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
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!");
}
}
}
其他回答
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);
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
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
编辑:提供了新的答案。 当我写这个问题的第一个答案时,我刚刚开始使用c#,事后我意识到我的“解决方案”是幼稚和低效的。
我最初的回答是: 我会选择更简单的版本:
' if(Enumerable.Range(1100).Contains(intInQuestion)){…DoStuff;} '
更好的方法
因为我还没有看到任何其他更有效的解决方案(至少根据我的测试),我将再试一次。
新的和更好的方法,也适用于负范围:
// Returns true if x is in range [min..max], else false
bool inRange(int x, int min=1, int max=100) => ((x - max)*(x - min) <= 0);
这可以用于正负范围,并且默认为
1 . . 100(包括)并使用x作为数字来检查,然后是由min和max定义的可选范围。
为好的措施添加例子
示例1:
// Returns true if x is in range [min..max], else false
bool inRange(int x, int min=1, int max=100) => ((x - max)*(x - min) <= 0);
Console.WriteLine(inRange(25));
Console.WriteLine(inRange(1));
Console.WriteLine(inRange(100));
Console.WriteLine(inRange(25, 30, 150));
Console.WriteLine(inRange(-25, -50, 0));
返回:
True
True
True
False
True
示例2: 使用100000个1到150之间的随机整数的列表
// Returns true if x is in range [min..max], else false
bool inRange(int x, int min=1, int max=100) => ((x - max)*(x - min) <= 0);
// Generate 100000 ints between 1 and 150
var intsToCheck = new List<int>();
var randGen = new Random();
for(int i = 0; i < 100000; ++i){
intsToCheck.Add(randGen.Next(150) + 1);
}
var counter = 0;
foreach(int n in intsToCheck) {
if(inRange(n)) ++counter;
}
Console.WriteLine("{0} ints found in range 1..100", counter);
返回:
66660 ints found in range 1..100
Execution Time: 0.016 second(s)
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。