Object源代碼及註釋
equals是Object的公有方法，那麼我們通常都會在自己的類中重寫這個equals方法，同時必須重寫hasCode方法，知道爲什麼重寫equals方法必須重寫hasCode方法呢？
/**
* 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);
}上面是Object對象中hasCode和equals的簽名和方法說明。
爲啥重寫equals方法
判斷一些對象是否等於當前對象，實現在非空對象映射的等價關係。this==obj,由此可知，這是一個引用對象的等價判斷，只有在this和obj是指向用一個應用時才返回true.這是判定實例的唯一性，我們通常使用equals時，其實並不想判斷實例的唯一性，而是想要比較裏面的幾個（自定義）屬性值是否相等，來判斷是否爲同一個"東西"。
契約：
reflexive(自反性)：x爲非null時，x.equals(x),一定返回true.
symmetric(對稱性):x,y都不爲null，如果x.equals(y)返回true,那麼y.equals(x).
transitive(傳遞性):x,y,z都不爲null,如果x.equals(y),y.equals(z)返回true,那麼x.equals(z)返回true.
consistent(一致性):對於任何非null的x\y，在沒有修改這些對象信息的前提下，多次調用下，x.equals(y)結果是一致的。
非null的x,x.equals(null)返回false.
下面是Book類的equals重寫實現，通過bookN、bookName確定是否爲同一本書（東西），而不是判斷兩個對象的引用是否爲同一個。
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Book book = (Book) o;
return Objects.equal(bookNo, book.bookNo) &&
Objects.equal(bookName, book.bookName);
}爲啥重寫hasCode方法
hasCode鍥約：
1.任何時候在沒有修改對象的前提下，多次調用同一個對象的hasCode方法，這個返回值必須是同一個Integer值。
2.如果兩個對象equals結果爲true，那麼hasCode的值必須相同。
3.如果兩個對象equals結果爲false,那麼hasCode就不一定不相同。
如果重寫equals方法時，不去重寫hasCode方法，那麼必然會違背第二條鍥約。所以每次hasCode必須跟隨equals重寫而重寫。
擴展知識
是不是有人就是問：如果就不重寫hasCode，不行嗎？就違背鍥約不可以嗎？只equals重寫有什麼問題嗎？
那麼這裏就會引申出hasCode在集合中的應用場景。
Java中的集合（Collection）有兩類，一類是List，再有一類是Set。前者集合內的元素是有序的，元素可以重複；後者元素無序，但元素不可重複。那麼這裏就有一個比較嚴重的問題了：要想保證元素不重複，可兩個元素是否重複應該依據什麼來判斷呢？這就是 Object.equals方法了。但是，如果每增加一個元素就檢查一次，那麼當元素很多時，後添加到集合中的元素比較的次數就非常多了。也就是說，如果集合中現在已經有1000個元素，那麼第1001個元素加入集合時，它就要調用1000次equals方法。這顯然會大大降低效率。
於是，Java採用了哈希表的原理。哈希算法也稱爲散列算法，是將數據依特定算法直接指定到一個地址上。可以這樣簡單理解，hashCode方法實際上返回的就是對象存儲位置的映像。
當集合中添加元素時，先調用hasCode方法，如果這個位置沒有元素，那麼就直接存儲在這裏，如果有元素，那麼再調用equals方法與新元素比較，如果相同不存了，如果不相同則說明發生碰撞（衝突），碰撞解決方法多樣，最終都會存儲在一個合適的位置。有了hasCode後，調用equals的次數大大降低，一般一兩次就搞定了。
擴展的內容有興趣可以自行深入學習。爲什麼系列採用簡短，易懂方式歸納總結一些工作中、面試中常常出現的問題進行解答。如果有錯誤請及時聯繫我，以免誤導他人。