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。
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
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;
}
}
其他回答
通过一些扩展方法的滥用,我们可以得到以下“优雅”的解决方案:
using System;
namespace Elegant {
public class Range {
public int Lower { get; set; }
public int Upper { get; set; }
}
public static class Ext {
public static Range To(this int lower, int upper) {
return new Range { Lower = lower, Upper = upper };
}
public static bool In(this int n, Range r) {
return n >= r.Lower && n <= r.Upper;
}
}
class Program {
static void Main() {
int x = 55;
if (x.In(1.To(100)))
Console.WriteLine("it's in range! elegantly!");
}
}
}
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
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;
}
}
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
当检查一个“数字”是否在一个范围内时,你必须清楚你的意思,两个数字相等意味着什么?一般来说,你应该把所有浮点数包装在一个所谓的“epsilon球”中,这是通过选择一个小的值来完成的,如果两个值如此接近,它们就是相同的。
private double _epsilon = 10E-9;
/// <summary>
/// Checks if the distance between two doubles is within an epsilon.
/// In general this should be used for determining equality between doubles.
/// </summary>
/// <param name="x0">The orgin of intrest</param>
/// <param name="x"> The point of intrest</param>
/// <param name="epsilon">The minimum distance between the points</param>
/// <returns>Returns true iff x in (x0-epsilon, x0+epsilon)</returns>
public static bool IsInNeghborhood(double x0, double x, double epsilon) => Abs(x0 - x) < epsilon;
public static bool AreEqual(double v0, double v1) => IsInNeghborhood(v0, v1, _epsilon);
有了这两个辅助,并假设任何数字都可以转换为double而不需要所需的精度。现在需要的是一个枚举和另一个方法
public enum BoundType
{
Open,
Closed,
OpenClosed,
ClosedOpen
}
另一种方法如下:
public static bool InRange(double value, double upperBound, double lowerBound, BoundType bound = BoundType.Open)
{
bool inside = value < upperBound && value > lowerBound;
switch (bound)
{
case BoundType.Open:
return inside;
case BoundType.Closed:
return inside || AreEqual(value, upperBound) || AreEqual(value, lowerBound);
case BoundType.OpenClosed:
return inside || AreEqual(value, upperBound);
case BoundType.ClosedOpen:
return inside || AreEqual(value, lowerBound);
default:
throw new System.NotImplementedException("You forgot to do something");
}
}
现在,这可能远远超过了您想要的,但它使您不必一直处理舍入问题,并试图记住一个值是否被舍入到哪个位置。如果你需要,你可以很容易地将它扩展到任意的情况并允许变化。