为了弄清为什么重写equals()方法时,必须重写hashCode()方法,我们首先需要明确Object实现hashCode()返回值是什么?
/**
* 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();
哈希码的通用约定如下:
- 在java程序执行过程中,在一个对象没有被改变的前提下,无论这个对象被调用多少次,hashCode()方法都会返回相同的整数值。对象的哈希码没有必要在不同的程序中保持相同的值。
- 如果两个对象使用equals()方法进行比较是等价的,那么这两个对象的hashCode()方法必须返回相同的值。
- 如果两个对象使用equals()方法进行比较是不等价的,那么这两个对象的hashCode()方法返回值不必相同。但是,不等价的对象的hashCode()值不同的话可以提高哈希表的性能。
hashCode()方法是一个native方法,它返回的是由对象存储地址转化得到的值。
若重写equals()方法而不重写hashCode()方法,当两个对象使用equals()方法进行比较是等价的时,两个对象的hashCode()方法返回值是不想等的(两个对象的存储地址是不相同的),违背了哈希码通用约定第二条。
下面的代码中,新建了两个等价的对象person1和person2,并将person1作为key存入HashMap中。我们希望将这两个对象当成一个key,在HashMap中取person2的时候可以返回键person1的value。由于只重写了equals()方法,而未重写hashCode()方法,实际返回值为null。
public class Person {
private int id;
private String name;
public Person(int id, String name) {
this.id = id;
this.name = name;
}
@Override
public boolean equals(Object obj) {
if (this.id == ((Person)obj).id)
return true;
return false;
}
}
public class Test {
public static void main(String[] args) {
Person person1 = new Person(1, "Lisa");
Person person2 = new Person(1, "Lisa");
HashMap<Person, Integer> persons = new HashMap<>();
persons.put(person1, 10);
System.out.println(persons.get(person2));
}
}
null
String同时重写了equals()方法和hashCode()方法
public class Test {
public static void main(String[] args) {
HashMap<String, Integer> strs = new HashMap<>();
String str1 = new String("ddd");
String str2 = new String("ddd");
strs.put(str1, 10);
System.out.println(strs.get(str2));
}
}
10
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