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

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。

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

只是为了增加这里的噪音，你可以创建一个扩展方法:

public static bool IsWithin(this int value, int minimum, int maximum)
{
    return value >= minimum && value <= maximum;
}

这样你就能做…

int val = 15;

bool foo = val.IsWithin(5,20);

话虽如此，当检查本身只有一行时，这样做似乎是一件愚蠢的事情。

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