基于返回类型重载方法

Question

全部 - 我的一个课程中有这两种方法:

public static void DeSerializeFromXml(out cDBData db, string pathToXML)
{
    using (XmlTextReader reader = new XmlTextReader(pathToXML))
    {
        XmlSerializer ser = new XmlSerializer(typeof(cDBData));
        db = (cDBData)ser.Deserialize(reader);
    }
}

public static void DeSerializeFromXml(out cDatabases db, string pathToXML)
{
    using (XmlTextReader reader = new XmlTextReader(pathToXML))
    {
        XmlSerializer ser = new XmlSerializer(typeof(cDatabases));
        db = (cDatabases)ser.Deserialize(reader);
    }
}

它们工作正常,但我想知道为什么我不能根据方法的返回类型创建方法重载.我以为我读到这可能在某个地方,但我显然是错的,因为它不起作用:

public static cDBData DeSerializeFromXml(string pathToXML)
{
    cDBData db;

    using (XmlTextReader reader = new XmlTextReader(pathToXML))
    {
        XmlSerializer ser = new XmlSerializer(typeof(cDBData));
        db = (cDBData)ser.Deserialize(reader);
    }
    return db;
}

public static cDatabases DeSerializeFromXml(string pathToXML)
{
    cDatabases db;

    using (XmlTextReader reader = new XmlTextReader(pathToXML))
    {
        XmlSerializer ser = new XmlSerializer(typeof(cDatabases));
        db = (cDatabases)ser.Deserialize(reader);
    }
    return db;
}

感谢您的深思熟虑的回复

感谢dlev​​,这是最终的解决方案

    public static T DeSerializeFromXml<T>(string pathToXML)
    {
        T db;

        using (XmlTextReader reader = new XmlTextReader(pathToXML))
        {
            XmlSerializer ser = new XmlSerializer(typeof(T));
            db = (T)ser.Deserialize(reader);
        }
        return db;
    }

    public static void SerializeToXml<T>(T db, string pathToXML)
    {
        using (var fileStream = new FileStream(pathToXML, FileMode.Create))
        {
            var ser = new XmlSerializer(typeof(T));
            ser.Serialize(fileStream, db);
        }
    }

我想在搜索警察关闭这个问题之前发布这些内容.

Answer 1

你不能这样做,因为C#根本不支持那种重载.您可以通过使方法通用并返回泛型参数类型来实现类似的效果:

public static T DeSerializeFromXml<T>(string pathToXML)
{
    T db;

    using (XmlTextReader reader = new XmlTextReader(pathToXML))
    {
        XmlSerializer ser = new XmlSerializer(typeof(T));
        db = (T)ser.Deserialize(reader);
    }
    return db;
}

然后你可以调用这样的函数:

DeSerializeFromXml<cDatabases>(pathToXml);

虽然这不是你想要的,但它确实有一个好处,只需要一个方法.