如何最好地表达这种用于序列化/反序列化类型的扩展方法

How to best express this extension method for serializing / deserializing types

我有以下两种扩展方法:

public static object DeserializeFromXmlFile(this Type type, string FileName)
{
        try
        {
            using (TextReader reader = new StreamReader(FileName))
            {
                XmlSerializer ser = new XmlSerializer(type);
                return ser.Deserialize(reader);
            }
        }
        catch (Exception ex)
        {
            return null;
        }
}

public static bool SerializeToXmlFile(this object obj, string FileName)
{
        try
        {
            using (TextWriter writer = new StreamWriter(FileName))
            {
                XmlSerializer ser = new XmlSerializer(obj.GetType());
                ser.Serialize(writer, obj);
            }
            return true;
        }
        catch (Exception ex)
        {
            return false;
        }
}

然后是使用它们的代码:

SaveToFile(string LogFile, MyClass c)
{
    c.SerializeToXmlFile(LogFile);
}

MyClass LoadFromFile(string LogFile)
{
    return (MyClass)typeof(MyClass).(LogFile);
}

我不喜欢 LoadFromFile 使用 Deserialize 方法的方式,我可以让它像 Serialize 用法一样好(在 SaveToFile() 中)吗?

我不确定 Type 的扩展方法是否适用于此处。你不是真正的 "extending" Type class。您可以改为通用:

public static T DeserializeFromXmlFile<T>(string FileName) where T:class
{
    try
    {
        using (TextReader reader = new StreamReader(FileName))
        {
            XmlSerializer ser = new XmlSerializer(typeof(T));
            return ser.Deserialize(reader) as T;
        }
    }
    catch (Exception ex)
    {
        return default(T); // is this really the right approach?  Just ignore the error and silently return null?
    }
}

那么用法就是

MyClass LoadFromFile(string LogFile)
{
    return DeserializeFromXmlFile<MyClass>.(LogFile);

    // or return {classname}.DeserializeFromXmlFile<MyClass>.(LogFile);
    // if the static method is defined in a different class.
}

请注意,您必须在调用中包含 <MyClass>,因为编译器无法使用 return 类型推断通用类型参数。

你只需要简化界面,使两个调用的签名相同即可。这将使代码更加清晰。

MyClass LoadFromFile(string LogFile, MyClass c)
{
    return (MyClass) c.DeserializeFromXmlFile(LogFile);            
}

并获取执行必要烟花的方法而不是调用者。

public static object DeserializeFromXmlFile(this object obj, string FileName)
{
    try
    {
        using (TextReader reader = new StreamReader(FileName))
        {
            XmlSerializer ser = new XmlSerializer(obj.GetType());
            return ser.Deserialize(reader);
        }
    }
    catch (Exception ex)
    {
        return null;
    }
}