计算C#中HybridDictionary中的键/值对的数量
要计算HybridDictionary中的键/值对的数量，代码如下-
示例
using System;
using System.Collections;
using System.Collections.Specialized;
public class Demo {
public static void Main(){
HybridDictionary dict1 = new HybridDictionary();
dict1.Add("A", "SUV");
dict1.Add("B", "MUV");
dict1.Add("C", "AUV");
Console.WriteLine("HybridDictionary1 elements...");
foreach(DictionaryEntry d in dict1){
Console.WriteLine(d.Key + " " + d.Value);
}
Console.WriteLine("Count of Key/value pairs in Dictionary1 = "+dict1.Count);
HybridDictionary dict2 = new HybridDictionary();
dict2.Add("1", "One");
dict2.Add("2", "Two");
dict2.Add("3", "Three");
dict2.Add("4", "Four");
dict2.Add("5", "Five");
dict2.Add("6", "Six");
Console.WriteLine("\nHybridDictionary2 elements...");
foreach(DictionaryEntry d in dict2){
Console.WriteLine(d.Key + " " + d.Value);
}
Console.WriteLine("Count of Key/value pairs in Dictionary2 = "+dict1.Count);
dict2.Clear();
Console.WriteLine("Count of Key/value pairs in Dictionary2 (Updated) = "+dict2.Count);
}
}输出结果
这将产生以下输出-
HybridDictionary1 elements... A SUV B MUV C AUV Count of Key/value pairs in Dictionary1 = 3 HybridDictionary2 elements... 1 One 2 Two 3 Three 4 Four 5 Five 6 Six Count of Key/value pairs in Dictionary2 = 3 Count of Key/value pairs in Dictionary2 (Updated) = 0
示例
现在让我们来看另一个示例-
using System;
using System.Collections;
using System.Collections.Specialized;
public class Demo {
public static void Main(){
HybridDictionary dict = new HybridDictionary();
dict.Add("A", "SUV");
dict.Add("B", "MUV");
dict.Add("C", "AUV");
Console.WriteLine("HybridDictionary elements...");
foreach(DictionaryEntry d in dict){
Console.WriteLine(d.Key + " " + d.Value);
}
Console.WriteLine("Count of Key/value pairs in Dictionary = "+dict.Count);
dict.Add("D", "Utility Vehicle");
dict.Add("E", "Convertible");
Console.WriteLine("Count of Key/value pairs in Dictionary (Updated) = "+dict.Count);
}
}输出结果
这将产生以下输出-
HybridDictionary elements... A SUV B MUV C AUV Count of Key/value pairs in Dictionary = 3 Count of Key/value pairs in Dictionary (Updated) = 5