The hashCode()<\/strong>\u00a0method of Object is used anytime we insert the\u00a0object into a HashMap, HashTable, or HashSet. The hashCode() method effectively generates an index to map objects to.<\/p>\n To over simplify<\/span>, think of a hashtable’s\u00a0main data structure as an array; then if we call hashTable.put(key<\/em>, \u00a0value<\/em>) the HashTable\u00a0will call key<\/em>.hashCode() to figure out the index to store the object key<\/em> in the array.<\/p>\n Just like the contract\u00a0for equals() that keep us honest, we have a contract\u00a0for hashCode() as well. But let’s pay careful attention to the wording here because it will help us understand what is really going on. I have truncated a couple of the rules so that we can focus on the meat of it. The truncated rules end with an ellipsis (…), the full rules are available here<\/a>:<\/p>\n If we look again at\u00a0the somewhat awkwardly worded #3, we see that it says “it is acceptable that two completely different objects produce the same hashcode”. This situation is called a hash collision and is completely legal. How is this possible? How do we find objects in the hashtable if different objects can map to the same location? To illustrate how this works, let’s extend the conceptual model of the hashtable we started above.<\/p>\n I previously over simplified a \u00a0hashtable’s main data structure as an array. Let’s continue with this simplification and say that each index of the array stores a Linked List of objects. This means that if we have 3 objects that map to index\u00a05 of the hashtable’s array, then index 5 of the array will have a Linked List with 3\u00a0nodes; one for each object mapped to that location.<\/p>\n We now have a hashtable construct in our minds that looks like the following- Notice that we have a hashcode collision and the Cat object mapped to the same location as the Dog objects:<\/p>\n\n
![\"hash_function2\"](\"http:\/\/jeffcardillo.com\/blog\/wp-content\/uploads\/2015\/01\/hash_function21.png\")
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