Reputation: 69379
Looking at the following class I've made:
public class FibonacciSupplier implements Iterator<Integer> {
private final IntPredicate hasNextPredicate;
private int beforePrevious = 0;
private int previous = 1;
private FibonacciSupplier(final IntPredicate hasNextPredicate) {
this.hasNextPredicate = hasNextPredicate;
}
@Override
public boolean hasNext() {
return hasNextPredicate.test(previous);
}
@Override
public Integer next() {
int result = beforePrevious + previous;
beforePrevious = previous;
previous = result;
return result;
}
public static FibonacciSupplier infinite() {
return new FibonacciSupplier(i -> true);
}
public static FibonacciSupplier finite(final IntPredicate predicate) {
return new FibonacciSupplier(predicate);
}
}
And the usage of it in:
public class Problem2 extends Problem<Integer> {
@Override
public void run() {
result = toList(FibonacciSupplier.finite(i -> (i <= 4_000_000)))
.stream()
.filter(i -> (i % 2 == 0))
.mapToInt(i -> i)
.sum();
}
@Override
public String getName() {
return "Problem 2";
}
private static <E> List<E> toList(final Iterator<E> iterator) {
List<E> list = new ArrayList<>();
while (iterator.hasNext()) {
list.add(iterator.next());
}
return list;
}
}
How would I be able to create an infinite Stream<E>
?
If I were to use Stream<Integer> infiniteStream = toList(FibonacciSupplier.infinite()).stream()
, I would, possibly surprisingly, never get an infinite stream.
Instead the code would loop forever in the creation of the list
in an underlying method.
This so far is purely theoretical, but I can definately understand the need for it if I would want to first skip the first x numbers from an infinite stream, and then limit it by the last y numbers, something like:
int x = MAGIC_NUMBER_X;
int y = MAGIC_NUMBER_y;
int sum = toList(FibonacciSupplier.infinite())
.stream()
.skip(x)
.limit(y)
.mapToInt(i -> i)
.sum();
The code would not ever return a result, how should it be done?
Upvotes: 29
Views: 14530
Reputation: 1004
You can use the low level stream support primitives and the Spliterators
library to make a stream out of an Iterator
.
The last parameter to StreamSupport.stream()
says that the stream is not parallel. Be sure to let it like that because your Fibonacci iterator depends on previous iterations.
return StreamSupport.stream( Spliterators.spliteratorUnknownSize( new Iterator<Node>()
{
@Override
public boolean hasNext()
{
// to implement
return ...;
}
@Override
public ContentVersion next()
{
// to implement
return ...;
}
}, Spliterator.ORDERED ), false );
Upvotes: 1
Reputation: 301
To add another answer, perhaps AbstractSpliterator is a better choice, especially given the example code. Generate is inflexible as there is no [good] way to give a stop condition except by using limit. Limit only accepts a number of items rather than a predicate, so we have to know how many items we want to generate - which might not be possible, and what if the generator is a black box passed to us?
AbstractSpliterator is a halfway house between having to write a whole spliterator, and using Iterator/Iterable. AbstractSpliterator lacks just the tryAdvance method where we first check our predicate for being done, and if not pass the generated value to an action. Here's an example of a Fibonacci sequence using AbstractIntSpliterator:
public class Fibonacci {
private static class FibonacciGenerator extends Spliterators.AbstractIntSpliterator
{
private IntPredicate hasNextPredicate;
private int beforePrevious = 0;
private int previous = 0;
protected FibonacciGenerator(IntPredicate hasNextPredicate)
{
super(Long.MAX_VALUE, 0);
this.hasNextPredicate = hasNextPredicate;
}
@Override
public boolean tryAdvance(IntConsumer action)
{
if (action == null)
{
throw new NullPointerException();
}
int next = Math.max(1, beforePrevious + previous);
beforePrevious = previous;
previous = next;
if (!hasNextPredicate.test(next))
{
return false;
}
action.accept(next);
return true;
}
@Override
public boolean tryAdvance(Consumer<? super Integer> action)
{
if (action == null)
{
throw new NullPointerException();
}
int next = Math.max(1, beforePrevious + previous);
beforePrevious = previous;
previous = next;
if (!hasNextPredicate.test(next))
{
return false;
}
action.accept(next);
return true;
}
}
public static void main(String args[])
{
Stream<Integer> infiniteStream = StreamSupport.stream(
new FibonacciGenerator(i -> true), false);
Stream<Integer> finiteStream = StreamSupport.stream(
new FibonacciGenerator(i -> i < 100), false);
// Print with a side-effect for the demo
infiniteStream.limit(10).forEach(System.out::println);
finiteStream.forEach(System.out::println);
}
}
For more details I've covered generators in Java 8 in my blog http://thecannycoder.wordpress.com/
Upvotes: 0
Reputation: 50104
In Java 8 there are no public, concrete classes implementing the interface Stream. However, there are some static factory methods. One of the most important is StreamSupport.stream. In particular, it is used in the default method Collection.stream –inherited by most collection classes:
default Stream<E> stream() {
return StreamSupport.stream(spliterator(), false);
}
The default implementation of this method creates a Spliterator by invoking spliterator()
, and passes the created object to the factory method. Spliterator is a new interface introduced with Java 8 to support parallel streams. It is similar to Iterator, but in contrast to the latter, a Spliterator can be divided into parts, that can be processed independently. See Spliterator.trySplit for details.
The default method Iterable.spliterator was also added in Java 8, so that every Iterable class automatically supports Spliterators. The implementation looks as follows:
default Spliterator<T> spliterator() {
return Spliterators.spliteratorUnknownSize(iterator(), 0);
}
The method creates the Spliterator from an arbitrary Iterator. If you combine these two steps, you can create a Stream from an arbitrary Iterator:
<T> Stream<T> stream(Iterator<T> iterator) {
Spliterator<T> spliterator
= Spliterators.spliteratorUnknownSize(iterator, 0);
return StreamSupport.stream(spliterator, false);
}
To get an impression of Spliterators, here is a very simple example without using collections. The following class implements Spliterator to iterate over a half-open interval of integers:
public final class IntRange implements Spliterator.OfInt {
private int first, last;
public IntRange(int first, int last) {
this.first = first;
this.last = last;
}
public boolean tryAdvance(IntConsumer action) {
if (first < last) {
action.accept(first++);
return true;
} else {
return false;
}
}
public OfInt trySplit() {
int size = last - first;
if (size >= 10) {
int temp = first;
first += size / 2;
return new IntRange(temp, first);
} else {
return null;
}
}
public long estimateSize() {
return Math.max(last - first, 0);
}
public int characteristics() {
return ORDERED | DISTINCT | SIZED | NONNULL
| IMMUTABLE | CONCURRENT | SUBSIZED;
}
}
Upvotes: 18
Reputation: 298499
Your mistake is to think that you need an Iterator
or a Collection
to create a Stream
. For creating an infinite stream, a single method providing one value after another is enough. So for your class FibonacciSupplier
the simplest use is:
IntStream s=IntStream.generate(FibonacciSupplier.infinite()::next);
or, if you prefer boxed values:
Stream<Integer> s=Stream.generate(FibonacciSupplier.infinite()::next);
Note that in this case the method does not have to be named next
nor fulfill the Iterator
interface. But it doesn’t matter if it does as with your class. Further, as we just told the stream to use the next
method as a Supplier
, the hasNext
method will never be called. It’s just infinite.
Creating a finite stream using your Iterator
is a bit more complicated:
Stream<Integer> s=StreamSupport.stream(
Spliterators.spliteratorUnknownSize(
FibonacciSupplier.finite(intPredicate), Spliterator.ORDERED),
false);
In this case if you want a finite IntStream
with unboxed int
values your FibonacciSupplier
should implement PrimitiveIterator.OfInt
.
Upvotes: 28