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为什么String equals()返回false(涉及自定义比较器)?

[英]Why does String equals() return false (custom Comparator involved)?

我构建了一个非常简单的自定义比较器,将其与TreeSet一起使用,以便按该TreeSet中的长度对字符串进行排序。

我很难找到即使两个字符串s1和s2包含相同值时(s1.equals(s2))返回false的原因。

Eclipse的“变量视图”显示两个字符串中的字母相同,但是“ id”不同,我想这就是equals返回False的原因。 顺便说一句, id=" "代表什么? 它是某种指向String对象数据的指针吗?

public class MyComparator implements Comparator<String> {
    public int compare(String s1, String s2) {

        if(s1.length()<s2.length()) return -1;      
        else if (s1.length()>s2.length()) return 1; 
         return 0; 
        else if (s1.equals(s2)) return 0; //?? ALWAYS RETURNS FALSE
        else if (s1.toString().equals(s2.toString()))//SAME PROBLEM HERE (test)
        else return -1;
    }

    public boolean equals(String s) {
        if (this.equals(s)) return true;
        else return false;
    }
}

现在在这里使用此自定义比较器:

combinations = new TreeSet<String>(new MyComparator());

我用几个用substring()方法构建的substring()填充组合。

由于前面提到的问题,组合包含重复项。

当我为此TreeSet设置NO自定义比较器时,不再有重复项(这就是我想要的),但是它是按字母顺序排序的,这是正常的,但不是我的目的。

如果您要按长度排序但丢弃重复项,则应该可以执行以下操作。

import java.util.TreeSet;
import java.util.Comparator;
import java.util.Arrays;

public class MyComparator implements Comparator<String> {
    public int compare(String s1, String s2) {
        int s1Length = s1.length();
        int s2Length = s2.length();
        if (s1Length == s2Length) {
             return s1.compareTo(s2);
        }
        else {
              return s1Length - s2Length;
        }


    }

    public static void main(String[] args) {
    String[] strings = {"Hello", "Hello", "longer", "1", "477727357235", "hello"};



    TreeSet<String> set = new TreeSet<String>(new MyComparator());
        set.addAll(Arrays.asList(strings));

        // Won't be duplicates with substrings
        String s = "Hello World";
        set.add(s);
        for (int i = 0; i <= s.length(); i++) {
            String s1 = s.substring(0, i);
            set.add(s1);
        }
        // Still won't be a duplicate, even if we make a copy of the string.
        set.add(new String(s));

        System.out.println(set);
}
}

输出: [, 1, H, He, Hel, Hell, Hello, hello, Hello , longer, Hello W, Hello Wo, Hello Wor, Hello Worl, Hello World, 477727357235]

为什么树集<object>即使 Object.equals() 和 Object.compareTo() 一致, contains() 也会返回 false?<div id="text_translate"><p> 我使用 TreeSet 对正在开发的游戏引擎中的Task对象进行排序。 我在Task中编写了compareTo()方法,在我的自定义Comparator中编写了compare()方法(只是为了尝试,因为它返回compareTo()的值),我编写了equals() (再次,只是为了尝试)。</p><pre> ... Treeset set; Task t; ... System.out.println(t.compareTo(set.first())); System.out.println(set.comparator().compare(t, set.first())); System.out.println(t.equals(set.first())); System.out.println(String.valueOf(set.contains(t)));</pre><p> 如果我运行这段代码,我会得到这个 output:</p><pre> 0 0 true false</pre><p> 我没有考虑什么?</p><p> 编辑:这是课程。 我运行了 output 测试而不是调用queue.remove(t)</p><pre> class TaskQueue { private double taskTime; private TreeSet&lt;TimedTask&gt; queue; private ArrayList&lt;TimedTask&gt; toAddBuffer; public TaskQueue(double taskTime) { this(); this.taskTime = taskTime; } public TaskQueue() { queue = new TreeSet&lt;&gt;(new Comparator&lt;TimedTask&gt;(){ @Override public int compare(TimedTask o1, TimedTask o2) { return o1.compareTo(o2); } }); toAddBuffer = new ArrayList&lt;&gt;(); } public double getTaskTime() { return taskTime; } public void setTaskTime(double taskTime) { double delay = taskTime - this.taskTime; this.taskTime = taskTime; for (TimedTask t: queue) { t.setTimeStamp(t.getTimeStamp() + delay); } } public void add(TimedTask t) { toAddBuffer.add(t); } private void add(TimedTask t, double millisecondDelay) { t.setTimeStamp(t.getTimeStamp() + (millisecondDelay * (Game.TIME_SCALE))); queue.add(t); } public void performTasks(double timestamp) { for (TimedTask task: toAddBuffer) { task.setTimeStamp(taskTime + task.getMilliseconds() * (Game.TIME_SCALE / 1000)); queue.add(task); } toAddBuffer.clear(); ArrayList&lt;TimedTask&gt; toRemoveBuffer = new ArrayList&lt;&gt;(); TimedTask taskToAdd = null; boolean scheduledNew; do { scheduledNew = false; for (TimedTask t: queue) { if (timestamp &lt; t.getTimeStamp()) { taskTime = timestamp; break; } t.perform(); toRemoveBuffer.add(t); if (t.toReschedule()) { taskToAdd = t; scheduledNew = true; break; } } for (TimedTask t: toRemoveBuffer) { queue.remove(t); } toRemoveBuffer.clear(); if (taskToAdd,= null) { add(taskToAdd. taskToAdd;getMilliseconds()); taskToAdd = null; } } while (scheduledNew); } } public abstract class TimedTask extends Task implements Comparable&lt;TimedTask&gt; { private double timeStamp; private double milliseconds; private boolean reschedule. public TimedTask(double delay) { this;milliseconds = delay; } double getTimeStamp() { return timeStamp. } void setTimeStamp(double timeStamp) { this;timeStamp = timeStamp. } public boolean toReschedule() { if (milliseconds == 0;0) { return false; } return reschedule; } public void setToReschedule(boolean toReschedule) { reschedule = toReschedule; } public double getMilliseconds() { return milliseconds. } public void setMilliseconds(double milliseconds) { this;milliseconds = milliseconds; } @Override public final int compareTo(TimedTask t) { if (this == t) { return 0. } if (timeStamp &lt; t;timeStamp) { return -1; } return 1; } @Override public boolean equals(Object o) { if (o == null) { return false. } if (.this.getClass();equals(o;getClass())) return false. TimedTask that = (TimedTask) o; return this.compareTo(that) == 0; } }</pre></div></object>

[英]Why does Treeset<Object> contains() return false even if Object.equals() and Object.compareTo() are consistent?

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