C＃的相当于Java的通配符 [英] C#'s equivalent of Java's wildcard

问题描述

如果它存在，什么是C＃相当于下面的Java代码：

If it exists, what is the C# equivalent of the following Java code:

new HashMap<Class<? extends BaseClass>, Integer>();

我目前使用新字典<类型，INT>（），这更像是
新的HashMap<类<>中整数方式>（）这显然是不一样的。

I currently use new Dictionary<Type, int>(), which is more like
new HashMap<Class<?>, Integer>() which is obviously not the same.

的（忽略的HashMap 之间的差异和词典）

(Ignore the differences between HashMap and Dictionary)

修改：要澄清，我不是试图定义一个新的类，只需创建一个实例的HashMap / 词典。

To clarify, I am not trying to define a new class, simply create an instance of HashMap/Dictionary.

推荐答案

没有在C＃中的Java通配符的当量。在Java中，类型的类型是类< T> ，其中 T 是类本身。在C＃中的等价物的类型是键入，其中的不是泛型的。如此看来，你能做的最好是有，如你所说，一个词典<类型，INT> ，如果它封装在一个类中，你可以限制你摆在字典中的代码（因此它只是一个运行时检查）：

There is no equivalent of the Java wildcard in C#. In Java, the type for types is Class<T> where T is the class itself. The equivalent in C# is the type Type, which is not generic. So it seems that the best you can do is to have, as you said, a Dictionary<Type, int>, and if it's encapsulated in a class you can restrict what you put in the dictionary in the code (so it will just be a runtime check):

private Dictionary<Type, int> myDictionary = new Dictionary<Type, int>();
public void Add(Type type, int number) {
   if (!typeof(BaseClass).IsAssignableFrom(type)) throw new Exception();
   myDictionary.Add(type, number);
}

您甚至可以实现自己的 的IDictionary 与逻辑。

You can even implement your own IDictionary with that logic.

更新

另外的运行的把戏我能想到的是使用包装类的类型：

Another runtime trick I can think of is to use a wrapper class for your types:

public class TypeWrapper<T>
{
    public Type Type { get; private set; }
    public TypeWrapper(Type t)
    {
        if (!typeof(T).IsAssignableFrom(t)) throw new Exception();
        Type = t;
    }
    public static implicit operator TypeWrapper<T>(Type t) {
        return new TypeWrapper<T>(t);
    }
}

（也实现等于和的GetHashCode ，只委托给键入）

然后你的字典里就变成了：

And then your dictionary becomes:

var d = new Dictionary<TypeWrapper<BaseClass>, int>();
d.Add(typeof(BaseClass), 2);
d.Add(typeof(Child), 3);

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