我如何用c#优雅地做到这一点?

例如,一个数字可以是1到100之间。

我知道一个简单的if (x >= 1 && x <= 100)就足够了;但是有很多语法糖和新特性不断添加到c# /。Net这个问题是关于更习惯的(一个可以称之为优雅的)写法。

性能不是问题,但请在非O(1)的解决方案中添加性能说明,因为人们可能会复制粘贴建议。


当前回答

2022年6月

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

其他回答

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

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);
    }

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。

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);

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;
    }
}