HashSet或HashMap沒有在新類中定義hashCode（）方法

Question

• 如果您設計了一個新類並嘗試將該類的對象插入到HashSet或HashMap而未定義hashCode（）方法，會發生什麼？

請保持說明容易。我正在學習考試，而且我仍然在Java中使用哈希來生鏽。謝謝。

Answer 1

不會發生任何事情:-)

的每個對象都有自己的hashCode從對象類繼承（）方法。所以，你的每一個新對象都是獨一無二的。通過自己，它們將被HashSet或HashMap識別爲唯一。

下面是官方的意見：

/** 
* 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&trade; 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(); 

Answer 2

一個HashMap將數據存儲到項（也稱爲桶或桶）的多單鏈表。所有列表都在Entry（Entry []數組）中註冊

下圖顯示了具有可爲空條目的數組的HashMap實例的內部存儲。每個條目可鏈接到另一個條目以形成鏈接列表。

當用戶調用put（K key，V value）或get（Object key）時，該函數將計算Entry應該在的存儲桶的索引。

桶的索引（鏈表）使用密鑰的哈希碼生成。 因此，如果您已經重寫了hashCode方法，它將使用重寫方法來計算存儲區的索引 否則將使用默認哈希碼，這是您的對象的內存地址。所以在那種情況下，即使你的對象是你的地圖上會有一個新的條目。所以即使你試圖存儲邏輯上相同的對象。他們將被重新編號爲不同的哈希映射。

儘管合理實用，類Object定義的hashCode方法確實爲不同的對象返回不同的整數。 （這一般是通過將該對象的內部地址轉換成一個整數來實現的，但不是由的JavaTM編程語言不需要這種實現技巧。）

例如：

MyObject a = new MyObject("a", 123,"something"); 
MyObject b = new MyObject("a", 123,"something"); 
a and b will have different hashcodes.