如何在Type变量中创建具有任何类型的Dictionary？ [英] How to create a Dictionary with any types in Type variables?

问题描述

当前我正面临这个问题，我需要创建一个包含任何类型案例的Dictionary实例。类型由方法的参数传递。
我不只是想创建一个动态或对象类型的Dictionary，因为使用我的库会给用户带来很多转换问题。
我也不能使用简单的构造函数，因为我的方法用数据（存储在文件中）实际填充字典。通过Type变量创建特定的词典很重要。
这就是我的想法：

Currently im facing the problem I need to create an instance of Dictionary with any type of its cases. The types are delivered by the arguments of the method. I do not simply want to create a dynamic or object type Dictionary, because this faces the user alot of convertion problems by using my library. I also cannot use a simple constructor, because my method fills the Dictionary with data (that is stored in a file) in real. It's important to create the specific dictionary by Type variables. This is what im thinking of:

public static Dictionary<dynamic, dynamic> createDict(Type type1, Type type2)
{
    // how to create the dictionary?
    // im filling my dictionary with any data ...
    return dict;
}

用户在这里调用我的方法：

Here the user calls my method:

Dictionary<string, int> dict = MyLib.createDict(typeof(string), typeof(int));
// or
MyOwnType myInstance = new MyOwnType();
Dictionary<string, MyOwnType> dict = MyLib.createDict(typeof(string), myInstance.GetType());

推荐答案

Type dictType = typeof(Dictionary<, >).MakeGenericType(Type1, Type2);
var dict = Activator.CreateInstance(dictType);

• 获取词典的类型


• 使用 MakeGenericType


• 创建具有指定类型的通用类型使用 Activator.CreateInstance


• Get the Type for Dictionary
• Create a generic type with the specified Types using MakeGenericType
• Create an instance of the generic type using Activator.CreateInstance

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