CODEWITHSUNDEEP
javacompiler-errorsiterator
MMCXCVII
MMCXCVII

Reputation: 23

Iterator as an integer

As far as I know, an Iterator is an object which type is defined during its declaration but which also comes with two other methods: hasNext() and next(). So, besides those two methods, if I write Iterator<Integer> iterator, then iterator is supposed to behave like an Integer object. However, when I try to use iterator.intValue(), I get an error.

Exception in thread "main" java.lang.Error: Unresolved compilation problems: 
    The method intValue() is undefined for the type Class<capture#1-of ? extends Iterator>
    Syntax error, insert "}" to complete Block

Here is the full code:

    Iterator<Integer> iterator = content.iterator(); //content is a HashSet<Integer> object
    System.out.println(iterator.intValue());

    while(iterator.hasNext())
    {
        iterator.next();
        System.out.println(iterator);

    }

Upvotes: 1

Views: 21728

Answers (3)

Peter Lawrey
Peter Lawrey

Reputation: 533500

An Iterator is not an Integer, though it might give you one if you call next() Instead of calling the Iterator directly is usually better to use a for-each loop which was added to Java 5.0 in 2006.

for(int n : content) //content is a HashSet<Integer> object
    System.out.println(n);

From Java 8 you can use the forEach method.

content.forEach(System.out::println);

Upvotes: 0

Greg Valcourt
Greg Valcourt

Reputation: 731

You need to get the Integer from the iterator, then print that Integer:

Integer i = iterator.next();
System.out.println(i);

An Iterator is an iterator for integers, it doesn't behave like an integer. The next method will return an Integer.

Upvotes: 3

Mureinik
Mureinik

Reputation: 311198

No, an Iterator<T> does not act like a T - its next() method, however, returns a T value. I.e.:

while(iterator.hasNext()) {
    Integer myInteger = iterator.next();
    int myInt = myInteger.intValue();
}

Upvotes: 4

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