编写高质量代码改善C#程序的157个建议——建议57：实现ISerializable的子类型应负责父类的序列化

Posted 2020-10-16

 

建议57：实现ISerializable的子类型应负责父类的序列化

我们将要实现的继承自ISerializable的类型Employee有一个父类Person，假设Person没有实现序列化，而现在子类Employee却需要满足序列化的场景。不过序列化器并没有默认处理Person类型对象，这些事情只能由我们自己做。

以下是一个不妥的实现，序列化器只发现和处理了Employee中Salary字段：

    class Program
    {
        static void Main()
        {
            Employee  liming = new Employee() { Name = "liming", Salary = 2000 };
            BinarySerializer.SerializeToFile(liming, @"c:\", "person.txt");
            Employee limingCopy = BinarySerializer.DeserializeFromFile<Employee>(@"c:\person.txt");
            Console.WriteLine(string.Format("姓名：{0}", limingCopy.Name));
            Console.WriteLine(string.Format("薪水：{0}", limingCopy.Salary));
        }
    }
    
    public class Person
    {
        public string Name { get; set; }
    }

    [Serializable]
    public class Employee : Person, ISerializable
    {
        public int Salary { get; set; }

        public Employee()
        {
        }

        protected Employee(SerializationInfo info, StreamingContext context)
        {
            Salary = info.GetInt32("Salary");
        }

        public override void GetObjectData(SerializationInfo info, StreamingContext context)
        {
            info.AddValue("Salary", Salary);
        }
    }    

序列化工具类：

    public class BinarySerializer
    {
        //将类型序列化为字符串
        public static string Serialize<T>(T t)
        {
            using (MemoryStream stream = new MemoryStream())
            {
                BinaryFormatter formatter = new BinaryFormatter();
                formatter.Serialize(stream, t);
                return System.Text.Encoding.UTF8.GetString(stream.ToArray());
            }
        }

        //将类型序列化为文件
        public static void SerializeToFile<T>(T t, string path, string fullName)
        {
            if (!Directory.Exists(path))
            {
                Directory.CreateDirectory(path);
            }
            string fullPath = Path.Combine(path, fullName);
            using (FileStream stream = new FileStream(fullPath, FileMode.OpenOrCreate))
            {
                BinaryFormatter formatter = new BinaryFormatter();
                formatter.Serialize(stream, t);
                stream.Flush();
            }
        }

        //将字符串反序列化为类型
        public static TResult Deserialize<TResult>(string s) where TResult : class
        {
            byte[] bs = System.Text.Encoding.UTF8.GetBytes(s);
            using (MemoryStream stream = new MemoryStream(bs))
            {
                BinaryFormatter formatter = new BinaryFormatter();
                return formatter.Deserialize(stream) as TResult;
            }
        }

        //将文件反序列化为类型
        public static TResult DeserializeFromFile<TResult>(string path) where TResult : class
        {
            using (FileStream stream = new FileStream(path, FileMode.Open))
            {
                BinaryFormatter formatter = new BinaryFormatter();
                return formatter.Deserialize(stream) as TResult;
            }
        }
    }

View Code

输出为：

姓名：
薪水：2000

看见，Name字段并没有正确处理。这需要我们修改类型Employee中受保护的构造方法GetObjectData方法，为它加入父类字段的处理：

    [Serializable]
    public class Employee : Person, ISerializable
    {
        public int Salary { get; set; }

        public Employee()
        {
        }

        protected Employee(SerializationInfo info, StreamingContext context)
        {
            Name = info.GetString("Name");
            Salary = info.GetInt32("Salary");
        }

        public void GetObjectData(SerializationInfo info, StreamingContext context)
        {
            info.AddValue("Name", Name);
            info.AddValue("Salary", Salary);
        }
    } 

修改后输出：

姓名：liming
薪水：2000

上面的例子中Person类未被设置成支持序列化。现在，假设Person类已经实现了ISerializable接口，那么这个问题处理起来会相对容易，在子类Employee中，我们只需要调用父类受保护的构造方法和GetObjectData方法就可以了。如下所示：

    [Serializable]
    public class Person : ISerializable
    {
        public string Name { get; set; }

        public Person()
        {
        }

        protected Person(SerializationInfo info, StreamingContext context)
        {
            Name = info.GetString("Name");
        }

        public virtual void GetObjectData(SerializationInfo info, StreamingContext context)
        {
            info.AddValue("Name", Name);
        }
    }

    [Serializable]
    public class Employee : Person, ISerializable
    {
        public int Salary { get; set; }

        public Employee()
        {
        }

        protected Employee(SerializationInfo info, StreamingContext context)
            : base(info, context)
        {
            Salary = info.GetInt32("Salary");
        }

        public override void GetObjectData(SerializationInfo info, StreamingContext context)
        {
            base.GetObjectData(info, context);
            info.AddValue("Salary", Salary);
        }
    }

 

 

 

 

转自：《编写高质量代码改善C#程序的157个建议》陆敏技