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

String类和包装器类的equals()和hashCode()方法实现与Object类不同。Object类的equals()方法比较对象的引用,而不是内容。Object类的hashCode()方法为每个对象返回不同的hashCode,无论内容是否相同。

It leads problem when you use Map collection and the key is of Persistent type, StringBuffer/builder type. Since they don't override equals() and hashCode() unlike String class, equals() will return false when you compare two different objects even though both have same contents. It will make the hashMap storing same content keys. Storing same content keys means it is violating the rule of Map because Map doesnt allow duplicate keys at all. Therefore you override equals() as well as hashCode() methods in your class and provide the implementation(IDE can generate these methods) so that they work same as String's equals() and hashCode() and prevent same content keys.

你必须重写hashCode()方法和equals(),因为equals()根据hashCode工作。

此外,与equals()一起重写hashCode()方法有助于完好无损equals()-hashCode()契约:“如果两个对象相等,那么它们必须具有相同的哈希码。”

什么时候需要为hashCode()编写自定义实现?

正如我们所知,HashMap的内部工作是基于哈希原理的。条目集存储在特定的存储桶中。您可以根据自己的需求定制hashCode()实现,以便相同的类别对象可以存储在相同的索引中。 当你使用put(k,v)方法将值存储到Map集合中时,put()的内部实现是:

put(k, v){
hash(k);
index=hash & (n-1);
}

意思是,它生成索引,索引是基于特定键对象的hashcode生成的。所以让这个方法根据你的需求生成哈希码,因为相同的哈希码条目集将存储在相同的桶或索引中。

就是这样!

其他回答

class A {
    int i;
    // Hashing Algorithm
    if even number return 0 else return 1
    // Equals Algorithm,
    if i = this.i return true else false
}

put('key','value')将使用hashCode()计算哈希值来确定 桶,并使用equals()方法查找该值是否已经 出现在桶里。如果不是,它将被添加,否则它将被替换为当前值 get('key')将使用hashCode()首先找到条目(桶) equals()来查找Entry中的值

如果两者都被覆盖,

地图<A>

Map.Entry 1 --> 1,3,5,...
Map.Entry 2 --> 2,4,6,...

If =没有被覆盖

地图<A>

Map.Entry 1 --> 1,3,5,...,1,3,5,... // Duplicate values as equals not overridden
Map.Entry 2 --> 2,4,6,...,2,4,..

如果hashCode没有被覆盖

地图<A>

Map.Entry 1 --> 1
Map.Entry 2 --> 2
Map.Entry 3 --> 3
Map.Entry 4 --> 1
Map.Entry 5 --> 2
Map.Entry 6 --> 3 // Same values are Stored in different hasCodes violates Contract 1
So on...

HashCode等价契约

根据equal方法,两个相等的键应该生成相同的hashCode 生成相同hashCode的两个key不需要相等(在上面的例子中,所有偶数生成相同的hashCode)

考虑在一个桶中收集所有黑色的球。你的工作是像下面这样给这些球上色,并将其用于适当的游戏,

对于网球-黄色,红色。 板球-白色

现在水桶有三种颜色的球黄色,红色和白色。只有你知道哪个颜色适合哪个游戏。

给球上色-哈希。 选择比赛的球-平等。

如果你给球上色,然后有人选了板球或网球,他们不会介意颜色的!!

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:-(

Joshua Bloch在Effective Java上说

必须在重写equals()的每个类中重写hashCode()。如果不这样做,将违反Object.hashCode()的一般约定,这将阻止您的类与所有基于哈希的集合(包括HashMap、HashSet和Hashtable)一起正常运行。

让我们通过一个例子来理解它,如果我们重写equals()而不重写hashCode()并尝试使用Map会发生什么。

假设我们有这样一个类,如果MyClass的两个对象的importantField相等(使用eclipse生成的hashCode()和equals()),则MyClass的两个对象相等

public class MyClass {
    private final String importantField;
    private final String anotherField;

    public MyClass(final String equalField, final String anotherField) {
        this.importantField = equalField;
        this.anotherField = anotherField;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result
                + ((importantField == null) ? 0 : importantField.hashCode());
        return result;
    }

    @Override
    public boolean equals(final Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        final MyClass other = (MyClass) obj;
        if (importantField == null) {
            if (other.importantField != null)
                return false;
        } else if (!importantField.equals(other.importantField))
            return false;
        return true;
    }
}

想象一下你有这个

MyClass first = new MyClass("a","first");
MyClass second = new MyClass("a","second");

只覆盖等于

如果只有equals被覆盖,那么当你调用myMap.put(first,someValue)首先将哈希到某个桶,当你调用myMap.put(second,someOtherValue)它将哈希到其他一些桶(因为它们有不同的hashCode)。所以,尽管它们是相等的,因为它们不散列到同一个桶,映射无法意识到这一点,它们都留在映射中。


虽然如果重写hashCode()就没有必要重写equals(),但让我们看看在这种特殊情况下会发生什么:我们知道MyClass的两个对象是相等的,如果它们的importantField相等,但我们没有重写equals()。

只覆盖hashCode

如果你只覆盖hashCode,那么当你调用myMap.put(first,someValue)时,它首先接受,计算它的hashCode并将其存储在给定的bucket中。然后,当您调用myMap.put(second,someOtherValue)时,它应该根据Map文档将first替换为second,因为它们是相等的(根据业务需求)。

但问题是,equals没有被重新定义,所以当map哈希second并遍历bucket时,寻找是否有一个对象k,使得second.equals(k)为真,它不会找到任何对象,因为second.equals(first)将为假。

希望这是清楚的

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.)