如何调用词典< K，V> .TryGetValue（）其中K：谓词< T>，V：枚举 [英] How to call Dictionary<K, V>.TryGetValue() where K : Predicate<T>, V : enum
问题描述
我词典<谓语<双>中SomeEnum> ：
var dic = new Dictionary<Predicate<double>, SomeEnum>
{
{ (d) => d < 10, SomeEnum.Foo },
{ (d) => d > 90, SomeEnum.Bar }
};
我要打电话 TryGetValue（K，走出V）反对这样的：
dic.TryGetValue(99)
和接受
SomeStruct.Bar
但 TryGetValue第一个参数（）是谓词< T> ，而不仅仅是 T 。 ？如何做我想做的。
But first param for TryGetValue() is Predicate<T>, not just T. How can I do what I want?
我发现只有一个肮脏的解决方法：
I found only a dirty workaround:
var kpv = dic.FirstOrDefault(p => p.Key(99));
if (kpv.Key != null)
var result = kpv.Value;
还有没有其他的办法呢？
Are there other ways?
< ？strong>或如何正确实现我的想法 - 声明的关键不是一个常数，而是像一个段
Or how to implement my idea properly? - declare a key not as a constant but like a segment.
推荐答案
有几件事情错在这里：
谓词<双> 不是合适的类型使用作为 TKEY的。对字典的关键是应该的确定的一个值，而不是计算值。
Predicate<double> is not an appropriate type to use as a TKey. The key for a dictionary is supposed to identify a value, not calculate a value.
这不会让使用lambda表达式或者任何意义。因为它们是匿名的，你就不会得到任何等价，并且不能使用的字典。双>
This wouldn't make any sense using lambdas either. Because they are anonymous, you wouldn't get any equivalence, and won't be able use a dictionary.
See this code sample for an illustration:有关的说明，请参阅此代码示例; fn_1 = D => ð== 34.0d;
谓<双> fn_2 = D => ð== 34.0d;
//注：有不等于
如果（fn_1 == fn_2）
Console.WriteLine（这些都是平等？）;
Predicate<double> fn_1 = d => d == 34.0d;
Predicate<double> fn_2 = d => d == 34.0d;
// Note: There are not equal
if (fn_1 == fn_2)
Console.WriteLine("These are Equal?");
如果任何东西，你可以使用委托列表并执行每一个找到匹配的那些，但在这一点上，你必须准备多个结果。如果你只希望得到一个结果，那么你必须考虑它的为了的谓词存储在您的列表中。
If anything, you could use a list of delegates and execute each one to find the ones that match, but at that point you must expect multiple results. If you only want to get a single result, then you have to consider which order the predicates are stored within your list.
不要滥用 KeyValuePair 作为不具有破解元组LT; T1，T2> 。这将是相当容易的创建一个既有谓词和SomeStruct的类。你看：
Don't misuse KeyValuePair as a hack for not having Tuple<T1,T2>. It would be fairly easy to create a class that has both a Predicate and a SomeStruct. Look:
public class MySegment
{
public Predicate<double> Predicate {get;set;}
public SomeStruct Result {get;set;}
}
要经过谓词的序列，找到匹配的人是这样的：
To go through a sequence of predicates, and find the matching ones would look like this:
...
List<MySegment> list = new List<MySegment>();
...
list.Add(new MySegment { Predicate = d => d < 10, Result = SomeStruct.Foo });
list.Add(new MySegment { Predicate = d => d > 90, Result = SomeStruct.Bar });
...
public IEnumerable<SomeStruct> GetResults(double input)
{
foreach (var item in list)
if (item.Predicate(input))
yield return item.Result;
}
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