【发布时间】:2014-11-01 08:31:11
【问题描述】:
我正在构建一个通用的Tree<T> 类,它支持子树的继承。但是我遇到了一些问题。请您帮帮我好吗?
说明
让我们定义Tree 类和BlueTree 类,其中BlueTree extends Tree。
让我们定义Leaf 类和RedLeaf 类,其中RedLeaf extends Leaf。它们被用作树包含的“数据”。
Tree<Leaf> 表示 Tree 类型的树,其“数据”类型为 Leaf。
对于继承(这不是正确的 Java 继承):
-
Tree<Leaf>可以有子类型-
Tree<Leaf>、Tree<RedLeaf>、BlueTree<Leaf>和BlueTree<RedLeaf>。
-
.
-
Tree<RedLeaf>可以有子类型-
Tree<RedLeaf>,和BlueTree<RedLeaf>, -
但不是
Tree<Leaf>,或BlueTree<Leaf>。
-
.
-
BlueTree<Leaf>可以有子类型-
BlueTree<Leaf>,和BlueTree<RedLeaf>, -
但不是
Tree<Leaf>,或Tree<RedLeaf>。
-
.
-
BlueTree<RedLeaf>可以有子类型-
BlueTree<RedLeaf>, -
但不是
Tree<Leaf>、Tree<RedLeaf>或BlueTree<Leaf>。
-
*这里的“子”是指树的树枝/树叶。
(有点复杂,这就是我分开行的原因。)
代码
(如果你有解决方案,你可能不需要阅读下面我尝试的详细说明。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,它可能会混淆他们。)
初审:(最简单的)
// 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
}
}
这个类结构满足描述中的大部分要求。除了,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
违反了
-
BlueTree<Leaf>不能拥有Tree<Leaf>类型的子级。
问题在于,在BlueTree<Leaf> 中,它的addChild() 方法签名仍然是
public void addChild(final Tree<? extends Leaf> subTree) {
// add the subTree to mChildren
}
理想的情况是,BlueTree<Leaf>.addChild() 方法签名被更改(自动,继承时)
public void addChild(final BlueTree<? extends Leaf> subTree) {
// add the subTree to mChildren
}
(请注意,此方法不能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,并在这种情况下抛出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
}
但是让它成为编译时错误比运行时错误要好得多。我想在编译时强制执行此行为。
二审
第一个试结构的问题是,方法addChild()中的参数类型Tree不是泛型类型参数。因此它不会在继承时更新。这一次,我们也尝试将其设为泛型类型参数。
首先,定义通用的Tree类。
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
}
}
然后是管理Tree 对象的TreeManager。
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());
}
}
TreeManager 初始化没有问题;行虽然有点长。它也符合描述中的规则。
但是,在TreeManager 中调用Tree.addChild() 时会出现编译错误,如上图所示。
第三次试用
为了修复第二次试用中的编译错误,我尝试更改类签名(更长)。现在mTree.addChild(subTree); 编译没有问题。
// 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
}
}
我已经使用与第二次试用非常相似的代码对其进行了测试。就像第二次试验一样,它没有任何问题。 (甚至更长。)
(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,当我尝试拨打 TreeManager3.managerAddChild() 时出现问题。
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。 TreeManager3.managerAddChild(NodeType) 表示TreeManager3.managerAddChild(Tree<T>),参数类型中没有通配符Tree<? extends T>,如初审中的Tree.addChild(final Tree<? extends T> subTree)。
请求您的帮助...
我的想法已经用完了。为了解决这个问题,我是否走错了方向?我花了很多时间输入这个问题,并尽我最大的努力使其更具可读性、更易于理解和遵循。我不得不说抱歉,它仍然很长而且很冗长。但是,如果您知道路,请您帮忙,或者请给我您的任何想法?非常感谢您的每一个输入。非常感谢!
编辑 #1(用于comment below)
基于一审,只允许mChildren被addChild()修改(以及其他带有isAssignableFrom()检查的方法),所以即使允许用户继承Tree和覆盖 addChild() 不会破坏树的完整性。
/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);
}
}
}
/Main.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.");
}
}
你怎么看?
【问题讨论】:
-
你认为你的对象是 Tree
,LEAF> 吗?所以在这种情况下,你也会有你的树的类型 -
感谢您的回复!!一旦我回到家,我必须尝试一下。如果这行得通,那就太好了!还有简单、漂亮、整洁的方式:-)
-
试试
public class Tree<TREE extends Tree<?,?>,LEAF extends Leaf> { } -
没问题,如果您对此有更多问题,请大声喊叫
-
是的,使用javadoc就足够了,如果这真的是问题,那么你可以在你的类中添加类参数,并使用静态工厂来构造对象。但正如我所说,我不认为这是真正的问题
标签: java generics inheritance tree type-parameter