通用的ArrayList冒泡问题 [英] Generic arraylist bubblesort problems
问题描述
我知道有内置函数，但是，作为一个学习者，我想用我自己的设备，进行排序和自排序是旧帽子，我决定尝试使自己的通用的排序例程我可以用数字或字符串，甚至日期，如果我弄清楚他们在Java中是如何工作的。
I know there are builtin routines, but as a learner, I want to sort using my own devices, and since sorting is old hat, I decided to try to make my own generic sort routine that I could use for numbers or strings and maybe even dates, if I ever figure out how they work in Java.
因此，这里是我已经，已经换了另一个是另一个错误到现在为止我只有在两个地方的错误（在**标记附后），与需要弄清楚如何比较。
So here's what I have, having traded one error for another for another until now I only have errors in two places (enclosed within "**" markers), with need to figure out how to compare.
package sort;
import java.util.ArrayList;
public abstract class Sort<E> implements Comparable<E> {
public void swap(ArrayList<E> a, int i, int j) {
E c = a.get(i);
a.set(i,a.get(j));// = a[j];
a.set(j, c);
}
public void bubbleSort(ArrayList<E> a) {
boolean inOrder = false;
while (!inOrder) {
inOrder = true;
for (int i = 1; i < a.size(); i++) {
**if( a.get(i - 1).compareTo(a.get(i)) > 0 )** {
//cannot find symbol: method compareTo(E); location: class Object
//where E is a type-variable: E extends Object declared in class Sort
inOrder = false;
swap(a, i, i - 1);
}
}
}
}
public static void main(String args[]) //hadda lose 'static' for 'setLayout' to work
{
ArrayList<Integer> ary = new ArrayList<>();
ary.add(2); ary.add(4); ary.add(7); ary.add(3);
**bubbleSort(ary)**;
//method bubbleSort in class Sort<E> cannot be applied to given types;
//required: ArrayList<E>
//found: ArrayList<Integer>
//reason: actual argument ArrayList<Integer> cannot be converted to ArrayList<E>
//by method invocation conversion where E is a type-variable:
//E extends Object declared in class Sort
for (int i = 0; i < ary.size(); i++) {
System.out.println(ary.get(i));
}
}
@Override
public int compareTo(E o) {
**return 0;** // fixing errors above may help this fall into place
}
}
我想学的东西，我的感觉的准备才发现，我没有完全准备好;密切，cigarless。
I'm trying to learn things that I feel ready for only to find that I'm not quite ready; close, cigarless.
推荐答案
这
public abstract class Sort<E> implements Comparable<E> {
意味着电子是一个任意的对象类型，而实例排序＆LT; E＆GT; 可以比电子的实例。 （所以你的错误消息，抱怨 E.compareTo 不存在，因为对象不具备这样的。法）你想要的是这样的：
means that E is an arbitrary object type, and that instances of Sort<E> can be compared to instances of E. (So your error message is complaining that E.compareTo doesn't exist, since Object doesn't have such a method.) What you want is this:
public abstract class Sort<E extends Comparable<E>> {
这意味着电子必须是一个类型，其实例可以相互比较。
which means that E must be a type whose instances can be compared to each other.
编辑补充：其实，SLaks共同指出，有一个为没有真正的理由排序是通用的;你只需要在冒泡方法是通用的。此外，作为MadProgrammer意味着，无论是排序应该是非 - 摘要（这样你就可以直接将它实例）或冒泡应静态（因此它可以被称为不实例化排序实例），或者两者兼而有之。例如：
Edited to add: Actually, as SLaks together point out, there's no real reason for Sort to be generic; you just need the bubbleSort method to be generic. Further, as MadProgrammer implies, either Sort should be non-abstract (so you can instantiate it directly) or bubbleSort should be static (so it can be called without instantiating a Sort instance) or both. For example:
public class Sort {
private static <E> void swap(ArrayList<E> a, int i, int j) {
...
}
private static <E extends Comparable<E>> void bubbleSort(ArrayList<E> a) {
...
}
...
}
更重要的是，排序可以用排序方法，以及冒泡的接口的.sort（...）只是它（而不是让排序特定冒泡法）。
Better yet, Sort can be an interface with a sort method, and BubbleSort.sort(...) is just an implementation of it (rather than giving Sort a specific bubbleSort method).
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