Java hashCode 和 equals()

1 Object中定义的hashCode()

public int hashCode()

Returns a hash code value for the object. This method is supported for the benefit of hash tables such as those provided by HashMap.

The general contract of hashCode is:

• Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in 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.
• If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.
• It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the 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.

JDK中的帮助文档，hashcode的使用约定是:

• 同一个java程序的一次执行过程中，返回值必须一个；不同次执行，返回值可以不同。//这里可以理解为地址了
• 如果两个对象equals()，那么hashcode()必须返回相同的值。//此时的equals()函数被覆盖了，当两个对象认为是同一个的时候，那么hascode必须相同。用在hashmap中。
• 如果两个对象的hashcode()相同，那么equals()可以不等。//如在hashmap中，两个不同的对象有相同的hashcode()，存在同一个槽位里面。
• 如果两个对象不等equals(), 那么hashcode()可以相同，就是上面这句反过来说。
• 如果两个对象的hashcode()不同，那么他们不能相等。//当然你可以重载为相同，但是hashmap无法使用了。

As much as is reasonably practical, the hashCode method defined by class 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^TM programming language.)

不覆盖Object的hashCode()，那么就可以理解为返回内存地址。

2 equals()

public boolean equals(Object obj)

Indicates whether some other object is "equal to" this one.

The equals method implements an equivalence relation on non-null object references:

• It is reflexive: for any non-null reference value x, x.equals(x) should return true.
• It is symmetric: for any non-null reference values x and y, x.equals(y) should return true if and only if y.equals(x) returns true.
• It is transitive: for any non-null reference values x, y, and z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.
• It is consistent: for any non-null reference values x and y, multiple invocations of x.equals(y) consistently return true or consistently return false, provided no information used in equals comparisons on the objects is modified.
• For any non-null reference value x, x.equals(null) should return false.
• 这个定义是在是太严谨，太科学了，以至于都不愿意看它。其实翻译成大白话多好理解啊。

The equals method for class Object implements the most discriminating possible equivalence relation on objects; that is, for any non-null reference values x and y, this method returns true if and only if x and y refer to the same object (x == y has the value true).

这里讲了equals和==的关系。 x==y的意思就是二者的地址相同，如果equals不覆盖，那么==和equals一样。如果equals覆盖了，通常==指地址相等，equals指的是内容相同。

如String。

String str = "hello"; String str2 = new String("hello"); 问二者什么关系？==？equals()? 详细解释见Java String

Note that it is generally necessary to override the hashCode method whenever this method is overridden, so as to maintain the general contract for the hashCode method, which states that equal objects must have equal hash codes.

这里讲了和hashCode的关系。其实很简单，hashCode是为hashmap用的。假如equals，必须hashCode相同，否则hashmap无法使用了。

在Object中，equals和hashcode都是和地址相关的. 如果重载equals，是的二者内容相同。如果不重载hashcode，那么二者的hashcode和地址相关而不是内容相关，那么hashcode为不同值。导致hashmap无法使用。所以必须把hashcode也改为相似的语义。

3 String的hashCode()和equals()定义

当入参anObject不是null，且类型为String实例，长度相同，value一模一样的时候才返回true
 public boolean equals(Object anObject) {
        if (this == anObject) {
            return true;
        }
        if (anObject instanceof String) {
            String anotherString = (String)anObject;
            int n = value.length;
            if (n == anotherString.value.length) {
                char v1[] = value;
                char v2[] = anotherString.value;
                int i = 0;
                while (n-- != 0) {
                    if (v1[i] != v2[i])
                        return false;
                    i++;
                }
                return true;
            }
        }
        return false;
    }

//hashcode = s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1] 

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;
    }