Esiste un modo rapido "computazionalmente" per ottenere il conteggio di un iteratore?
int i = 0;
for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();
... sembra uno spreco di cicli della CPU.
to provide an implementation-independent method for access, in which the user does not need to know whether the underlying implementation is some form of array or of linked list, and allows the user go through the collection without explicit indexing. penguin.ewu.edu/~trolfe/LinkedSort/Iterator.html