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);

使用内置的Range结构体(c# 8+)，我们可以创建一个扩展方法来检查索引是否在原始范围内。

public static bool IsInRangeOf(this Range range, Index index)
{
   return index.Value >= range.Start.Value && index.Value < range.End.Value;
}

由于Index覆盖隐式操作符，因此可以传递int型而不是Index结构体。

var range = new Range(1, 10);
var isInRange = range.IsInRangeOf(1); // true, 1..10 is inclusive min range index(1)
var isInRange = range.IsInRangeOf(10); // false, 1..10 exclusive on max range index (10).
var isInRange = range.IsInRangeOf(100); // false

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

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。

因为所有其他答案都不是我发明的，这里只是我的实现:

public enum Range
{
    /// <summary>
    /// A range that contains all values greater than start and less than end.
    /// </summary>
    Open,
    /// <summary>
    /// A range that contains all values greater than or equal to start and less than or equal to end.
    /// </summary>
    Closed,
    /// <summary>
    /// A range that contains all values greater than or equal to start and less than end.
    /// </summary>
    OpenClosed,
    /// <summary>
    /// A range that contains all values greater than start and less than or equal to end.
    /// </summary>
    ClosedOpen
}

public static class RangeExtensions
{
    /// <summary>
    /// Checks if a value is within a range that contains all values greater than start and less than or equal to end.
    /// </summary>
    /// <param name="value">The value that should be checked.</param>
    /// <param name="start">The first value of the range to be checked.</param>
    /// <param name="end">The last value of the range to be checked.</param>
    /// <returns><c>True</c> if the value is greater than start and less than or equal to end, otherwise <c>false</c>.</returns>
    public static bool IsWithin<T>(this T value, T start, T end) where T : IComparable<T>
    {
        return IsWithin(value, start, end, Range.ClosedOpen);
    }

    /// <summary>
    /// Checks if a value is within the given range.
    /// </summary>
    /// <param name="value">The value that should be checked.</param>
    /// <param name="start">The first value of the range to be checked.</param>
    /// <param name="end">The last value of the range to be checked.</param>
    /// <param name="range">The kind of range that should be checked. Depending on the given kind of range the start end end value are either inclusive or exclusive.</param>
    /// <returns><c>True</c> if the value is within the given range, otherwise <c>false</c>.</returns>
    public static bool IsWithin<T>(this T value, T start, T end, Range range) where T : IComparable<T>
    {
        if (value == null)
            throw new ArgumentNullException(nameof(value));

        if (start == null)
            throw new ArgumentNullException(nameof(start));

        if (end == null)
            throw new ArgumentNullException(nameof(end));

        switch (range)
        {
            case Range.Open:
                return value.CompareTo(start) > 0
                       && value.CompareTo(end) < 0;
            case Range.Closed:
                return value.CompareTo(start) >= 0
                       && value.CompareTo(end) <= 0;
            case Range.OpenClosed:
                return value.CompareTo(start) > 0
                       && value.CompareTo(end) <= 0;
            case Range.ClosedOpen:
                return value.CompareTo(start) >= 0
                       && value.CompareTo(end) < 0;
            default:
                throw new ArgumentException($"Unknown parameter value {range}.", nameof(range));
        }
    }
}

然后你可以这样使用它:

var value = 5;
var start = 1;
var end = 10;

var result = value.IsWithin(start, end, Range.Closed);

你在寻找[1..100]?这只是帕斯卡。