好吧，我会配合的。已经有这么多答案了，但也许还有一些其他新奇的空间:

(显然你根本不用这些)

    var num = 7;
    const int min = 5;
    const int max = 10;
    var inRange = Math.Clamp(num, min, max) == num;

Or

    var num = 7;
    const int min = 5;
    const int max = 10;
    var inRange = num switch { < min => false, > max => false, _ => true };

Or

    var num = 7;
    const int min = 5;
    const int max = 10;
    var inRange = num is >= min and <= max;

好吧，也许你可以用最后一个。

好的，再来一个

    var num = 7;
    const int min = 5;
    const int max = 10;
    var inRange = Enumerable.Range(min, max-min).Contains(num);

2022年6月

int id = 10;
if(Enumerable.Range(1, 100).Select(x => x == id).Any()) // true

有很多选择:

int x = 30;
if (Enumerable.Range(1,100).Contains(x))  //true

实际上，基本的，如果更优雅的话，可以在第一张支票中用倒序写:

if (1 <= x && x <= 100)   //true

此外，查看这篇SO帖子的正则表达式选项。

注:

LINQ solution is strictly for style points - since Contains iterates over all items its complexity is O(range_size) and not O(1) normally expected from a range check. More generic version for other ranges (notice that second argument is count, not end): if (Enumerable.Range(start, end - start + 1).Contains(x) There is temptation to write if solution without && like 1 <= x <= 100 - that look really elegant, but in C# leads to a syntax error "Operator '<=' cannot be applied to operands of type 'bool' and 'int'"

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

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))
{

}

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