被问到这个问题的时候,压根是根本没去看这玩意,没关系,直接说不知道,不用在意,之后我们直接去JDK文档中直接看源码,10分钟必懂。
一、源码
所有类的祖宗(超类)Object
:
/**
* Returns a hash code value for the object. This method is
* supported for the benefit of hash tables such as those provided by
* {@link java.util.HashMap}.
* <p>
* The general contract of {@code hashCode} is:
* <ul>
* <li>Whenever it is invoked on the same object more than once during
* an execution of a Java application, the {@code hashCode} method
* must consistently return the same integer, provided no information
* used in {@code equals} comparisons on the object is modified.
* This integer need not remain consistent from one execution of an
* application to another execution of the same application.
* <li>If two objects are equal according to the {@code equals(Object)}
* method, then calling the {@code hashCode} method on each of
* the two objects must produce the same integer result.
* <li>It is <em>not</em> required that if two objects are unequal
* according to the {@link java.lang.Object#equals(java.lang.Object)}
* method, then calling the {@code hashCode} method on each of the
* two objects must produce distinct integer results. However, the
* programmer should be aware that producing distinct integer results
* for unequal objects may improve the performance of hash tables.
* </ul>
* <p>
* As much as is reasonably practical, the hashCode method defined by
* class {@code Object} does return distinct integers for distinct
* objects. (This is typically implemented by converting the internal
* address of the object into an integer, but this implementation
* technique is not required by the
* Java™ programming language.)
*
* @return a hash code value for this object.
* @see java.lang.Object#equals(java.lang.Object)
* @see java.lang.System#identityHashCode
*/
public native int hashCode();
/**
* Indicates whether some other object is "equal to" this one.
* <p>
* The {@code equals} method implements an equivalence relation
* on non-null object references:
* <ul>
* <li>It is <i>reflexive</i>: for any non-null reference value
* {@code x}, {@code x.equals(x)} should return
* {@code true}.
* <li>It is <i>symmetric</i>: for any non-null reference values
* {@code x} and {@code y}, {@code x.equals(y)}
* should return {@code true} if and only if
* {@code y.equals(x)} returns {@code true}.
* <li>It is <i>transitive</i>: for any non-null reference values
* {@code x}, {@code y}, and {@code z}, if
* {@code x.equals(y)} returns {@code true} and
* {@code y.equals(z)} returns {@code true}, then
* {@code x.equals(z)} should return {@code true}.
* <li>It is <i>consistent</i>: for any non-null reference values
* {@code x} and {@code y}, multiple invocations of
* {@code x.equals(y)} consistently return {@code true}
* or consistently return {@code false}, provided no
* information used in {@code equals} comparisons on the
* objects is modified.
* <li>For any non-null reference value {@code x},
* {@code x.equals(null)} should return {@code false}.
* </ul>
* <p>
* The {@code equals} method for class {@code Object} implements
* the most discriminating possible equivalence relation on objects;
* that is, for any non-null reference values {@code x} and
* {@code y}, this method returns {@code true} if and only
* if {@code x} and {@code y} refer to the same object
* ({@code x == y} has the value {@code true}).
* <p>
* Note that it is generally necessary to override the {@code hashCode}
* method whenever this method is overridden, so as to maintain the
* general contract for the {@code hashCode} method, which states
* that equal objects must have equal hash codes.
*
* @param obj the reference object with which to compare.
* @return {@code true} if this object is the same as the obj
* argument; {@code false} otherwise.
* @see #hashCode()
* @see java.util.HashMap
*/
public boolean equals(Object obj) {
return (this == obj);
}
看完注释,巴拉巴拉一大片,最终重写以上方法的时候需要满足以下通用契约:
设此处有非空引用x
,y
,z
1.hashCode()
- 返回此引用对象的
哈希码值
(通过本地方法将对象地址转化为整数) - 幂等性执行(同一个环境程序中执行多次,始终如一返回相同整数)
- 对于
x.equals(y)
返回true
,两个引用对象的hashCode
也必须产生相同的结果 - 对于
x.equals(y)
返回false
,两个运用对象的hashCode
不必一定产生不同的结果;但是为了哈希表的性能,开发者最好需要去产生不同的hashCode
(这里后面会讲到)
2.equals(Object obj)
- 通过
==
来比较两个引用是否引用的同一个对象(此比较符比较的是对象真正的内存地址) -
reflexive(反身性):
x.equals(x)
需要返回true
-
symmetric(对称性):
x.equals(y)
返回true
,则y.equals(x)
需要返回true
-
transitive(传递性):
x.equals(y)
返回true
,y.equals(z)
返回true
,则x.equals(z)
返回true
-
consistent(一致性):
x.equals(y)
返回true
或false
,多次调用,结果不会发生改变。(引用的对象属性不会发生改变的情况下) -
x.equals(null)
必须返回false - 当
x
,y
引用的是同一个对象的时候,即x==y
为true
,则必须返回true - 如果
x.equals(y)
返回true
,那么hashCode
值必须相等
二、为什么重写equals就需要重写hashCode
通过以上的了解,equals
方法默认是比较是否是引用的同一个内存地址,从而判断true
或false
,所以当我们自定义类出现的时候,我们需要去重写equlas
才能满足业务中对于对象是否属于同一个的判定:
public class Person{
private String idCard;
private String name;
public Person(String idCard, String name){
this.idCard = idCard;
this.name = name;
}
}
Person zhangsan1 = new Person("123","张三");
Person zhangsan2 = new Person("123","张三");
System.out.println(zhangsan1.equlas(zhangsan2));
在没有重写equals
的情况下,结果显而易见false
。但是按照业务上来说,这两个对象是属于同一个的,他们的身份证和姓名一样,即他们就是同一个人,所以我们要重写equals
:
public boolean equals(Object obj) {
if(this == obj) return true;
if(this.idCard.equlas(obj.idCard) && this.name.equlas(obj.name)){
return ture;
}
}
那好,重点来了,这时候已经可以判定对象是否相等了,为啥还要重写hashCode
呢?
请注意看源代码,其中hashCode
为hashMap
提供了支持哈希表
。
这里详情就不说了,可以查看HashMap,大概的意思就是通过hashCode
计算每个元素数组的下标值(因为可以通过哈希码值
实现精准定位),所以需要重写来避免业务上对象相同但是默认方法哈希码值
值却不相同而造成对象不在同一个位置上的错误(这一点也是符合源码规范定义的,也是需要重写的原因体现)。
三、总结
重写equlas
方法是业务需要,重写hashCode
方法是代码逻辑上的必须,是它处使用到了hashCode
和equals
要按照规范来实现,使其结果保证规范上的一致性正确性。
大白话就是说:这两个方法就是为了判断对象是否相等用的,代码怎么实现随意,但是结果必须要保证和源码上注释所描述的规范要一样。
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