C# 3.0：在集合中填充具有不同行为的对象

发布于 2024-07-16 13:54:39 字数 1016 浏览 3 评论 0原文

我想填充组合框中的项目，每个项目都有不同的行为。是的，我知道我可以简单地创建 3 个从基类派生的类。但我的问题是“还有其他方法吗”和“什么是可能的”。在 Java 中，我们可以执行“new MyClass(){public void overriddenmethod(){...} }”，但在 C# 中我们不能，不是吗？

现在我使用 lambda 动态定义一个方法，但问题是我稍后想要 new XxxFormatter() 作为该对象的实例变量。由于 XxxFormatters 不共享公共基类，因此我无法将它们作为 SerializingHelper 类中的单个字段。

你有什么想法？

public delegate void SerializingHandler(Stream s, object o);

class SerializingHelper
{
    public string Name { get; set; }
    public SerializingHandler Serializer { get; set; }
}

comboFormat.Items.AddRange(new object[] 
{ 
            new SerializingHelper{ Name = "Binary", 
Serializer = (s,o)=>new BinaryFormatter().Serialize(s,o),

            new SerializingHelper{ Name = "Soap", 
Serializer = (s,o)=>new SoapFormatter().Serialize(s,o), 

            new SerializingHelper{ Name = "Xml", 
Serializer = (s,o)=>new XmlSerializer(typeof(KontaktpartnerData), new Type[]   
 {typeof(ArrayList), typeof(KontaktPartner)}).Serialize(s,o), }

});

原文

I want to fill items in a combobox, each of them has different behaviour.
Yes I know I could simply create 3 classes deriving from a base class. But my question is kind of "is there another way" and "what is possible".
In Java one can do "new MyClass(){public void overriddenmethod(){...} }" but in C# we can not, can we?

Now I use a lambda to define a method on the fly but the problem is that I later want the new XxxFormatter() as instance variable of that object. Since the XxxFormatters share no common base class I cannot put them as a single field in the SerializingHelper class.

Do you have any Ideas?

public delegate void SerializingHandler(Stream s, object o);

class SerializingHelper
{
    public string Name { get; set; }
    public SerializingHandler Serializer { get; set; }
}

comboFormat.Items.AddRange(new object[] 
{ 
            new SerializingHelper{ Name = "Binary", 
Serializer = (s,o)=>new BinaryFormatter().Serialize(s,o),

            new SerializingHelper{ Name = "Soap", 
Serializer = (s,o)=>new SoapFormatter().Serialize(s,o), 

            new SerializingHelper{ Name = "Xml", 
Serializer = (s,o)=>new XmlSerializer(typeof(KontaktpartnerData), new Type[]   
 {typeof(ArrayList), typeof(KontaktPartner)}).Serialize(s,o), }

});

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相思故 2024-07-23 13:54:39

如果您只是想防止每次实例化新的序列化程序实例，您可以在 lambda 之外实例化它们：

var binaryFormatter = new BinaryFormatter();

comboFormat.Items.AddRange(new object[]
{ 
    new SerializingHelper
    {
        Name = "Binary",
        Serializer = binaryFormatter.Serialize
    }

    ...
});

如果您确实需要将格式化程序存储为字段，您可以执行以下操作：

delegate void SerializingHandler<TFormatter>(TFormatter formatter,
                                             Stream stream,
                                             object graph);

interface ISerializingHelper
{
    void Serialize(Stream stream, object graph);
}

class SerializingHelper<TFormatter> : ISerializingHelper
{
    private readonly SerializingHandler<TFormatter> handler;
    private readonly TFormatter formatter;

    public SerializingHelper(SerializingHandler<TFormatter> handler,
                             TFormatter formatter)
    {
        this.handler = handler;
        this.formatter = formatter;
    }

    public TFormatter Formatter
    {
        get { return this.formatter; }
    }

    public void Serialize(Stream stream, object graph)
    {
        this.handler(this.formatter, stream, graph);
    }
}

If you just want to prevent instantiating a new serializer instance each time, you can instantiate them outside of the lambda:

var binaryFormatter = new BinaryFormatter();

comboFormat.Items.AddRange(new object[]
{ 
    new SerializingHelper
    {
        Name = "Binary",
        Serializer = binaryFormatter.Serialize
    }

    ...
});

If you really need to store the formatter as a field, you could do something like this:

delegate void SerializingHandler<TFormatter>(TFormatter formatter,
                                             Stream stream,
                                             object graph);

interface ISerializingHelper
{
    void Serialize(Stream stream, object graph);
}

class SerializingHelper<TFormatter> : ISerializingHelper
{
    private readonly SerializingHandler<TFormatter> handler;
    private readonly TFormatter formatter;

    public SerializingHelper(SerializingHandler<TFormatter> handler,
                             TFormatter formatter)
    {
        this.handler = handler;
        this.formatter = formatter;
    }

    public TFormatter Formatter
    {
        get { return this.formatter; }
    }

    public void Serialize(Stream stream, object graph)
    {
        this.handler(this.formatter, stream, graph);
    }
}

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