No, filter does not scan the whole stream. It's an intermediate operation, which returns a lazy stream (actually all intermediate operations return a lazy stream). To convince you, you can simply do the following test:
List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
int a = list.stream()
.peek(num -> System.out.println("will filter " + num))
.filter(x -> x > 5)
.findFirst()
.get();
System.out.println(a);
Which outputs:
will filter 1
will filter 10
10
You see that only the two first elements of the stream are actually processed.
So you can go with your approach which is perfectly fine.
Answer from Alexis C. on Stack OverflowNo, filter does not scan the whole stream. It's an intermediate operation, which returns a lazy stream (actually all intermediate operations return a lazy stream). To convince you, you can simply do the following test:
List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
int a = list.stream()
.peek(num -> System.out.println("will filter " + num))
.filter(x -> x > 5)
.findFirst()
.get();
System.out.println(a);
Which outputs:
will filter 1
will filter 10
10
You see that only the two first elements of the stream are actually processed.
So you can go with your approach which is perfectly fine.
However this seems inefficient to me, as the filter will scan the whole list
No it won't - it will "break" as soon as the first element satisfying the predicate is found. You can read more about laziness in the stream package javadoc, in particular (emphasis mine):
Many stream operations, such as filtering, mapping, or duplicate removal, can be implemented lazily, exposing opportunities for optimization. For example, "find the first String with three consecutive vowels" need not examine all the input strings. Stream operations are divided into intermediate (Stream-producing) operations and terminal (value- or side-effect-producing) operations. Intermediate operations are always lazy.
There is StreamEx library that extends standard Java's Stream API. Using StreamEx.of(Iterator) and peekFirst :
StreamEx.of(itemIter.iterator())
.peekFirst(e -> System.out.println(e.getClass().getSimpleName()))
.forEach(System.out::println);
Native solution: Stream in Java is not reusable. This means, that consuming stream can be done only once. If you get the first element from a stream, you can iterate over it one more time.
Workaround would be to create another stream same as the first one or getting the first item and then creating a stream, something like that:
Stream<E> stream = StreamSupport.stream(Spliterators.spliteratorUnknownSize(sourceIterator, Spliterator.ORDERED), false);
E firstElement = itemIter.next();
stream.foreach(...);
Edit
There is not really any way to "copy" a stream, you need to keep an iterator / collection. More about it here. When it comes to memory once stream is exhausted, it can be collected by garbage collector as there is no use from it. Stream itself does not take more space than iterator it originates from. Bear in mind, that streams can be potentially infinite. Elements currently manipulated are stored in memory.