构建具有继承性的通用树描述代码乞求你的帮助...编辑#1（对于下面的评论）)

inheritance java tree generics type-parameter

2022-09-03 16:16:58

我正在构建一个泛型类，它支持子树的继承。但是我遇到了一些问题。请你帮帮我吗？Tree<T>

描述

让我们定义类和类，其中 .TreeBlueTreeBlueTree extends Tree

让我们定义类和类，其中 .它们被用作树包含的“数据”。LeafRedLeafRedLeaf extends Leaf

A 表示类型的树，其“数据”类型为。Tree<Leaf>TreeLeaf

对于继承（这不是正确的 Java 继承）：

a 可以有类型的子项Tree<Leaf>
- Tree<Leaf>和。Tree<RedLeaf>BlueTree<Leaf>BlueTree<RedLeaf>

.

a 可以有类型的子项Tree<RedLeaf>
- Tree<RedLeaf>和BlueTree<RedLeaf>
- 但不是 ，或 .Tree<Leaf>BlueTree<Leaf>

.

a 可以有类型的子项BlueTree<Leaf>
- BlueTree<Leaf>和BlueTree<RedLeaf>
- 但不是 ，或 .Tree<Leaf>Tree<RedLeaf>

.

a 可以有类型的子项BlueTree<RedLeaf>
- BlueTree<RedLeaf>,
- 但不是 、或。Tree<Leaf>Tree<RedLeaf>BlueTree<Leaf>

*在这里，“孩子”是指树的树枝/叶子。

（有点复杂，这就是为什么我把线条分开。

代码

（如果你有一个解决方案，你可能不需要阅读下面我尝试的详细插图。如果你想一起找出解决方案，我的代码可能会给你一些想法 - 或者，它可能会让他们感到困惑。

第一次审判：（简单的一次）

// This is the focus of this question, the class signature
public class Tree<T> {
    // some fields, but they are not important in this question
    private Tree<? super T> mParent;
    private T mData;
    private ArrayList<Tree<? extends T>> mChildren;

    // This is the focus of this question, the addChild() method signature
    public void addChild(final Tree<? extends T> subTree) {
        // add the subTree to mChildren
    }
}

此类结构满足描述中的大多数要求。除非，它允许

class BlueTree<T> extends Tree<T> { }
class Leaf { }
class RedLeaf extends Leaf { }

Tree<Leaf> tree_leaf = new Tree<Leaf>();
BlueTree<Leaf> blueTree_leaf = new BlueTree<Leaf>();

blueTree_leaf.addChild(tree_leaf);    // should be forbidden

这违反了

a 不能有类型的子项。BlueTree<Leaf>Tree<Leaf>

问题在于，在中，它的方法签名仍然是BlueTree<Leaf>addChild()

public void addChild(final Tree<? extends Leaf> subTree) {
     // add the subTree to mChildren
}

理想的情况是，方法签名（在继承时自动更改为）BlueTree<Leaf>.addChild()

public void addChild(final BlueTree<? extends Leaf> subTree) {
     // add the subTree to mChildren
}

（请注意，此方法无法通过继承覆盖上述方法，因为参数类型不同。

有一个解决方法。我们可以添加一个类继承检查，并针对这种情况进行抛出：RuntimeException

public void addChild(final Tree<? extends Leaf> subTree) {
    if (this.getClass().isAssignableFrom(subTree.getClass()))
        throw new RuntimeException("The parameter is of invalid class.");
    // add the subTree to mChildren
}

但是，使其成为编译时错误远比运行时错误好得多。我想在编译时强制执行此行为。

第二次审判

第一个试验结构中的问题是，方法中的参数类型不是泛型类型参数。因此，它不会在继承时更新。这一次，让我们尝试也使其成为泛型类型参数。TreeaddChild()

首先，定义通用类。Tree

public class Tree<T> {
    private Tree<? super T> mParent;
    private T mData;
    private ArrayList<Tree<? extends T>> mChildren;

    /*package*/ void addChild(final Tree<? extends T> subTree) {
        // add the subTree to mChildren
    }
}

然后，哪个管理一个对象。TreeManagerTree

public final class TreeManager<NodeType extends Tree<? super DataType>, DataType> {
    private NodeType mTree;

    public TreeManager(Class<NodeType> ClassNodeType) {
        try {
            mTree = ClassNodeType.newInstance();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public void managerAddChild(final NodeType subTree) {
        mTree.addChild(subTree);
        // compile error: The method addChild(Tree<? extends capture#1-of ? super DataType>)
        //                in the type Tree<capture#1-of ? super DataType>
        //                is not applicable for the arguments (NodeType)
    }

    // for testing
    public static void main(String[] args) {
        @SuppressWarnings("unchecked")
        TreeManager<Tree    <Leaf>   , Leaf>    tm_TreeLeaf_Leaf           = new TreeManager<Tree    <Leaf>,    Leaf>   ((Class<Tree    <Leaf>>)    new Tree    <Leaf>   ().getClass());
        TreeManager<Tree    <RedLeaf>, RedLeaf> tm_TreeRedLeaf_RedLeaf     = new TreeManager<Tree    <RedLeaf>, RedLeaf>((Class<Tree    <RedLeaf>>) new Tree    <RedLeaf>().getClass());
        TreeManager<BlueTree<Leaf>   , Leaf>    tm_BlueTreeLeaf_Leaf       = new TreeManager<BlueTree<Leaf>,    Leaf>   ((Class<BlueTree<Leaf>>)    new BlueTree<Leaf>   ().getClass());
        TreeManager<BlueTree<RedLeaf>, RedLeaf> tm_BlueTreeRedLeaf_RedLeaf = new TreeManager<BlueTree<RedLeaf>, RedLeaf>((Class<BlueTree<RedLeaf>>) new BlueTree<RedLeaf>().getClass());

        System.out.println(tm_TreeLeaf_Leaf          .mTree.getClass());    // class Tree
        System.out.println(tm_TreeRedLeaf_RedLeaf    .mTree.getClass());    // class Tree
        System.out.println(tm_BlueTreeLeaf_Leaf      .mTree.getClass());    // class BlueTree
        System.out.println(tm_BlueTreeRedLeaf_RedLeaf.mTree.getClass());    // class BlueTree

        @SuppressWarnings("unchecked")
        TreeManager<Tree    <Leaf>   , RedLeaf> tm_TreeLeaf_RedLeaf     = new TreeManager<Tree    <Leaf>,    RedLeaf>((Class<Tree    <Leaf>>)    new Tree    <Leaf>   ().getClass());
        TreeManager<BlueTree<Leaf>   , RedLeaf> tm_BlueTreeLeaf_RedLeaf = new TreeManager<BlueTree<Leaf>,    RedLeaf>((Class<BlueTree<Leaf>>)    new BlueTree<Leaf>   ().getClass());

        System.out.println(tm_TreeLeaf_RedLeaf       .mTree.getClass());    // class Tree
        System.out.println(tm_BlueTreeLeaf_RedLeaf   .mTree.getClass());    // class BlueTree

        // the following two have compile errors, which is good and expected.
        TreeManager<Tree    <RedLeaf>, Leaf>    tm_TreeRedLeaf_Leaf     = new TreeManager<Tree    <RedLeaf>, Leaf>   ((Class<Tree    <RedLeaf>>) new Tree    <RedLeaf>().getClass());
        TreeManager<BlueTree<RedLeaf>, Leaf>    tm_BlueTreeRedLeaf_Leaf = new TreeManager<BlueTree<RedLeaf>, Leaf>   ((Class<BlueTree<RedLeaf>>) new BlueTree<RedLeaf>().getClass());
    }
}

初始化没有问题;虽然线条有点长。它也符合描述中的规则。TreeManager

但是，在内部调用时会出现编译错误，如上图所示。Tree.addChild()TreeManager

第三次审判

为了修复第二次试验中的编译错误，我尝试更改类签名（甚至更长）。现在编译没有问题。mTree.addChild(subTree);

// T is not used in the class. T is act as a reference in the signature only
public class TreeManager3<T, NodeType extends Tree<T>, DataType extends T> {
    private NodeType mTree;

    public TreeManager3(Class<NodeType> ClassNodeType) {
        try {
            mTree = ClassNodeType.newInstance();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public void managerAddChild(final NodeType subTree) {
        mTree.addChild(subTree);    // compile-error is gone
    }
}

我已经用与第二次试验非常相似的代码对其进行了测试。它创建没有任何问题，就像第二次试验一样。（只是更长。

（您可以跳过下面的代码块，因为它只是在逻辑上重复。

public static void main(String[] args) {
    @SuppressWarnings("unchecked")
    TreeManager3<Leaf   , Tree    <Leaf>   , Leaf>    tm_TreeLeaf_Leaf           = new TreeManager3<Leaf   , Tree    <Leaf>,    Leaf>   ((Class<Tree    <Leaf>>)    new Tree    <Leaf>   ().getClass());
    TreeManager3<RedLeaf, Tree    <RedLeaf>, RedLeaf> tm_TreeRedLeaf_RedLeaf     = new TreeManager3<RedLeaf, Tree    <RedLeaf>, RedLeaf>((Class<Tree    <RedLeaf>>) new Tree    <RedLeaf>().getClass());
    TreeManager3<Leaf   , BlueTree<Leaf>   , Leaf>    tm_BlueTreeLeaf_Leaf       = new TreeManager3<Leaf   , BlueTree<Leaf>,    Leaf>   ((Class<BlueTree<Leaf>>)    new BlueTree<Leaf>   ().getClass());
    TreeManager3<RedLeaf, BlueTree<RedLeaf>, RedLeaf> tm_BlueTreeRedLeaf_RedLeaf = new TreeManager3<RedLeaf, BlueTree<RedLeaf>, RedLeaf>((Class<BlueTree<RedLeaf>>) new BlueTree<RedLeaf>().getClass());

    System.out.println(tm_TreeLeaf_Leaf          .mTree.getClass());    // class Tree
    System.out.println(tm_TreeRedLeaf_RedLeaf    .mTree.getClass());    // class Tree
    System.out.println(tm_BlueTreeLeaf_Leaf      .mTree.getClass());    // class BlueTree
    System.out.println(tm_BlueTreeRedLeaf_RedLeaf.mTree.getClass());    // class BlueTree

    @SuppressWarnings("unchecked")
    TreeManager3<Leaf   , Tree    <Leaf>   , RedLeaf> tm_TreeLeaf_RedLeaf     = new TreeManager3<Leaf   , Tree    <Leaf>,    RedLeaf>((Class<Tree    <Leaf>>)    new Tree    <Leaf>   ().getClass());
    TreeManager3<Leaf   , BlueTree<Leaf>   , RedLeaf> tm_BlueTreeLeaf_RedLeaf = new TreeManager3<Leaf   , BlueTree<Leaf>,    RedLeaf>((Class<BlueTree<Leaf>>)    new BlueTree<Leaf>   ().getClass());

    System.out.println(tm_TreeLeaf_RedLeaf       .mTree.getClass());    // class Tree
    System.out.println(tm_BlueTreeLeaf_RedLeaf   .mTree.getClass());    // class BlueTree

    // the following two have compile errors, which is good and expected.
    TreeManager3<RedLeaf, Tree    <RedLeaf>, Leaf>    tm_TreeRedLeaf_Leaf     = new TreeManager3<RedLeaf, Tree    <RedLeaf>, Leaf>   ((Class<Tree    <RedLeaf>>) new Tree    <RedLeaf>().getClass());
    TreeManager3<RedLeaf, BlueTree<RedLeaf>, Leaf>    tm_BlueTreeRedLeaf_Leaf = new TreeManager3<RedLeaf, BlueTree<RedLeaf>, Leaf>   ((Class<BlueTree<RedLeaf>>) new BlueTree<RedLeaf>().getClass());
}

但是，当我尝试调用.TreeManager3.managerAddChild()

tm_TreeLeaf_Leaf.managerAddChild(new Tree<Leaf>());
tm_TreeLeaf_Leaf.managerAddChild(new Tree<RedLeaf>());      // compile error: managerAddChild(Tree<RedLeaf>) cannot cast to managerAddChild(Tree<Leaf>)
tm_TreeLeaf_Leaf.managerAddChild(new BlueTree<Leaf>());
tm_TreeLeaf_Leaf.managerAddChild(new BlueTree<RedLeaf>());  // compile error: managerAddChild(BlueTree<RedLeaf>) cannot cast to managerAddChild(BlueTree<Leaf>)

这是可以理解的。 means，并且参数类型中没有通配符，就像在第一次试验中一样。TreeManager3.managerAddChild(NodeType)TreeManager3.managerAddChild(Tree<T>)Tree<? extends T>Tree.addChild(final Tree<? extends T> subTree)

乞求你的帮助...

我已经没有想法了。我是否在解决问题方面走错了方向？我花了很多时间输入这个问题，并尽我最大的努力使它更具可读性，更易于理解和遵循。我不得不说对不起，它仍然很长很长很冗长。但是，如果您知道方法，请您提供帮助，或者请给我任何您的想法吗？非常感谢您的每一个输入。多谢！

编辑#1（对于下面的评论）)

基于第一次试验，只允许通过（和其他方法与检查）修改，所以即使允许用户继承和覆盖也不会破坏树的完整性。mChildrenaddChild()isAssignableFrom()TreeaddChild()

/developer/util/Tree.java

package developer.util;

import java.util.ArrayList;

public class Tree<T> {

    private Tree<? super T> mParent;
    private final ArrayList<Tree<? extends T>> mChildren = new ArrayList<Tree<? extends T>>();

    public int getChildCount() { return mChildren.size(); }
    public Tree<? extends T> getLastChild() { return mChildren.get(getChildCount()-1); }

    public void addChild(final Tree<? extends T> subTree) {
        if (this.getClass().isAssignableFrom(subTree.getClass()) == false)
            throw new RuntimeException("The child (subTree) must be a sub-class of this Tree.");

        subTree.mParent = this;
        mChildren.add(subTree);
    }
}

/user/pkg/BinaryTree.java

package user.pkg;

import developer.util.Tree;

public class BinaryTree<T> extends Tree<T> {
    @Override
    public void addChild(final Tree<? extends T> subTree) {
        if (getChildCount() < 2) {
            super.addChild(subTree);
        }
    }
}

/主要.java

import user.pkg.BinaryTree;
import developer.util.Tree;

public class Main {

    public static void main(String[] args) {
        Tree<Integer> treeOfInt = new Tree<Integer>();
        BinaryTree<Integer> btreeOfInt = new BinaryTree<Integer>();

        treeOfInt.addChild(btreeOfInt);
        System.out.println(treeOfInt.getLastChild().getClass());
        // class user.pkg.BinaryTree

        try {
            btreeOfInt.addChild(treeOfInt);
        } catch (Exception e) {
            System.out.println(e);
            // java.lang.RuntimeException: The child (subTree) must be a sub-class of this Tree.
        }

        System.out.println("done.");
    }
}

你觉得怎么样？

答案 1

在我看来，这个问题没有完美的解决方案。这基本上是由于类型擦除。泛型方法的擦除一文解释了您的函数将成为一个函数。因此，即使您可以以某种方式拥有通用参数（不是有效的语法！），它也会在编译时被删除。不过，添加运行时测试将起作用，因此您所做的编辑将完成这项工作。addChild(final Tree<? extends Leaf> subTree)addChild(final Tree subTree)<TreeType extends Tree<? extends Leaf>> addChild(final TreeType subTree)addChild(final Tree subTree)

答案 2

我认为你需要的是以下几点

class Tree<LT extends Leaf>{
//have your generic add/delete/traverse methods here.
}

class BlueTree<LT extends Leaf> extends Tree<LT>{
//have your blue tree specific add/delete/traverse methods here.
}

class Leaf {
//have basic data members here
}
class BlueLeaf extends Leaf{
//have blue leaf specific data members here
}