没有重复的 C# 不可变和可变类型

Question

鉴于以下可变和不可变类型的实现，有没有办法避免重复代码(主要是重复属性)？

我希望默认使用不可变类型，除非需要可变类型(例如，绑定(bind)到 UI 元素时)。

我们正在使用 .NET Framework 4.0，但计划很快切换到 4.5。

public class Person {
    public string Name { get; private set; }
    public List<string> Jobs { get; private set; } // Change to ReadOnlyList<T>
    public Person() {}
    public Person(Mutable m) {
        Name = m.Name;
    }
    public class Mutable : INotifyPropertyChanged {
        public string Name { get; set; }
        public List<string> Jobs { get; set; }
        public Mutable() {
            Jobs = new List<string>();
        }
        public Mutable(Person p) {
            Name = p.Name;
            Jobs = new List<string>(p.Jobs);
        }
        public event PropertyChangedEventHandler PropertyChanged;
        protected void OnPropertyChanged(string propertyName) {
            // TODO: implement
        }
    }
}

public class Consumer {
    public Consumer() {
        // We can use object initializers :)
        Person.Mutable m = new Person.Mutable {
            Name = "M. Utable"
        };
        // Consumers can happily mutate away....
        m.Name = "M. Utated";
        m.Jobs.Add("Herper");
        m.Jobs.Add("Derper");

        // But the core of our app only deals with "realio-trulio" immutable types.

        // Yey! Have constructor with arity of one as opposed to
        // new Person(firstName, lastName, email, address, im, phone)
        Person im = new Person(m);
    }
}

Answer 1

我做了一些东西来满足你最近的要求(使用 T4 模板)，所以这是绝对可能的:

https://github.com/xaviergonz/T4Immutable

例如，给定这个:

[ImmutableClass(Options = ImmutableClassOptions.IncludeOperatorEquals)]
class Person {
  private const int AgeDefaultValue = 18;

  public string FirstName { get; }
  public string LastName { get; }
  public int Age { get; }

  [ComputedProperty]
  public string FullName {
    get {
      return FirstName + " " + LastName;
    }
  }
}

它会在单独的部分类文件中自动为您生成以下内容:

• 一个构造函数，例如 public Person(string firstName, stringlastName, int age = 18) 将初始化值。
• Equals(object other) 和 Equals(Person other) 的有效实现。它还会为您添加 IEquatable 接口(interface)。在职的operator== 和 operator!= 的实现
• GetHashCode() 的有效实现更好的 ToString() 输出如“Person { FirstName=John, LastName=Doe, Age=21 }”
• Person With(...) 方法，可用于生成具有 0 个或多个属性更改的新不可变克隆(例如 var janeDoe = johnDoe.With(firstName: "Jane", age: 20)

所以它会生成这个(排除一些冗余属性):

using System;

partial class Person : IEquatable<Person> {
  public Person(string firstName, string lastName, int age = 18) {
    this.FirstName = firstName;
    this.LastName = lastName;
    this.Age = age;
    _ImmutableHashCode = new { this.FirstName, this.LastName, this.Age }.GetHashCode();
  }

  private bool ImmutableEquals(Person obj) {
    if (ReferenceEquals(this, obj)) return true;
    if (ReferenceEquals(obj, null)) return false;
    return T4Immutable.Helpers.AreEqual(this.FirstName, obj.FirstName) && T4Immutable.Helpers.AreEqual(this.LastName, obj.LastName) && T4Immutable.Helpers.AreEqual(this.Age, obj.Age);
  }

  public override bool Equals(object obj) {
    return ImmutableEquals(obj as Person);
  }

  public bool Equals(Person obj) {
    return ImmutableEquals(obj);
  }

  public static bool operator ==(Person a, Person b) {
    return T4Immutable.Helpers.AreEqual(a, b);
  }

  public static bool operator !=(Person a, Person b) {
    return !T4Immutable.Helpers.AreEqual(a, b);
  }

  private readonly int _ImmutableHashCode;

  private int ImmutableGetHashCode() {
    return _ImmutableHashCode;
  }

  public override int GetHashCode() {
    return ImmutableGetHashCode();
  }

  private string ImmutableToString() {
    var sb = new System.Text.StringBuilder();
    sb.Append(nameof(Person) + " { ");

    var values = new string[] {
      nameof(this.FirstName) + "=" + T4Immutable.Helpers.ToString(this.FirstName),
      nameof(this.LastName) + "=" + T4Immutable.Helpers.ToString(this.LastName),
      nameof(this.Age) + "=" + T4Immutable.Helpers.ToString(this.Age),
    };

    sb.Append(string.Join(", ", values) + " }");
    return sb.ToString();
  }

  public override string ToString() {
    return ImmutableToString();
  }

  private Person ImmutableWith(T4Immutable.WithParam<string> firstName = default(T4Immutable.WithParam<string>), T4Immutable.WithParam<string> lastName = default(T4Immutable.WithParam<string>), T4Immutable.WithParam<int> age = default(T4Immutable.WithParam<int>)) {
    return new Person(
      !firstName.HasValue ? this.FirstName : firstName.Value,
      !lastName.HasValue ? this.LastName : lastName.Value,
      !age.HasValue ? this.Age : age.Value
    );
  }

  public Person With(T4Immutable.WithParam<string> firstName = default(T4Immutable.WithParam<string>), T4Immutable.WithParam<string> lastName = default(T4Immutable.WithParam<string>), T4Immutable.WithParam<int> age = default(T4Immutable.WithParam<int>)) {
    return ImmutableWith(firstName, lastName, age);
  }

}

项目页面中还介绍了一些其他功能。

PS:如果你想要一个包含其他不可变对象(immutable对象)列表的属性，只需添加:

public ImmutableList<string> Jobs { get; }