I have 2 different instances of HashMap
I want to merge the keysets of both HashMaps;
Code:
Set<String> mySet = hashMap1.keySet();
mySet.addAll(hashMap2.keySet());
Exception:
java.lang.UnsupportedOperationException
at java.util.AbstractCollection.add(AbstractCollection.java:238)
at java.util.AbstractCollection.addAll(AbstractCollection.java:322)
I don't get a compile warning or error.
From java doc this should work. Even if the added collection is also a set:
boolean addAll(Collection c)
Adds all of the elements in the specified collection to this set if they're not already present (optional operation). If the specified collection is also a set, the addAll operation effectively modifies this set so that its value is the union of the two sets. The behavior of this operation is undefined if the specified collection is modified while the operation is in progress.
If you look at the docs of the HashMap#keySet()
method, you'll get your answer(emphasis mine).
Returns a Set view of the keys contained in this map. The set is backed by the map, so changes to the map are reflected in the set, and vice-versa. If the map is modified while an iteration over the set is in progress (except through the iterator's own remove operation), the results of the iteration are undefined. The set supports element removal, which removes the corresponding mapping from the map, via the Iterator.remove, Set.remove, removeAll, retainAll, and clear operations. It does not support the add or addAll operations.
Therefore, you need to create a new set and add all the elements to it, instead of adding the elements to the Set
returned by the keySet()
.
The result of keySet()
does not support adding elements to it.
If you are not trying to modify hashMap1
but just want a set containing the union of the two maps' keys, try:
Set<String> mySet = new HashSet<String>();
mySet.addAll(hashMap1.keySet());
mySet.addAll(hashMap2.keySet());
Doesn't support by nature of Set which is from map.keySet(). It supports only remove, removeAll, retainAll, and clear operations.
Please read documentation
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