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Java Forum / General / December 2007about equals() and hashCode()?
the java API says that:"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." why?i wonder the root cause of that. > the java API says that:"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." > why?i wonder the root cause of that. Because hashCode() is used in HashMap and Hashtable to store the element in a bucket. If two elements are considered equal, then adding one to a HashMap and then removing the other, should remove the first one added (because, as far as HashMap knows, they are the same object). If they have different hashCode values, then the remove fails to find the bucket where the first object added is stored, and cannot remove it. /L  SignatureLasse Reichstein Nielsen - lrn@hotpop.com DHTML Death Colors: <URL:http://www.infimum.dk/HTML/rasterTriangleDOM.html> 'Faith without judgement merely degrades the spirit divine.' > the java API says that:"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." > why?i wonder the root cause of that. Because it is needed to make hashCode useful. Take a look at the source code for e.g. HashMap. It assigns the keys to buckets based on their hash codes. During a get call, it calculates the bucket for the probe key's hash code, and only looks there. If two objects with different hash codes could be equal, it would have to look in every bucket. Patricia >> the java API says that:"If two objects are equal according to the >> equals(Object) method, then calling the hashCode method on each of the [quoted text clipped - 8 lines] > objects with different hash codes could be equal, it would have to look > in every bucket. Looking at what Patricia said from a different angle - the very purpose of hashCode() is to speed up determination of equality. If you break the connection between hashCode() and equals(), you do not have a useful hashCode(). So what would be the point of breaking it? SignatureLew > the java API says that:"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." > why?i wonder the root cause of that. When checking a collection to find an element equal to another, the system uses hashCode to quickly eliminate elements that can't be equal. It is legal to set hashCode to return a constant, say 1. If so your collections will exhibit poorer performance. Note the inverse is not necessarily true; if the hashCodes are the same, you still need to check for equality. Java has contracts like this that are not checked by a compiler; it is up to the programmer to ensure that when you over-ride equals, you make sure hashCode is acceptable (usually when you over-ride one, you over-ride the other). -Wayne Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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