Develop Reference

Java-Stream - mapMulti() with Infinite Streams

I thought that all stream pipelines written using flatMap() can be converted to use mapMulti. Looks like I was wrong when the flatMap() or mapMulti() returns/operates on an infinite stream.
Note: this is for educational purpose only
When we map an element to an infinite stream inside a flatMap() followed by a limit(), then the stream pipeline is lazy and evaluates as per the required number of elements.
list.stream()
.flatMap(element -> Stream.generate(() -> 1))
.limit(3)
.forEach(System.out::println);
Output:
1
1
1
But when doing the same in a mapMulti(), the pipeline is still lazy i.e., it doesn't consume the infinite stream. But when running this in IDE (Intellij), it hangs and doesn't terminate (I guess waiting for other elements consumption) and doesn't come out of the stream pipeline execution.
With a mapMulti(),
list.stream()
.mapMulti((element, consumer) -> {
Stream.generate(() -> 1)
.forEach(consumer);
})
.limit(3)
.forEach(System.out::println);
System.out.println("Done"); //Never gets here
Output:
1
1
1
But the last print (Done) doesn't get executed.
Is this the expected behaviour?
I couldn't find any warning or points on infinite stream and mapMulti() in Javadoc.

The advantage of mapMulti() is that it consumes new elements which became a part of the stream, replacing the initial element (opposed to flatMap() which internally generates a new stream for each element). If you're generating a fully-fledged stream with a terminal operation inside the mapMulti() it should be executed. And you've created an infinite stream which can't terminate (as #Lino has pointed out in the comment).
On the contrary, flatMap() expects a function producing a stream, i.e. function only returns it not processes.
Here's a quote from the API note that emphasizes the difference between the two operations:
API Note:
This method is similar to flatMap in that it applies a one-to-many
transformation to the elements of the stream and flattens the result
elements into a new stream. This method is preferable to flatMap in
the following circumstances:
When replacing each stream element with a small (possibly zero) number of elements. Using this method avoids the overhead of creating
a new Stream instance for every group of result elements, as required
by flatMap.

Do I need to add extra stream terminal operation to trigger Future task execution before calling Future::get [duplicate]

I have a question on the intermediate stages sequential state - are the operations from a stage applied to all the input stream (items) or are all the stages / operations applied to each stream item?
I'm aware the question might not be easy to understand, so I'll give an example. On the following stream processing:
List<String> strings = Arrays.asList("Are Java streams intermediate stages sequential?".split(" "));
strings.stream()
.filter(word -> word.length() > 4)
.peek(word -> System.out.println("f: " + word))
.map(word -> word.length())
.peek(length -> System.out.println("m: " + length))
.forEach(length -> System.out.println("-> " + length + "\n"));
My expectation for this code is that it will output:
f: streams
f: intermediate
f: stages
f: sequential?
m: 7
m: 12
m: 6
m: 11
-> 7
-> 12
-> 6
-> 11
Instead, the output is:
f: streams
m: 7
-> 7
f: intermediate
m: 12
-> 12
f: stages
m: 6
-> 6
f: sequential?
m: 11
-> 11
Are the items just displayed for all the stages, due to the console output? Or are they also processed for all the stages, one at a time?
I can further detail the question, if it's not clear enough.

This behaviour enables optimisation of the code. If each intermediate operation were to process all elements of a stream before proceeding to the next intermediate operation then there would be no chance of optimisation.
So to answer your question, each element moves along the stream pipeline vertically one at a time (except for some stateful operations discussed later), therefore enabling optimisation where possible.
Explanation
Given the example you've provided, each element will move along the stream pipeline vertically one by one as there is no stateful operation included.
Another example, say you were looking for the first String whose length is greater than 4, processing all the elements prior to providing the result is unnecessary and time-consuming.
Consider this simple illustration:
List<String> stringsList = Arrays.asList("1","12","123","1234","12345","123456","1234567");
int result = stringsList.stream()
.filter(s -> s.length() > 4)
.mapToInt(Integer::valueOf)
.findFirst().orElse(0);
The filter intermediate operation above will not find all the elements whose length is greater than 4 and return a new stream of them but rather what happens is as soon as we find the first element whose length is greater than 4, that element goes through to the .mapToInt which then findFirst says "I've found the first element" and execution stops there. Therefore the result will be 12345.
Behaviour of stateful and stateless intermediate operations
Note that when a stateful intermediate operation as such of sorted is included in a stream pipeline then that specific operation will traverse the entire stream. If you think about it, this makes complete sense as in order to sort elements you'll need to see all the elements to determine which elements come first in the sort order.
The distinct intermediate operation is also a stateful operation, however, as #Holger has mentioned unlike sorted, it does not require traversing the entire stream as each distinct element can get passed down the pipeline immediately and may fulfil a short-circuiting condition.
stateless intermediate operations such as filter , map etc do not have to traverse the entire stream and can freely process one element at a time vertically as mentioned above.
Lastly, but not least it's also important to note that, when the terminal operation is a short-circuiting operation the terminal-short-circuiting methods can finish before traversing all the elements of the underlying stream.
reading: Java 8 stream tutorial

Your answer is loop fusion. What we see is that the four
intermediate operations filter() – peek() – map() – peek() – println using forEach() which is a kinda terminal operation have been logically
joined together to constitute a single pass. They are executed in
order for each of the individual element. This joining
together of operations in a single pass is an optimization technique
known as loop fusion.
More for reading: Source

An intermediate operation is always lazily executed. That is to say
they are not run until the point a terminal operation is reached.
A few of the most popular intermediate operations used in a stream
filter – the filter operation returns a stream of elements that
satisfy the predicate passed in as a parameter to the operation. The
elements themselves before and after the filter will have the same
type, however the number of elements will likely change
map – the map operation returns a stream of elements after they have
been processed by the function passed in as a parameter. The
elements before and after the mapping may have a different type, but
there will be the same total number of elements.
distinct – the distinct operation is a special case of the filter
operation. Distinct returns a stream of elements such that each
element is unique in the stream, based on the equals method of the
elements
.java-8-streams-cheat-sheet

Apart from optimisation, the order of processing you'd describe wouldn't work for streams of indeterminate length, like this:
DoubleStream.generate(Math::random).filter(d -> d > 0.9).findFirst();
Admittedly this example doesn't make much sense in practice, but the point is that rather than backed by a fixed-size collection,DoubleStream.generate() creates a potentially infinite stream. The only way to process this is element by element.

Make Stream Parallel on the Result of flatMap

Consider the following simple code:
Stream.of(1)
.flatMap(x -> IntStream.range(0, 1024).boxed())
.parallel() // Moving this before flatMap has the same effect because it's just a property of the entire stream
.forEach(x -> {
System.out.println("Thread: " + Thread.currentThread().getName());
});
For a long time, I thought that Java would have parallel execution for elements even after flatMap. But the above code prints all "Thread: main", which proves my thought wrong.
A simple way to make it parallel after flatMap would be to collect and then stream again:
Stream.of(1)
.flatMap(x -> IntStream.range(0, 1024).boxed())
.parallel() // Moving this before flatMap has the same effect because it's just a property of the entire stream
.collect(Collectors.toList())
.parallelStream()
.forEach(x -> {
System.out.println("Thread: " + Thread.currentThread().getName());
});
I was wondering whether there is a better way, and about the design choice of flatMap that only parallelizes the stream before the call, but not after the call.
========= More Clarification about the Question ========
From some answers, it seems that my question is not fully conveyed. As #Andreas said, if I start with a Stream of 3 elements, there could be 3 threads running.
But my question really is: Java Stream uses a common ForkJoinPool that has a default size equal to one less than the number of cores, according to this post. Now suppose I have 64 cores, then I expect my above code would see many different threads after flatMap, but in fact, it sees only one (or 3 in Andreas' case). By the way, I did use isParallel to observe that the stream is parallel.
To be honest, I wasn't asking this question for pure academic interest. I ran into this problem in a project that presents a long chain of stream operations for transforming a dataset. The chain starts with a single file, and explodes to a lot of elements through flatMap. But apparently, in my experiment, it does NOT fully exploit my machine (which has 64 cores), but only uses one core (from observation of the cpu usage).

I was wondering [...] about the design choice of flatMap that only parallelizes the stream before the call, but not after the call.
You're mistaken. All steps both before and after the flatMap are run in parallel, but it only splits the original stream between threads. The flatMap operation is then handled by one such thread, and its stream isn't split.
Since your original stream only has 1 element, it cannot be split, and hence parallel has no effect.
Try changing to Stream.of(1, 2, 3), and you will see that the forEach, which is after the flatMap, is actually run in 3 different threads.

The documentation for forEach specifies:
For any given element, the action may be performed at whatever time and in whatever thread the library chooses.
In particular, "execute all the operations on the invoking thread" seems like a good broadly-safe implementation.
Note that your attempt to parallelize the stream does not require any specific parallelism, but you'd be much more likely to see an effect with this:
IntStream.range(0, 1024).boxed()
.parallel()
.map(i -> "Thread: " + Thread.currentThread().getName())
.forEach(System.out::println);

For anyone like me, who has a dire need to parallelize flatMap and needs some practical solution, not only history and theory. And for those who doesn't consider collecting all the items in between before parallelizing them.
The simplest solution I came up with is to do flattening by hand, basically by replacing it with map + reduce(Stream::concat).
I've already answered to the same question in another thread, see details at https://stackoverflow.com/a/66386078/3606820

Is the skip() method a short circuiting-operation?

I am reading about Java streams' short-circuiting operations and found in some articles that skip() is a short-circuiting operation.
In another article they didn't mention skip() as a short-circuiting operation.
Now I am confused; is skip() a short-circuiting operation or not?

From the java doc under the "Stream operations and pipelines" section :
An
intermediate operation is short-circuiting if, when presented with
infinite input, it may produce a finite stream as a result. A terminal
operation is short-circuiting if, when presented with infinite input,
it may terminate in finite time.
Emphasis mine.
if you were to call skip on an infinite input it won't produce a finite stream hence not a short-circuiting operation.
The only short-circuiting intermediate operation in JDK8 is limit as it allows computations on infinite streams to complete in finite time.
Example:
if you were to execute this program with the use of skip:
String[] skip = Stream.generate(() -> "test") // returns an infinite stream
.skip(20)
.toArray(String[]::new);
it will not produce a finite stream hence you would eventually end up with something along the lines of "java.lang.OutOfMemoryError: Java heap space".
whereas if you were to execute this program with the use of limit, it will cause the computation to finish in a finite time:
String[] limit = Stream.generate(() -> "test") // returns an infinite stream
.limit(20)
.toArray(String[]::new);

Just want to add my two cents here, this idea in general of a short-circuiting a stream is infinitely complicated (at least to me and at least in the sense that I have to scratch my head twice usually). I will get to skip at the end of the answer btw.
Let's take this for example:
Stream.generate(() -> Integer.MAX_VALUE);
This is an infinite stream, we can all agree on this. Let's short-circuit it via an operation that is documented to be as such (unlike skip):
Stream.generate(() -> Integer.MAX_VALUE).anyMatch(x -> true);
This works nicely, how about adding a filter:
Stream.generate(() -> Integer.MAX_VALUE)
.filter(x -> x < 100) // well sort of useless...
.anyMatch(x -> true);
What will happen here? Well, this never finishes, even if there is a short-circuiting operation like anyMatch - but it's never reached to actually short-circuit anything.
On the other hand, filter is not a short-circuiting operation, but you can make it as such (just as an example):
someList.stream()
.filter(x -> {
if(x > 3) throw AssertionError("Just because");
})
Yes, it's ugly, but it's short-circuiting... That's how we (emphases on we, since lots of people, disagree) implement short-circuiting reduce - throw an Exception that has no stack traces.
In java-9 there was an addition of another intermediate operation that is short-circuiting: takeWhile that acts sort of like limit but for a certain condition.
And to be fair, the bulk of the answer about skip was an already give by Aomine, but the most simple answer is that it is not documented as such. And in general (there are cases when documentation is corrected), but that is the number one indication you should look at. See limit and takeWhile for example that clearly says:
This is a short-circuiting stateful intermediate operation

Java 8 stream peek and limit interaction

Why this code in java 8:
IntStream.range(0, 10)
.peek(System.out::print)
.limit(3)
.count();
outputs:
012
I'd expect it to output 0123456789, because peek preceeds limit.
It seems to me even more peculiar because of the fact that this:
IntStream.range(0, 10)
.peek(System.out::print)
.map(x -> x * 2)
.count();
outputs 0123456789 as expected (not 02481012141618).
P.S.: .count() here is used just to consume stream, it can be replaced with anything else

The most important thing to know about streams are that they do not contain elements themselves (like collections) but are working like a pipe whose values are lazily evaluated. That means that the statements that build up a stream - including mapping, filtering, or whatever - are not evaluated until the terminal operation runs.
In your first example, the stream tries to count from 0 to 9, one at each time doing the following:
print out the value
check whether 3 values are passed (if yes, terminate)
So you really get the output 012.
In your second example, the stream again counts from 0 to 9, one at each time doing the following:
print out the value
maping x to x*2, thus forwarding the double of the value to the next step
As you can see the output comes before the mapping and thus you get the result 0123456789. Try to switch the peek and the map calls. Then you will get your expected output.

From the docs:
limit() is a short-circuiting stateful intermediate operation.
map() is an intermediate operation
Again from the docs what that essentially means is that limit() will return a stream with x values from the stream it received.
An intermediate operation is short-circuiting if, when presented with infinite input, it may produce a finite stream as a result.

Streams are defined to do lazy processing. So in order to complete your count() operation it doesn’t need to look at the other items. Otherwise, it would be broken, as limit(…) is defined to be a proper way of processing infinite streams in a finite time (by not processing more than limit items).
In principle, it would be possible to complete your request without ever looking at the int values at all, as the operation chain limit(3).count() doesn’t need any processing of the previous operations (other than verifying whether the stream has at least 3 items).

Streams use lazy evaluation, the intermediate operations, i.e. peek() are not executed till the terminal operation runs.
For instances, the following code will just print 1 .In fact, as soon as the first element of the stream,1, will reach the terminal operation, findAny(), the stream execution will be ended.
Arrays.asList(1,2,3)
.stream()
.peek(System.out::print)
.filter((n)->n<3)
.findAny();
Viceversa, in the following example, will be printed 123. In fact the terminal operation, noneMatch(), needs to evaluate all the elements of the stream in order to make sure there is no match with its Predicate: n>4
Arrays.asList(1, 2, 3)
.stream()
.peek(System.out::print)
.noneMatch(n -> n > 4);

For future readers struggling to understand how the count method doesn't execute the peek method before it, I thought I add this additional note:
As per Java 9, the Java documentation for the count method states that:
An implementation may choose to not execute the stream pipeline
(either sequentially or in parallel) if it is capable of computing the
count directly from the stream source.
This means terminating the stream with count is no longer enough to ensure the execution of all previous steps, such as peek.