为什么重写 equals() 要重写 hashCode()? hashCode 值相等,两个对象不一定相等?
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2024-03-22 17:15:22
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为什么重写 equals() 要重写 hashCode()? hashCode 值相等,两个对象不一定相等?
hashCode 方法,如果不重写的话,返回的实际上是该对象在 jvm 的堆上的地址,而不同对象的地址肯定不同,所以这个 hashCode 也就肯定不同了。如果重写了的话,由于采用的算法的问题,有可能导致两个不同对象的 hashCode 相同。
equals 方法,如果不重写的话,是严格判断一个对象是否相等的方法(object1 == object2)
看了一些博客,感觉不靠谱,虽然我的可能也会出现小问题,但会尽量追求证实,来看源码。
java.lang.Object.equals() 方法
原版:
/**
* 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);
}
翻译版:
/**
* 表明其他对象是否“等于”此对象.
* <p>
* equals 方法对非空对象引用实现等价关系:
* <ul>
* <li>
* 自反的: 对于任何非空引用值 x, 对于 x.equals(x) 应该 return true
* <li>
* 对称的: 对于任何非空引用值 x, y, 对于 x.equals(y) 语句 return true, 当且仅当 y.equals(x) 也 returns true 是成立时,
* <li>
* 可传递的: 对于任何非空引用值 x, y, z, 如果
* x.equals(y) returns true 并且 y.equals(z) returns true,则
* x.equals(z) 应该 return true
* <li>
* 一致的: 对于任何非空引用值 x, y, 多次调用 x.equals(y) 始终如一地 return true
* 或者始终如一地 return false, 如果在 equals 方法在对象比较中未使用任何信息,则将修改这些对象的比较
* <li>
* 对于任何非空引用值 x, 对于 x.equals(null) 应该 return false
* </ul>
* <p>
* Object 类的 equals 方法在对象上实现最具辨别力的可能等价关系;即, 对于任何非空引用值 x, y, 该方法 returns true 当且仅当 x, y 指向相同的 object。此时 x == y 的值也是 true
* <p>
* !!!请注意,每当重写此 equals() 方法时, 重写 hashCode() 方法通常是必要的,
* 以保持 hashcode() 方法的常规的约定,其中规定相等的对象必须具有相等的哈希代码
*
* @param obj 要与之比较的引用对象
* @return 如果是同一个对象返回 true 否则返回 false
* @see #hashCode()
* @see java.util.HashMap
*/
public boolean equals(Object obj) {
return (this == obj);
}
最后一句话说明了
java.lang.Object.equals() 方法
原版:
/**
* 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();
翻译版:
/**
* 返回对象的哈希代码值。此方法支持哈希表,例如 java.util.HashMap
* <p>
* 关于 hashCode() 常规的约定是:
* <ul>
* <li>
* 每当在 Java 应用程序执行过程中在同一对象上多次调用它时, hashCode() 必须始终如一地 return 相同的整数,
* 如果在 equals() 在对象比较中未使用任何信息,则该对象的比较将被修改。
* 从应用程序的一次执行到同一应用程序的另一次执行,这个整数不需要保持一致。
* <li>
* 如果两个对象根据 equals(object) 方法相等,则对两个对象中的每个对象调用 hashcode() 方法
* 都必须产生相同的整数结果。
* <li>如果 java.lang.Object#equals(java.lang.Object) 方法中两个对象不相等,则调用两个对象,然后调用两个对象中的每一个对象的 HASCODE() 方法必须产生不同的整数结果。但是,程序员应该知道,为不同的对象生成不同的整数结果可能会提高哈希表的性能。
* </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();