How does set ensures uniqueness

[Raj Kumar Bindal]

I tried looking code for Set and Hashset ,but from that i could not find out how set ensures duplicates elements are not present in it.Can someone tell.

[Taariq San]

Generally the javadoc is easier to read than the code.

add

public boolean add(E e)

Adds the specified element to this set if it is not already present. More formally, adds the specified element e to this set if this set contains no element e2 such that (e==null ? e2==null : e.equals(e2)). If this set already contains the element, the call leaves the set unchanged and returns false.

That's from HashSet.
Because HashSet allows a null element, it checks if there's already a null element if the new one is null. If it's not null, it uses the equals() method found in Object. That implementation returns true if the 2 references are the same and typically a class would override this method and check if the 2 instances are meaningfully equal.

Say your class SomeObject has a String or 2, it might compare each of those Strings using String's .equals() implementation.

[Campbell Ritchie]

HashSet uses a Map implementation, which I think is a HashMap; it inserts your value as the "K" in the Map and a "dummy" Object instance as the "V"; I think it is the same "V" throughout.
All you have to do in this instance is go through the Map and see whether it already contains the "K". If it hasn't, add it and return true; if it's already there return false.

[Ilja Preuss]

And the relevant code from HashMap.put is this (as of JDK 1.6):

if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
  V oldValue = e.value;
  e.value = value;
  e.recordAccess(this);
  return oldValue;
}

[Campbell Ritchie]

Originally posted by Ilja Preuss:
And the relevant code from HashMap.put is this . . .

You can see it doesn't alter the key; in the case of a HashSet, the value is always a reference to the same "dummy" Object, so that doesn't actually change either. Its correct operation is however dependent on the equals() and hashCode() methods being overridden correctly.