBetween(this int theNumber, int low, int high, Bounds boundDef)
    {
        bool result;
        switch (boundDef)
        {
            case Bounds.INCLUSIVE_INCLUSIVE:
                result = ((low <= theNumber) && (theNumber <= high));
                break;
            case Bounds.INCLUSIVE_EXCLUSIVE:
                result = ((low <= theNumber) && (theNumber < high));
                break;
            case Bounds.EXCLUSIVE_INCLUSIVE:
                result = ((low < theNumber) && (theNumber <= high));
                break;
            case Bounds.EXCLUSIVE_EXCLUSIVE:
                result = ((low < theNumber) && (theNumber < high));
                break;
            default:
                throw new System.ArgumentException("Invalid boundary definition argument");
        }
        return result;
    }
}

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。

你的意思是?

if(number >= 1 && number <= 100)

or

bool TestRange (int numberToCheck, int bottom, int top)
{
  return (numberToCheck >= bottom && numberToCheck <= top);
}

In C, if time efficiency is crucial and integer overflows will wrap, one could do if ((unsigned)(value-min) <= (max-min)) .... If 'max' and 'min' are independent variables, the extra subtraction for (max-min) will waste time, but if that expression can be precomputed at compile time, or if it can be computed once at run-time to test many numbers against the same range, the above expression may be computed efficiently even in the case where the value is within range (if a large fraction of values will be below the valid range, it may be faster to use if ((value >= min) && (value <= max)) ... because it will exit early if value is less than min).

不过，在使用这样的实现之前，请先对目标机器进行基准测试。在某些处理器上，由两部分组成的表达式可能在所有情况下都更快，因为两个比较可能是独立完成的，而在减法和比较方法中，减法必须在比较执行之前完成。

新花样的老最爱:

public bool IsWithinRange(int number, int topOfRange, int bottomOfRange, bool includeBoundaries) {
    if (includeBoundaries)
        return number <= topOfRange && number >= bottomOfRange;
    return number < topOfRange && number > bottomOfRange;
}

我不知道，但我用这个方法:

    public static Boolean isInRange(this Decimal dec, Decimal min, Decimal max, bool includesMin = true, bool includesMax = true ) {

    return (includesMin ? (dec >= min) : (dec > min)) && (includesMax ? (dec <= max) : (dec < max));
}

这是我使用它的方式:

    [TestMethod]
    public void IsIntoTheRange()
    {
        decimal dec = 54;

        Boolean result = false;

        result = dec.isInRange(50, 60); //result = True
        Assert.IsTrue(result);

        result = dec.isInRange(55, 60); //result = False
        Assert.IsFalse(result);

        result = dec.isInRange(54, 60); //result = True
        Assert.IsTrue(result);

        result = dec.isInRange(54, 60, false); //result = False
        Assert.IsFalse(result);

        result = dec.isInRange(32, 54, false, false);//result = False
        Assert.IsFalse(result);

        result = dec.isInRange(32, 54, false);//result = True
        Assert.IsTrue(result);
    }