You cannot do it this way. Both t
and a
will have different hashCode()
values because the the java.lang.Array.hashCode()
method is inherited from Object
, which uses the reference to compute the hash-code (default implementation). Hence the hash code for arrays is reference-dependent, which means that you will get a different hash-code value for t
and a
. Furthermore, equals
will not work for the two arrays because that is also based on the reference.
The only way you can do this is to create a custom class that keeps the boolean
array as an internal member. Then you need to override equals
and hashCode
in such a way that ensures that instances that contain arrays with identical values are equal and also have the same hash-code.
An easier option might be to use List<Boolean>
as the key. Per the documentation the hashCode()
implementation for List
is defined as:
int hashCode = 1;
Iterator<E> i = list.iterator();
while (i.hasNext()) {
E obj = i.next();
hashCode = 31*hashCode + (obj==null ? 0 : obj.hashCode());
}
As you can see, it depends on the values inside your list and not the reference, and so this should work for you.