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   從Object類(lèi)的hashCode()和equals()方法講起：   

      最近看了Object類(lèi)的源碼，對hashCode() 和equals()方法有了更深的認識。重寫(xiě)equals()方法就必須重寫(xiě)hashCode()方法的原因，從源頭Object類(lèi)講起就更好理解了。

先來(lái)看Object關(guān)于hashCode()和equals()的源碼：

    public native int hashCode();

    public boolean equals(Object obj) {
        return (this == obj);
    }

/**
     * 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();

1.在 Java 應用程序執行期間，在對同一對象多次調用 hashCode 方法時(shí)，必須一致地返回相同的整數，前提是將對象進(jìn)行 equals 比較時(shí)所用的信息沒(méi)有被修改。從某一應用程序的一次執行到同一應用程序的另一次執行，該整數無(wú)需保持一致。    
2.如果根據 equals(Object) 方法，兩個(gè)對象是相等的，那么對這兩個(gè)對象中的每個(gè)對象調用 hashCode 方法都必須生成相同的整數結果。    
3.如果根據 equals(java.lang.Object) 方法，兩個(gè)對象不相等，那么對這兩個(gè)對象中的任一對象上調用 hashCode 方法不 要求一定生成不同的整數結果。但是，程序員應該意識到，為不相等的對象生成不同整數結果可以提高哈希表的性能。

     再簡(jiǎn)單的翻譯一下第二三點(diǎn)就是：hashCode()和equals()保持一致，如果equals方法返回true，那么兩個(gè)對象的hasCode()返回值必須一樣。如果equals方法返回false，hashcode可以不一樣，但是這樣不利于哈希表的性能，一般我們也不要這樣做。重寫(xiě)equals()方法就必須重寫(xiě)hashCode()方法的原因也就顯而易見(jiàn)了。

   假設兩個(gè)對象，重寫(xiě)了其equals方法，其相等條件是屬性相等，就返回true。如果不重寫(xiě)hashcode方法，其返回的依然是兩個(gè)對象的內存地址值，必然不相等。這就出現了equals方法相等，但是hashcode不相等的情況。這不符合hashcode的規則。下邊，會(huì )介紹在集合框架中，這種情況會(huì )導致的嚴重問(wèn)題。

    重寫(xiě)的作用：

     用HashSet來(lái)驗證兩個(gè)需都重寫(xiě)的必要性

public class A {
    @Override
    public boolean equals(Object obj) {
        return true;
    }
}

public class B {
    @Override
    public int hashCode() {
        return 1;
    }
}

public class C {
    @Override
    public int hashCode() {
        return 2;
    }
    @Override
    public boolean equals(Object obj) {
        return true;
    }
}

public class HashSetTest {
    public static void main(String[] args) {
        HashSet hashSet = new HashSet();
        hashSet.add(new A());
        hashSet.add(new A());
        hashSet.add(new B());
        hashSet.add(new B());
        hashSet.add(new C());
        hashSet.add(new C());
        for (Object hs : hashSet) {
            System.out.println(hs);
        }
        //HashSet重寫(xiě)了toString()方法
//        System.out.println(hashSet);
    }
}

cn.edu.uestc.collection.B@1
cn.edu.uestc.collection.B@1
cn.edu.uestc.collection.C@2
cn.edu.uestc.collection.A@3f84246a
cn.edu.uestc.collection.A@18a9fa9c
Process finished with exit code 0

     重寫(xiě)hashCode()的原則

//f是Filed屬性
boolean    hashCode=(f?0:1)
(byte,short,char,int)      hashCode=(int)f
long       hashCode=(int)(f^(f>>>32))
float       hashCode=Float.floatToIntBits(f)
double   hashCode=(int)(1^(1>>>32))
普通引用類(lèi)型    hashCode=f.hashCode()

return  f1.hashCode()*17+(int)f2.13

  /**
     * Returns a hash code for this string. The hash code for a
     * <code>String</code> object is computed as
     * <blockquote><pre>
     * s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
     * </pre></blockquote>
     * using <code>int</code> arithmetic, where <code>s[i]</code> is the
     * <i>i</i>th character of the string, <code>n</code> is the length of
     * the string, and <code>^</code> indicates exponentiation.
     * (The hash value of the empty string is zero.)
     *
     * @return  a hash code value for this object.
     */
    public int hashCode() {
        int h = hash;
        if (h == 0 && value.length > 0) {
            char val[] = value;
            for (int i = 0; i < value.length; i++) {
                h = 31 * h + val[i];
            }
            hash = h;
        }
        return h;
    }