blocks|key|750922|text|Comparable是一个带有泛型的接口，因此它应该是Comparable<Employee>。(因此，您可以确定将员工与员工进行比较，而不是将员工与数字进行比较。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|750923|Employee+emp+=+(Employee)+emp;|code-block|syntax|javascript|750924|讲得通?|750925|将emp转换为Employee，并将其放入emp中。也许你是说(Employee)+o|750926|return+this.id+-+o.id;|750927|你使用的是o，也就是Object，而Object没有.id，也许你是说emp.id|750928|entityMap^0|0|A|R|K|0|0|0|1|3|7|8|L|3|V|C|0|0|5|1|A|6|I|6|Q|3|Z|6|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|T|$]]

<code>Comparable</code> is an interface with a generic, so it should be <code>Comparable&lt;Employee&gt;</code>. (So you are sure that you compare an Employee with an Employee and not an Employee with a number.)

<pre><code>Employee emp = (Employee) emp;
</code></pre>

Make sense?

You cast <code>emp</code> to <code>Employee</code> and put that in <code>emp</code>. Maybe you mean <code>(Employee) o</code>

<pre><code>return this.id - o.id;
</code></pre>

You use <code>o</code> which is <code>Object</code>, and <code>Object</code> don't have <code>.id</code> Maybe you mean <code>emp.id</code>?

blocks|key|750961|text|如果您确定id是唯一，且不存在整数溢出(这将需要负ids)，则代码应按预期工作。另一方面，如果两个Emp+(被认为是不同的)可以有相同的id+(因为它不是我们期望的名称中的标识符)，那么它可能导致这样一种情况，即两个Emp被认为是相等的，而实际上它们不是。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|750962|entityMap^0|5|2|P|2|1D|3|1W|2|30|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

If you are sure that <code>id</code> is unique, and there is not integer overflow (which would require negative <code>id</code>s), that code should work as expected. On the other hand, if two <code>Emp</code> (considered different) can have the same <code>id</code> (because it is not the identifier we expect it to be by its name), then it could lead to the situation where two <code>Emp</code> are considered equal when in fact they are not.

blocks|key|970565|text|您没有使用emp实例来做任何事情...如果您尝试将Employee+(this)与始终为Employee的Object+o进行比较，则需要使用return+this.id+-+emp.id;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|970566|entityMap^0|P|8|Z|4|18|8|1H|8|1Z|O|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

You're not using the emp instance to do anything with... If you're trying to compare an <code>Employee</code> (<code>this</code>) to <code>Object o</code> that is always an <code>Employee</code>, then you want to <code>return this.id - emp.id;</code>

blocks|key|4534754|text|public+int+compareTo(Employee+o){
+++return+this.id+-+o.id;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4534755|如果你真的是个固执的人：|unstyled|4534756|public+int+compareTo(Object+o){
+++long+thisId+=+this.id;
+++long+oId+=+o.id;
+++long+result+=+thisId+-+oId;
+++if+(result+>+Integer.MAX_VALUE)+result+=+Integer.MAX_VALUE;
+++if+(result+<+Integer.MIN_VALUE)+result+=+Integer.MIN_VALUE;
+++return+result;
}|4534757|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>public int compareTo(Employee o){
 return this.id - o.id;
}
</code></pre>

If you're really a stickler:

<pre><code>public int compareTo(Object o){
 long thisId = this.id;
 long oId = o.id;
 long result = thisId - oId;
 if (result &gt; Integer.MAX_VALUE) result = Integer.MAX_VALUE;
 if (result &lt; Integer.MIN_VALUE) result = Integer.MIN_VALUE;
 return result;
}
</code></pre>

blocks|key|970578|text|当你不能保证id是正数还是负数时，问题就变得棘手了。棘手的是，如果id变为负数，那么减法可能会溢出并产生不正确的结果|type|unstyled|depth|inlineStyleRanges|entityRanges|data|970579|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It becomes tricky when you can't guaranteed that id is positive or negative. tricky part is, If id becomes negative than subtraction may overflow and produce incorrect result

blocks|key|976815|text|class+Employee+implements+Comparable<Employee>+{

++++public+int+compareTo(Object+o)+{
++++++++if+(o+instanceof+Employee)+{
++++++++++++Employee+emp+=+(Employee)+o;
++++++++++++return+this.id+-+emp.id;
++++++++}else+{
++++++++++++throw+new+IllegalArgumentException("Parameter+not+of+type+Employee!");
++++++++}
++++++++return+null;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|976816|我想这没问题。如果id为负值会有什么问题呢？|unstyled|976817|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>class Employee implements Comparable&lt;Employee&gt; {

 public int compareTo(Object o) {
 if (o instanceof Employee) {
 Employee emp = (Employee) o;
 return this.id - emp.id;
 }else {
 throw new IllegalArgumentException("Parameter not of type Employee!");
 }
 return null;
 }
}
</code></pre>

I think thats ok. What's the problem if id is negative?

blocks|key|976774|text|compareTo()也是比较字符串时使用的保留函数。所以，也许他想让你注意到，同名的方法可能会导致问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|976775|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

compareTo() is also a reserved function usesd when comparing strings. So maybe he wants you to note that methods with the same name can cause issues.

I am relatively new to Java and trying to solve some basic questions. One of my friends asked me to explain what can go wrong with this piece of code.
To me it looks perfectly fine.
I even tried it out and it works.

Can you guys help me out with it? 

<pre><code>public int compareTo(Object o){
 Employee emp = (Employee) o;
 return this.id - o.id;
}
</code></pre>

I could not find any good explanations for it.
Also id is an integer.

Java compareTo method basics

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我对Java比较陌生，正在尝试解决一些基本问题。我的一个朋友请我解释一下这段代码会出什么问题。对我来说，它看起来非常好。我甚至试过了，它很管用。你们能帮帮我吗？public int compareTo(Object o){   Employee emp = (Employee) o;   return this.id - o.id;}我找不到任何好的解释。id也是一个整数。

问Java compareTo方法基础
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