CODEWITHSUNDEEP
javaiterator
黄兴海
黄兴海

Reputation: 11

Does the 'hasNext' must be called before the 'next' if the size is known in java?

If the size of list is known, only call the 'next' without 'hasNext'. Is it right?

final List<Integer> list = [1, 2, 3];
final Iterator<Integer> iter = list.iterator();

for(int i = 0; i < list.size(); ++i){
   System.out.println(iter.next());
}

Upvotes: 1

Views: 1201

Answers (2)

jacobm
jacobm

Reputation: 14025

Yes, that's allowed. In your example, there isn't much point to it (you'd be better off either using list.get(i) or using hasNext()/next()), but it doesn't break anything.

One place where you do see next() without hasNext() is to get an arbitrary item out of a collection that is known not to be empty. This idiom comes up sometimes:

Collection<T> myCollection = ...;
if (!myCollection.isEmpty()) {
  return myCollection.iterator().next();
}

Upvotes: 3

Kartik
Kartik

Reputation: 7917

If there is no next element and you still call next(), you'll get NoSuchElementException. To protect against this, you need to do a pre-check using hasNext().

In your example, you already know the size and the condition i < list.size() is guarding you against trying to jump after the last element, so there is no point calling hasNext().

We generally do:

while (iterator.hasNext()) { //protection against jumping after the last element
    //call next()
}

You have done a similar thing, just the "protection" is a bit different (but valid):

for(... i < list.size() ...) { //"i < list.size()" is providing that protection
    //call next()
}

So no need to use hasNext() here.

Upvotes: 3

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