c# 多键字典，"KeyNotFoundException" 用自己的可变 class 替换元组时

Question

我声明了一个可变的 class 来替代元组作为字典的键。原因是序列化。序列化工作得很好。出现以下问题，我在使用此 class 时得到 "KeyNotFoundException"，但前提是此 class 的新实例用于查找。为了更清楚地说明这一点，请参阅以下 class 定义：

    public class STuple<T1, T2>    {
    public T1 Item1 { get; set; }
    public T2 Item2 { get; set; }


    public static implicit operator Tuple<T1, T2>(STuple<T1, T2> st)
    {
        return Tuple.Create(st.Item1, st.Item2);
    }

    public static implicit operator STuple<T1, T2>(Tuple<T1, T2> t)
    {
        return new STuple<T1, T2>()
        {
            Item1 = t.Item1,
            Item2 = t.Item2,
        };
    }

    public STuple()
    {
    }
    public STuple(T1 t1, T2 t2) : this()
    {
        Item1 = t1;
        Item2 = t2;
    }
}

这里是示例程序：

        Dictionary<Tuple<string, string>, double> TupleDic = new Dictionary<Tuple<string, string>, double>();
        TupleDic.Add(new Tuple<string, string>("Name1", "Name2"), 5);
        TupleDic.Add(new Tuple<string, string>("Name3", "Name4"), 10);

        Console.WriteLine("dict-Entry 1: {0}", TupleDic[new Tuple<string, string>("Name1", "Name2")]);
        Console.WriteLine("dict-Entry 2: {0}", TupleDic[new Tuple<string, string>("Name3", "Name4")]);


        Dictionary<STuple<string, string>, double> STupleDic = new Dictionary<STuple<string, string>, double>();
        STuple<string, string> STuple1 = new STuple<string, string>("Name1", "Name2");
        STuple<string, string> STuple2 = new STuple<string, string>("Name3", "Name4");
        STupleDic.Add(STuple1, 5);
        STupleDic.Add(STuple2, 10);


        //Still working
        Console.WriteLine();
        Console.WriteLine("Sdict-Entry 1: {0}", STupleDic[STuple1]);
        Console.WriteLine("Sdict-Entry 2: {0}", STupleDic[STuple2]);

        //Not working
        STuple<string, string> STuple3 = new STuple<string, string>("Name1", "Name2");
        STuple<string, string> STuple4 = new STuple<string, string>("Name3", "Name4");
        Console.WriteLine();
        Console.WriteLine("Sdict-Entry 1: {0}", STupleDic[STuple3]);
        Console.WriteLine("Sdict-Entry 2: {0}", STupleDic[STuple4]);

        Console.ReadKey();

使用普通元组的示例工作得很好，但是当我使用我自己的 class STuple 时，如果我使用与添加时使用的完全相同的键（相同的实例），它才有效。我是一个初学者，是否可能因为值类型和引用类型的混淆而出现问题？

在我看来真的很奇怪，使用 foreach 进行查找仍然有效：

            Console.WriteLine();
        foreach (KeyValuePair<STuple<string, string>, double> s in STupleDic)
        {
            Console.WriteLine("Sdict-Entry 1: {0}", s.Value);
        }

Answer 1

警告：在 mutable 结构上实施 GetHashCode 会导致灾难。哈希码只有一个目的，那就是方便存储在 hash-tables 中。在 hash-tables 中用作键的项目不应发生变异（用于计算哈希的任何属性），因为更改哈希码会对 hash-table.[=20 造成不可恢复的损坏=]

为了让项目在 hash-table like 集合中工作，它们必须实现相等性和哈希码成员。因此，您可以（感谢 Resharper）：

public class STuple<T1, T2>
{
    public STuple()
    {
    }

    public STuple(T1 t1, T2 t2)
        : this()
    {
        Item1 = t1;
        Item2 = t2;
    }

    public T1 Item1 { get; set; }
    public T2 Item2 { get; set; }

    protected bool Equals(STuple<T1, T2> other)
    {
        return EqualityComparer<T1>.Default.Equals(Item1, other.Item1) &&
               EqualityComparer<T2>.Default.Equals(Item2, other.Item2);
    }

    public override bool Equals(object obj)
    {
        if (ReferenceEquals(null, obj)) return false;
        if (ReferenceEquals(this, obj)) return true;
        if (obj.GetType() != GetType()) return false;
        return Equals((STuple<T1, T2>) obj);
    }

    public override int GetHashCode()
    {
        unchecked
        {
            return (EqualityComparer<T1>.Default.GetHashCode(Item1)*397) ^
                   EqualityComparer<T2>.Default.GetHashCode(Item2);
        }
    }

    public static bool operator ==(STuple<T1, T2> left, STuple<T1, T2> right)
    {
        return Equals(left, right);
    }

    public static bool operator !=(STuple<T1, T2> left, STuple<T1, T2> right)
    {
        return !Equals(left, right);
    }


    public static implicit operator Tuple<T1, T2>(STuple<T1, T2> st)
    {
        return Tuple.Create(st.Item1, st.Item2);
    }

    public static implicit operator STuple<T1, T2>(Tuple<T1, T2> t)
    {
        return new STuple<T1, T2>
        {
            Item1 = t.Item1,
            Item2 = t.Item2
        };
    }
}