相等性和多态性

Question

有两个不可变的类 Base 和 Derived（派生自 Base）我想定义 Equality 以便

等式总是多态的——也就是说((Base)derived1).Equals((Base)derived2)会调用Derived.Equals
运算符==和!=将调用Equals而不是ReferenceEquals（值相等）

我做了什么：

class Base: IEquatable<Base> {
  public readonly ImmutableType1 X;
  readonly ImmutableType2 Y;

  public Base(ImmutableType1 X, ImmutableType2 Y) { 
    this.X = X; 
    this.Y = Y; 
  }

  public override bool Equals(object obj) {
    if (object.ReferenceEquals(this, obj)) return true;
    if (obj is null || obj.GetType()!=this.GetType()) return false;

    return obj is Base o 
      && X.Equals(o.X) && Y.Equals(o.Y);
  }

  public override int GetHashCode() => HashCode.Combine(X, Y);

  // boilerplate
  public bool Equals(Base o) => object.Equals(this, o);
  public static bool operator ==(Base o1, Base o2) => object.Equals(o1, o2);
  public static bool operator !=(Base o1, Base o2) => !object.Equals(o1, o2);    }

这里一切都以 Equals(object) 结束，它总是多态的，所以两个目标都实现了。

然后我这样推导：

class Derived : Base, IEquatable<Derived> {
  public readonly ImmutableType3 Z;
  readonly ImmutableType4 K;

  public Derived(ImmutableType1 X, ImmutableType2 Y, ImmutableType3 Z, ImmutableType4 K) : base(X, Y) {
    this.Z = Z; 
    this.K = K; 
  }

  public override bool Equals(object obj) {
    if (object.ReferenceEquals(this, obj)) return true;
    if (obj is null || obj.GetType()!=this.GetType()) return false;

    return obj is Derived o
      && base.Equals(obj) /* ! */
      && Z.Equals(o.Z) && K.Equals(o.K);
  }

  public override int GetHashCode() => HashCode.Combine(base.GetHashCode(), Z, K);

  // boilerplate
  public bool Equals(Derived o) => object.Equals(this, o);
}

这基本上是一样的，除了一个陷阱 - 当调用 base.Equals 时，我调用 base.Equals(object) 而不是 base.Equals(Derived)（这将导致无休止的递归）。

另外 Equals(C) 将在此实现中做一些 boxing/unboxing 但这对我来说是值得的。

我的问题是 -

首先这是正确的吗？我的 (testing) 似乎暗示它是这样，但是由于 C# 在平等方面如此困难，我只是不确定了..在任何情况下这是错误的吗？

其次 - 这样好吗？有没有更好更清洁的方法来实现这一目标？

Answer 1

嗯，我想你的问题有两个部分：

在嵌套级别执行 equals
限制为同一类型

这行得通吗？ https://dotnetfiddle.net/eVLiMZ （我不得不使用一些较旧的语法，否则它不会在 dotnetfiddle 中编译）

using System;


public class Program
{
    public class Base
    {
        public string Name { get; set; }
        public string VarName { get; set; }

        public override bool Equals(object o)
        {
            return object.ReferenceEquals(this, o) 
                || o.GetType()==this.GetType() && ThisEquals(o);
        }

        protected virtual bool ThisEquals(object o)
        {
            Base b = o as Base;
            return b != null
                && (Name == b.Name);
        }

        public override string ToString()
        {
            return string.Format("[{0}@{1} Name:{2}]", GetType(), VarName, Name);
        }

        public override int GetHashCode()
        {
            return Name.GetHashCode();
        }
    }

    public class Derived : Base
    {
        public int Age { get; set; }

        protected override bool ThisEquals(object o)
        {
            var d = o as Derived;
            return base.ThisEquals(o)
                && d != null
                && (d.Age == Age);
        }

        public override string ToString()
        {
            return string.Format("[{0}@{1} Name:{2} Age:{3}]", GetType(), VarName, Name, Age);
        }

        public override int GetHashCode()
        {
            return base.GetHashCode() ^ Age.GetHashCode();
        }
    }

    public static void Main()
    {
        var b1 = new Base { Name = "anna", VarName = "b1" };
        var b2 = new Base { Name = "leo", VarName = "b2" };
        var b3 = new Base { Name = "anna", VarName = "b3" };
        var d1 = new Derived { Name = "anna", Age = 21, VarName = "d1" };
        var d2 = new Derived { Name = "anna", Age = 12, VarName = "d2" };
        var d3 = new Derived { Name = "anna", Age = 21, VarName = "d3" };

        var all = new object [] { b1, b2, b3, d1, d2, d3 };

        foreach(var a in all) 
        {
            foreach(var b in all)
            {
                Console.WriteLine("{0}.Equals({1}) => {2}", a, b, a.Equals(b));
            }
        }
    }
}

Answer 2

可以结合使用扩展方法和一些样板代码来简化代码。这消除了几乎所有的痛苦，让类专注于比较它们的实例，而不必处理所有特殊的边缘情况：

namespace System {
  public static partial class ExtensionMethods {
    public static bool Equals<T>(this T inst, object obj, Func<T, bool> thisEquals) where T : IEquatable<T> =>
      object.ReferenceEquals(inst, obj) // same reference ->  equal
      || !(obj is null) // this is not null but obj is -> not equal
      && obj.GetType() == inst.GetType() // obj is more derived than this -> not equal
      && obj is T o // obj cannot be cast to this type -> not equal
      && thisEquals(o);
  }
}

我现在可以做到：

class Base : IEquatable<Base> {
    public SomeType1 X;
    SomeType2 Y;
    public Base(SomeType1 X, SomeType2 Y) => (this.X, this.Y) = (X, Y);

    public bool ThisEquals(Base o) => (X, Y) == (o.X, o.Y);

    // boilerplate
    public override bool Equals(object obj) => this.Equals(obj, ThisEquals);
    public bool Equals(Base o) => object.Equals(this, o);
    public static bool operator ==(Base o1, Base o2) => object.Equals(o1, o2);
    public static bool operator !=(Base o1, Base o2) => !object.Equals(o1, o2);
}


class Derived : Base, IEquatable<Derived> {
    public SomeType3 Z;
    SomeType4 K;
    public Derived(SomeType1 X, SomeType2 Y, SomeType3 Z, SomeType4 K) : base(X, Y) => (this.Z, this.K) = (Z, K);

    public bool ThisEquals(Derived o) => base.ThisEquals(o) && (Z, K) == (o.Z, o.K);

    // boilerplate
    public override bool Equals(object obj) => this.Equals(obj, ThisEquals);
    public bool Equals(Derived o) => object.Equals(this, o);
}

这很好，没有转换或 null 检查，所有实际工作在 ThisEquals 中清楚地分开了。
(testing)

对于不可变类，可以通过缓存哈希码并在 Equals 中使用它来进一步优化，如果哈希码不同，则可以实现快捷方式相等：

namespace System.Immutable {
  public interface IImmutableEquatable<T> : IEquatable<T> { };

  public static partial class ExtensionMethods {
    public static bool ImmutableEquals<T>(this T inst, object obj, Func<T, bool> thisEquals) where T : IImmutableEquatable<T> =>
      object.ReferenceEquals(inst, obj) // same reference ->  equal
      || !(obj is null) // this is not null but obj is -> not equal
      && obj.GetType() == inst.GetType() // obj is more derived than this -> not equal
      && inst.GetHashCode() == obj.GetHashCode() // optimization, hash codes are different -> not equal
      && obj is T o // obj cannot be cast to this type -> not equal
      && thisEquals(o);

    public static int GetHashCode<T>(this T inst, ref int? hashCache, Func<int> thisHashCode) where T : IImmutableEquatable<T> {
      if (hashCache is null) hashCache = thisHashCode();
      return hashCache.Value;
    }
  }
}

我现在可以做到：

class Base : IImmutableEquatable<Base> {
    public readonly SomeImmutableType1 X;
    readonly SomeImmutableType2 Y;
    public Base(SomeImmutableType1 X, SomeImmutableType2 Y) => (this.X, this.Y) = (X, Y);

    public bool ThisEquals(Base o) => (X, Y) == (o.X, o.Y);
    public int ThisHashCode() => (X, Y).GetHashCode();


    // boilerplate
    public override bool Equals(object obj) => this.ImmutableEquals(obj, ThisEquals);
    public bool Equals(Base o) => object.Equals(this, o);
    public static bool operator ==(Base o1, Base o2) => object.Equals(o1, o2);
    public static bool operator !=(Base o1, Base o2) => !object.Equals(o1, o2);
    protected int? hashCache;
    public override int GetHashCode() => this.GetHashCode(ref hashCache, ThisHashCode);
}


class Derived : Base, IImmutableEquatable<Derived> {
    public readonly SomeImmutableType3 Z;
    readonly SomeImmutableType4 K;
    public Derived(SomeImmutableType1 X, SomeImmutableType2 Y, SomeImmutableType3 Z, SomeImmutableType4 K) : base(X, Y) => (this.Z, this.K) = (Z, K);

    public bool ThisEquals(Derived o) => base.ThisEquals(o) && (Z, K) == (o.Z, o.K);
    public new int ThisHashCode() => (base.ThisHashCode(), Z, K).GetHashCode();


    // boilerplate
    public override bool Equals(object obj) => this.ImmutableEquals(obj, ThisEquals);
    public bool Equals(Derived o) => object.Equals(this, o);
    public override int GetHashCode() => this.GetHashCode(ref hashCache, ThisHashCode);
}

这还不错 - 有更多的复杂性，但它只是样板，我只是剪切和粘贴..逻辑在 ThisEquals 和 ThisHashCode

中明显分开

(testing)

Answer 3

这种使用Reflection的比较方法，比扩展方法更简单。它还使私有成员保持私有。

所有逻辑都在 IImmutableExtensions class 中。它只是查看哪些字段是只读的并将它们用于比较。

您不需要基类或派生类 class 中的方法来进行对象比较。覆盖 ==、!= 和 Equals() 时，只需调用扩展方法 ImmutableEquals。与哈希码相同。

public class Base : IEquatable<Base>, IImmutable
{
    public readonly ImmutableType1 X;
    readonly ImmutableType2 Y;

    public Base(ImmutableType1 X, ImmutableType2 Y) => (this.X, this.Y) = (X, Y);

    // boilerplate
    public override bool Equals(object obj) => this.ImmutableEquals(obj);
    public bool Equals(Base o) => this.ImmutableEquals(o);
    public static bool operator ==(Base o1, Base o2) => o1.ImmutableEquals(o2);
    public static bool operator !=(Base o1, Base o2) => !o1.ImmutableEquals(o2);
    private int? _hashCache;
    public override int GetHashCode() => this.ImmutableHash(ref _hashCache);
}

public class Derived : Base, IEquatable<Derived>, IImmutable
{
    public readonly ImmutableType3 Z;
    readonly ImmutableType4 K;

    public Derived(ImmutableType1 X, ImmutableType2 Y, ImmutableType3 Z, ImmutableType4 K) : base(X, Y) => (this.Z, this.K) = (Z, K);

    public bool Equals(Derived other) => this.ImmutableEquals(other);
}

和 IImmutableExtensions class:

public static class IImmutableExtensions
{
    public static bool ImmutableEquals(this IImmutable o1, object o2)
    {
        if (ReferenceEquals(o1, o2)) return true;
        if (o2 is null || o1.GetType() != o2.GetType() || o1.GetHashCode() != o2.GetHashCode()) return false;

        foreach (var tProp in GetImmutableFields(o1))
        {
            var test = tProp.GetValue(o1)?.Equals(tProp.GetValue(o2));
            if (test is null) continue;
            if (!test.Value) return false;
        }
        return true;
    }

    public static int ImmutableHash(this IImmutable o, ref int? hashCache)
    {
        if (hashCache is null)
        {
            hashCache = 0;

            foreach (var tProp in GetImmutableFields(o))
            {
                hashCache = HashCode.Combine(hashCache.Value, tProp.GetValue(o).GetHashCode());
            }
        }
        return hashCache.Value;
    }

    private static IEnumerable<FieldInfo> GetImmutableFields(object o)
    {
        var t = o.GetType();
        do
        {
            var fields = t.GetFields(BindingFlags.DeclaredOnly | BindingFlags.Instance | BindingFlags.NonPublic | BindingFlags.Public).Where(field => field.IsInitOnly);

            foreach(var field in fields)
            {
                yield return field;
            }
        }
        while ((t = t.BaseType) != typeof(object));
    }
}

旧答案：（留作参考）

根据你所说的必须强制转换为 object，我想到从派生的 class.

这对我说，逻辑应该从两个 classes 中移出，转移到一种能更好地描述我们意图的方法。

等式将保持多态性，因为基 class 中的 ImmutableEquals 将调用重写的 ValuesEqual。您可以在此处决定每个派生 class 如何比较相等性。

这是您按照该目标重构的代码。

修改后的答案：

我突然想到，如果我们简单地提供一个包含我们想要比较的不可变字段的元组，我们在 IsEqual() 和 GetHashCode() 中的所有逻辑都会起作用。这避免了在每个 class.

中重复这么多代码

由创建派生 class 的开发人员来覆盖 GetImmutableTuple()。不使用反射（见其他答案），我觉得这是所有罪恶中最少的。

public class Base : IEquatable<Base>, IImmutable
{
    public readonly ImmutableType1 X;
    readonly ImmutableType2 Y;

    public Base(ImmutableType1 X, ImmutableType2 Y) => 
      (this.X, this.Y) = (X, Y);

    protected virtual IStructuralEquatable GetImmutableTuple() => (X, Y);

    // boilerplate
    public override bool Equals(object o) => IsEqual(o as Base);
    public bool Equals(Base o) => IsEqual(o);
    public static bool operator ==(Base o1, Base o2) => o1.IsEqual(o2);
    public static bool operator !=(Base o1, Base o2) => !o1.IsEqual(o2);
    public override int GetHashCode() => hashCache is null ? (hashCache = GetImmutableTuple().GetHashCode()).Value : hashCache.Value;
    protected bool IsEqual(Base obj) => ReferenceEquals(this, obj) || !(obj is null) && GetType() == obj.GetType() && GetHashCode() == obj.GetHashCode() && GetImmutableTuple() != obj.GetImmutableTuple();
    protected int? hashCache;
}

public class Derived : Base, IEquatable<Derived>, IImmutable
{
    public readonly ImmutableType3 Z;
    readonly ImmutableType4 K;

    public Derived(ImmutableType1 X, ImmutableType2 Y, ImmutableType3 Z, ImmutableType4 K) : base(X, Y) => 
      (this.Z, this.K) = (Z, K);

    protected override IStructuralEquatable GetImmutableTuple() => (base.GetImmutableTuple(), K, Z);

    // boilerplate
    public bool Equals(Derived o) => IsEqual(o);
}

Answer 4

另一种方法是使用反射来自动比较所有字段和属性。您只需用 Immutable 属性装饰它们，AutoCompare() 会处理其余的事情。

这也会根据你Immutable装饰的字段和属性使用Reflection构建一个HashCode，然后缓存它来优化对象比较。

public class Base : ComparableImmutable, IEquatable<Base>, IImmutable
{
    [Immutable]
    public ImmutableType1 X { get; set; }

    [Immutable]
    readonly ImmutableType2 Y;

    public Base(ImmutableType1 X, ImmutableType2 Y) => (this.X, this.Y) = (X, Y);

    public bool Equals(Base o) => AutoCompare(o);
}

public class Derived : Base, IEquatable<Derived>, IImmutable
{
    [Immutable]
    public readonly ImmutableType3 Z;

    [Immutable]
    readonly ImmutableType4 K;

    public Derived(ImmutableType1 X, ImmutableType2 Y, ImmutableType3 Z, ImmutableType4 K)
        : base(X, Y)
        => (this.Z, this.K) = (Z, K);

    public bool Equals(Derived o) => AutoCompare(o);
}

[AttributeUsage(validOn: AttributeTargets.Field | AttributeTargets.Property)]
public class ImmutableAttribute : Attribute { }

public abstract class ComparableImmutable
{
    static BindingFlags flags = BindingFlags.Public | BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.DeclaredOnly;

    protected int? hashCache;

    public override int GetHashCode()
    {
        if (hashCache is null)
        {
            hashCache = 0;
            var type = GetType();

            do
            {
                foreach (var field in type.GetFields(flags).Where(field => Attribute.IsDefined(field, typeof(ImmutableAttribute))))
                    hashCache = HashCode.Combine(hashCache, field.GetValue(this));

                foreach (var property in type.GetProperties(flags).Where(property => Attribute.IsDefined(property, typeof(ImmutableAttribute))))
                    hashCache = HashCode.Combine(hashCache, property.GetValue(this));

                type = type.BaseType;
            }
            while (type != null);
        }

        return hashCache.Value;
    }

    protected bool AutoCompare(object obj2)
    {
        if (ReferenceEquals(this, obj2)) return true;

        if (obj2 is null
            || GetType() != obj2.GetType()
            || GetHashCode() != obj2.GetHashCode())
            return false;

        var type = GetType();

        do
        {
            foreach (var field in type.GetFields(flags).Where(field => Attribute.IsDefined(field, typeof(ImmutableAttribute))))
            {
                if (field.GetValue(this) != field.GetValue(obj2))
                {
                    return false;
                }
            }

            foreach (var property in type.GetProperties(flags).Where(property => Attribute.IsDefined(property, typeof(ImmutableAttribute))))
            {
                if (property.GetValue(this) != property.GetValue(obj2))
                {
                    return false;
                }
            }

            type = type.BaseType;
        }
        while (type != null);

        return true;
    }

    public override bool Equals(object o) => AutoCompare(o);
    public static bool operator ==(Comparable o1, Comparable o2) => o1.AutoCompare(o2);
    public static bool operator !=(Comparable o1, Comparable o2) => !o1.AutoCompare(o2);
}

相等性和多态性

Equality and polymorphism

c#

c#-7.0

c#-7.3