最近我读了这篇文章 开发人员工作文件。
该文档是关于有效和正确地定义hashCode()和equals(),但我无法弄清楚为什么我们需要覆盖这两个方法。
我如何决定有效地实现这些方法?
最近我读了这篇文章 开发人员工作文件。
该文档是关于有效和正确地定义hashCode()和equals(),但我无法弄清楚为什么我们需要覆盖这两个方法。
我如何决定有效地实现这些方法?
当前回答
它在使用值对象时很有用。以下摘自Portland Pattern Repository:
Examples of value objects are things like numbers, dates, monies and strings. Usually, they are small objects which are used quite widely. Their identity is based on their state rather than on their object identity. This way, you can have multiple copies of the same conceptual value object. So I can have multiple copies of an object that represents the date 16 Jan 1998. Any of these copies will be equal to each other. For a small object such as this, it is often easier to create new ones and move them around rather than rely on a single object to represent the date. A value object should always override .equals() in Java (or = in Smalltalk). (Remember to override .hashCode() as well.)
其他回答
你必须重写hashCode()在每个 重写equals()的类。失败 这样做会导致违反 总合同 Object.hashCode(),它将防止 你的类不能正常运行 结合所有基于哈希的 集合,包括HashMap, HashSet和Hashtable。 摘自Joshua Bloch的《Effective Java》
通过一致地定义equals()和hashCode(),可以提高类作为基于散列的集合中的键的可用性。正如hashCode的API文档所解释的那样:“支持此方法是为了受益于诸如java.util.Hashtable所提供的哈希表。”
关于如何有效地实现这些方法的问题,最好的答案是建议你阅读《Effective Java》的第3章。
Bah -“你必须在每个重写equals()的类中重写hashCode()。”
[出自Joshua Bloch的《Effective Java》?]
Isn't this the wrong way round? Overriding hashCode likely implies you're writing a hash-key class, but overriding equals certainly does not. There are many classes that are not used as hash-keys, but do want a logical-equality-testing method for some other reason. If you choose "equals" for it, you may then be mandated to write a hashCode implementation by overzealous application of this rule. All that achieves is adding untested code in the codebase, an evil waiting to trip someone up in the future. Also writing code you don't need is anti-agile. It's just wrong (and an ide generated one will probably be incompatible with your hand-crafted equals).
他们肯定应该在被写来用作键的对象上强制设置一个接口吗?无论如何,Object永远不应该提供默认的hashCode()和equals() imho。它可能鼓励了许多破碎的散列集合。
但无论如何,我认为“规则”是前后颠倒的。与此同时,我将继续避免使用“等号”进行相等性测试方法:-(
如果重写equals()而不是hashcode(),则不会发现任何问题,除非您或其他人在HashSet等散列集合中使用该类类型。 在我之前的人已经清楚地解释了很多次文献理论,我只是在这里提供一个非常简单的例子。
考虑一个类,它的equals()需要表示自定义的东西:-
public class Rishav {
private String rshv;
public Rishav(String rshv) {
this.rshv = rshv;
}
/**
* @return the rshv
*/
public String getRshv() {
return rshv;
}
/**
* @param rshv the rshv to set
*/
public void setRshv(String rshv) {
this.rshv = rshv;
}
@Override
public boolean equals(Object obj) {
if (obj instanceof Rishav) {
obj = (Rishav) obj;
if (this.rshv.equals(((Rishav) obj).getRshv())) {
return true;
} else {
return false;
}
} else {
return false;
}
}
@Override
public int hashCode() {
return rshv.hashCode();
}
}
现在考虑这个主类:-
import java.util.HashSet;
import java.util.Set;
public class TestRishav {
public static void main(String[] args) {
Rishav rA = new Rishav("rishav");
Rishav rB = new Rishav("rishav");
System.out.println(rA.equals(rB));
System.out.println("-----------------------------------");
Set<Rishav> hashed = new HashSet<>();
hashed.add(rA);
System.out.println(hashed.contains(rB));
System.out.println("-----------------------------------");
hashed.add(rB);
System.out.println(hashed.size());
}
}
这将产生以下输出:-
true
-----------------------------------
true
-----------------------------------
1
我对结果很满意。但是如果我没有覆盖hashCode(),它将导致噩梦,因为具有相同成员内容的Rishav对象将不再被视为唯一的hashCode将是不同的,因为由默认行为生成,这里将是输出:-
true
-----------------------------------
false
-----------------------------------
2
Java提出了一个规则
如果使用Object类的equals方法,两个对象相等,那么hashcode方法应该为这两个对象提供相同的值。
因此,如果在我们的类中重写equals(),我们也应该重写hashcode()方法来遵循此规则。 例如,equals()和hashcode()这两个方法都用于Hashtable中以键值对的形式存储值。如果我们覆盖其中一个而不是另一个,如果我们使用这样的对象作为键,哈希表可能不会像我们想要的那样工作。
The methods equals and hashcode are defined in the object class. By default if the equals method returns true, then the system will go further and check the value of the hash code. If the hash code of the 2 objects is also same only then the objects will be considered as same. So if you override only equals method, then even though the overridden equals method indicates 2 objects to be equal , the system defined hashcode may not indicate that the 2 objects are equal. So we need to override hash code as well.