Thread: Why don't I get all the elements in my TreeSet?

 
 public void createPersons()
    {
        Random rand = new Random();
        for(int index = 0; index < persons.length; index++) {
            Person person = new Person(rand.nextInt(maxAge), persons[index]);
            setOfPersons.add(person);
        }
    }
 
public void printPersons()
    {
        for(Iterator<Person> r = setOfPersons.iterator(); r.hasNext();) {
            System.out.println(r.next().getName() + " is " + r.next().getAge() + " years old");
        }
        System.out.println(setOfPersons.size());
    }

setOfPersons is a TreeSet, persons[] is an array of strings that contains 10 names. Why printPersons() prints only the half names of setOfPersons?
Note. I do get the size of the TreeSet as 10...

Typically, a while() loop is used with an iterator:

Iterator<Person> r = setOfPersons.iterator();
while ( r.hasNext() ) { // code here }

Why the for() loop?

Because you call "next()" twice in your loop body.
Everytime you call "next()" of an iterator you will get one additional element. If you have 10 elements you can call this method exactly 10 times. If you call it twice for each iteration of your loop you can only have 5 iterations.
If you would put 11 elements into your set you would get an exception thrown.

Instead, do a for-each loop:

for (Person p : setOfPersons) {...}

Good catch, Cornix. I didn't look past the for() loop.

Thank you Cornix.... Greg I used the for loop because I saw it in an example in my book and I just liked the fact you can declare the Iterator inside the condition...I have seen that is more common to use a while loop for that, and until I see the example I did use only while loop. But i realized one more time that the best way to Iterate in case you dont remove elements from a list is the for each...

Something else I am struggling with: do you have any idea how how my setOfPersons will sort the Person objects by age?
from younger to older?

A set is usually not sorted (because thats not part of its definition) but perhaps there are some sorted set variations that I dont know of.
The best solution would be to use a List if you need order. You could create your own List which is also a Set if you specifically need the uniqueness of elements.