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;
    }
}

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。

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

我正在寻找一种优雅的方式来做它的边界可能被切换(即。不确定值的顺序)。

这只适用于存在?:的新版本的c#

bool ValueWithinBounds(float val, float bounds1, float bounds2)
{
    return bounds1 >= bounds2 ?
      val <= bounds1 && val >= bounds2 : 
      val <= bounds2 && val >= bounds1;
}

显然，您可以根据自己的需要更改=号。也可以用类型转换。我只需要在边界内(或等于)返回一个浮点数

您可以使用模式匹配以最优雅的方式实现这一点:

int i = 5;
if(i is (>0 and <=10))
{

}

if (value > 1 && value < 100)
{
    // do work
}
else
{
    // handle outside of range logic
}