c# - 什么是变形，它可以在 C# 3.0 中实现吗？

Question

我正在尝试了解变质现象，并且我已经阅读了 the Wikipedia article和第一对帖子在the series of the topic for F#在 Inside F# 博客上。

我理解它是折叠的概括(即，将许多值的结构映射到一个值，包括值列表到另一个列表)。我认为折叠列表和折叠树是一个典型的例子。

是否可以使用 LINQ 的 Aggregate 运算符或其他一些高阶方法在 C# 中证明这一点？

Answer 1

LINQ 的 Aggregate() 仅适用于 IEnumerables。变形一般指的是任意数据类型的折叠模式。所以 Aggregate() 之于 IEnumerables 就像 FoldTree(下)之于 Trees(下)；两者都是各自数据类型的变形。

我翻译了part 4 of the series中的一些代码进入 C#。代码如下。请注意，等效的 F# 使用三个小于字符(用于泛型类型参数注释)，而此 C# 代码使用超过 60 个。这就是为什么没有人在 C# 中编写此类代码的证据 - 类型注释太多。我提供代码以防它帮助了解 C# 而不是 F# 的人玩这个。但是 C# 中的代码非常密集，很难理解。

给定二叉树的以下定义:

using System;
using System.Collections.Generic;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Shapes;

class Tree<T>   // use null for Leaf
{
    public T Data { get; private set; }
    public Tree<T> Left { get; private set; }
    public Tree<T> Right { get; private set; }
    public Tree(T data, Tree<T> left, Tree<T> rright)
    {
        this.Data = data;
        this.Left = left;
        this.Right = right;
    }

    public static Tree<T> Node<T>(T data, Tree<T> left, Tree<T> right)
    {
        return new Tree<T>(data, left, right);
    }
}

可以折叠树木，例如测量两棵树是否有不同的节点:

class Tree
{
    public static Tree<int> Tree7 =
        Node(4, Node(2, Node(1, null, null), Node(3, null, null)),
                Node(6, Node(5, null, null), Node(7, null, null)));

    public static R XFoldTree<A, R>(Func<A, R, R, Tree<A>, R> nodeF, Func<Tree<A>, R> leafV, Tree<A> tree)
    {
        return Loop(nodeF, leafV, tree, x => x);
    }

    public static R Loop<A, R>(Func<A, R, R, Tree<A>, R> nodeF, Func<Tree<A>, R> leafV, Tree<A> t, Func<R, R> cont)
    {
        if (t == null)
            return cont(leafV(t));
        else
            return Loop(nodeF, leafV, t.Left, lacc =>
                   Loop(nodeF, leafV, t.Right, racc =>
                   cont(nodeF(t.Data, lacc, racc, t))));
    }

    public static R FoldTree<A, R>(Func<A, R, R, R> nodeF, R leafV, Tree<A> tree)
    {
        return XFoldTree((x, l, r, _) => nodeF(x, l, r), _ => leafV, tree);
    }

    public static Func<Tree<A>, Tree<A>> XNode<A>(A x, Tree<A> l, Tree<A> r)
    {
        return (Tree<A> t) => x.Equals(t.Data) && l == t.Left && r == t.Right ? t : Node(x, l, r);
    }

    // DiffTree: Tree<'a> * Tree<'a> -> Tree<'a * bool> 
    // return second tree with extra bool 
    // the bool signifies whether the Node "ReferenceEquals" the first tree 
    public static Tree<KeyValuePair<A, bool>> DiffTree<A>(Tree<A> tree, Tree<A> tree2)
    {
        return XFoldTree((A x, Func<Tree<A>, Tree<KeyValuePair<A, bool>>> l, Func<Tree<A>, Tree<KeyValuePair<A, bool>>> r, Tree<A> t) => (Tree<A> t2) =>
            Node(new KeyValuePair<A, bool>(t2.Data, object.ReferenceEquals(t, t2)),
                 l(t2.Left), r(t2.Right)),
            x => y => null, tree)(tree2);
    }
}

在第二个例子中，另一棵树的重建方式不同:

class Example
{
    // original version recreates entire tree, yuck 
    public static Tree<int> Change5to0(Tree<int> tree)
    {
        return Tree.FoldTree((int x, Tree<int> l, Tree<int> r) => Tree.Node(x == 5 ? 0 : x, l, r), null, tree);
    }

    // here it is with XFold - same as original, only with Xs 
    public static Tree<int> XChange5to0(Tree<int> tree)
    {
        return Tree.XFoldTree((int x, Tree<int> l, Tree<int> r, Tree<int> orig) =>
            Tree.XNode(x == 5 ? 0 : x, l, r)(orig), _ => null, tree);
    }
}

在第三个示例中，折叠树用于绘图:

class MyWPFWindow : Window 
{
    void Draw(Canvas canvas, Tree<KeyValuePair<int, bool>> tree)
    {
        // assumes canvas is normalized to 1.0 x 1.0 
        Tree.FoldTree((KeyValuePair<int, bool> kvp, Func<Transform, Transform> l, Func<Transform, Transform> r) => trans =>
        {
            // current node in top half, centered left-to-right 
            var tb = new TextBox();
            tb.Width = 100.0; 
            tb.Height = 100.0;
            tb.FontSize = 70.0;
                // the tree is a "diff tree" where the bool represents 
                // "ReferenceEquals" differences, so color diffs Red 
            tb.Foreground = (kvp.Value ? Brushes.Black : Brushes.Red);
            tb.HorizontalContentAlignment = HorizontalAlignment.Center;
            tb.VerticalContentAlignment = VerticalAlignment.Center;
            tb.RenderTransform = AddT(trans, TranslateT(0.25, 0.0, ScaleT(0.005, 0.005, new TransformGroup())));
            tb.Text = kvp.Key.ToString();
            canvas.Children.Add(tb);
            // left child in bottom-left quadrant 
            l(AddT(trans, TranslateT(0.0, 0.5, ScaleT(0.5, 0.5, new TransformGroup()))));
            // right child in bottom-right quadrant 
            r(AddT(trans, TranslateT(0.5, 0.5, ScaleT(0.5, 0.5, new TransformGroup()))));
            return null;
        }, _ => null, tree)(new TransformGroup());
    }

    public MyWPFWindow(Tree<KeyValuePair<int, bool>> tree)
    {
        var canvas = new Canvas();
        canvas.Width=1.0;
        canvas.Height=1.0;
        canvas.Background = Brushes.Blue;
        canvas.LayoutTransform=new ScaleTransform(200.0, 200.0);
        Draw(canvas, tree);
        this.Content = canvas;
        this.Title = "MyWPFWindow";
        this.SizeToContent = SizeToContent.WidthAndHeight;
    }
    TransformGroup AddT(Transform t, TransformGroup tg) { tg.Children.Add(t); return tg; }
    TransformGroup ScaleT(double x, double y, TransformGroup tg) { tg.Children.Add(new ScaleTransform(x,y)); return tg; }
    TransformGroup TranslateT(double x, double y, TransformGroup tg) { tg.Children.Add(new TranslateTransform(x,y)); return tg; }

    [STAThread]
    static void Main(string[] args)
    {
        var app = new Application();
        //app.Run(new MyWPFWindow(Tree.DiffTree(Tree.Tree7,Example.Change5to0(Tree.Tree7))));
        app.Run(new MyWPFWindow(Tree.DiffTree(Tree.Tree7, Example.XChange5to0(Tree.Tree7))));
    }
}