Mixing
Java 8 Stream API - Does any stateful intermediate operation guarantee a new source collection?
Clash Royale CLAN TAG#URR8PPP
up vote
21
down vote
favorite
Is the following statement true? (source and source - they seem to copy from each other or come from the same source).
The
sorted()operation is a “stateful intermediate operationâ€Â, which means that subsequent operations no longer operate on the backing collection, but on an internal state.
I have tested Stream::sorted as a snippet from sources above:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.sorted()
.forEach(list::remove);
System.out.println(list); // prints correctly [0, 1, 2, 3, 4, 5]
It works. I replaced Stream::sorted with Stream::distinct, Stream::limit and Stream::skip:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.distinct()
.forEach(list::remove);
For my surprise, the NullPointerException is thrown. All the tested methods follow the stateful intermediate operation characteristics. Yet, this unique behavior of Stream::sorted is not documented nor the Stream operations and pipelines part explains whether the stateful intermediate operations really guarantee a new source collection.
Where my confusion comes from and what is the explanation of the behavior above?
java java-8 java-stream
asked yesterday
Nikolas
10.9k52956
add a comment |Â
up vote
21
down vote
favorite
Is the following statement true? (source and source - they seem to copy from each other or come from the same source).
The
sorted()operation is a “stateful intermediate operationâ€Â, which means that subsequent operations no longer operate on the backing collection, but on an internal state.
I have tested Stream::sorted as a snippet from sources above:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.sorted()
.forEach(list::remove);
System.out.println(list); // prints correctly [0, 1, 2, 3, 4, 5]
It works. I replaced Stream::sorted with Stream::distinct, Stream::limit and Stream::skip:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.distinct()
.forEach(list::remove);
For my surprise, the NullPointerException is thrown. All the tested methods follow the stateful intermediate operation characteristics. Yet, this unique behavior of Stream::sorted is not documented nor the Stream operations and pipelines part explains whether the stateful intermediate operations really guarantee a new source collection.
Where my confusion comes from and what is the explanation of the behavior above?
java java-8 java-stream
asked yesterday
Nikolas
10.9k52956
add a comment |Â
up vote
21
down vote
favorite
up vote
21
down vote
favorite
Is the following statement true? (source and source - they seem to copy from each other or come from the same source).
The
sorted()operation is a “stateful intermediate operationâ€Â, which means that subsequent operations no longer operate on the backing collection, but on an internal state.
I have tested Stream::sorted as a snippet from sources above:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.sorted()
.forEach(list::remove);
System.out.println(list); // prints correctly [0, 1, 2, 3, 4, 5]
It works. I replaced Stream::sorted with Stream::distinct, Stream::limit and Stream::skip:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.distinct()
.forEach(list::remove);
For my surprise, the NullPointerException is thrown. All the tested methods follow the stateful intermediate operation characteristics. Yet, this unique behavior of Stream::sorted is not documented nor the Stream operations and pipelines part explains whether the stateful intermediate operations really guarantee a new source collection.
Where my confusion comes from and what is the explanation of the behavior above?
java java-8 java-stream
asked yesterday
Nikolas
10.9k52956
Is the following statement true? (source and source - they seem to copy from each other or come from the same source).
The
sorted()operation is a “stateful intermediate operationâ€Â, which means that subsequent operations no longer operate on the backing collection, but on an internal state.
I have tested Stream::sorted as a snippet from sources above:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.sorted()
.forEach(list::remove);
System.out.println(list); // prints correctly [0, 1, 2, 3, 4, 5]
It works. I replaced Stream::sorted with Stream::distinct, Stream::limit and Stream::skip:
final List<Integer> list = IntStream.range(0, 10).boxed().collect(Collectors.toList());
list.stream()
.filter(i -> i > 5)
.distinct()
.forEach(list::remove);
For my surprise, the NullPointerException is thrown. All the tested methods follow the stateful intermediate operation characteristics. Yet, this unique behavior of Stream::sorted is not documented nor the Stream operations and pipelines part explains whether the stateful intermediate operations really guarantee a new source collection.
Where my confusion comes from and what is the explanation of the behavior above?
java java-8 java-stream
java java-8 java-stream
asked yesterday
Nikolas
10.9k52956
asked yesterday
Nikolas
10.9k52956
edited yesterday
asked yesterday
Nikolas
10.9k52956
asked yesterday
Nikolas
10.9k52956
asked yesterday
Nikolas
10.9k52956
10.9k52956
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
27
down vote
accepted
The API documentation makes no such guarantee “that subsequent operations no longer operate on the backing collectionâ€Â, hence, you should never rely on such a behavior of a particular implementation.
Your example happens to do the desired thing by accident; there’s not even a guarantee that the List created by collect(Collectors.toList()) supports the remove operation.
To show a counter-example
Set<Integer> set = IntStream.range(0, 10).boxed()
.collect(Collectors.toCollection(TreeSet::new));
set.stream()
.filter(i -> i > 5)
.sorted()
.forEach(set::remove);
throws a ConcurrentModificationException. The reason is that the implementation optimizes this scenario, as the source is already sorted. In principle, it could do the same optimization to your original example, as forEach is explicitly performing the action in no specified order, hence, the sorting is unnecessary.
There are other optimizations imaginable, e.g. sorted().findFirst() could get converted to a “find the minimum†operation, without the need to copy the element into a new storage for sorting.
So the bottom line is, when relying on unspecified behavior, what may happen to work today, may break tomorrow, when new optimizations are added.
answered yesterday
Holger
150k20206390
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
add a comment |Â
up vote
6
down vote
Well sorted has to be a full copying barrier for the stream pipeline, after all your source could be not sorted; but this is not documented as such, thus do not rely on it.
This is not just about sorted per-se, but what other optimization can be done to the stream pipeline, so that sorted could be entirely skipped. For example:
List<Integer> sortedList = IntStream.range(0, 10)
.boxed()
.collect(Collectors.toList());
StreamSupport.stream(() -> sortedList.spliterator(), Spliterator.SORTED, false)
.sorted()
.forEach(sortedList::remove); // fails with CME, thus no copying occurred
Of course, sorted needs to be a full barrier and stop to do an entire sort, unless, of course, it can be skipped, thus the documentation makes no such promises, so that we don't run in weird surprises.
distinct on the other hand does not have to be a full barrier, all distinct does is check one element at a time, if it is unique; so after a single element is checked (and it is unique) it is passed to the next stage, thus without being a full barrier. Either way, this is not documented also...
answered yesterday
Eugene
59.7k980139
add a comment |Â
up vote
2
down vote
You shouldn't have brought up the cases with a terminal operation forEach(list::remove) because list::remove is an interfering function and it violates the "non-interference" principle for terminal actions.
It's vital to follow the rules before wondering why an incorrect code snippet causes unexpected (or undocumented) behaviour.
I believe that list::remove is the root of the problem here. You wouldn't have noticed the difference between the operations for this scenario if you'd written a proper action for forEach.
answered yesterday
Andrew Tobilko
20.5k83873
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed thatlist::removeis inappropriate here. I wanted to highlight it.
– Andrew Tobilko
yesterday
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
27
down vote
accepted
The API documentation makes no such guarantee “that subsequent operations no longer operate on the backing collectionâ€Â, hence, you should never rely on such a behavior of a particular implementation.
Your example happens to do the desired thing by accident; there’s not even a guarantee that the List created by collect(Collectors.toList()) supports the remove operation.
To show a counter-example
Set<Integer> set = IntStream.range(0, 10).boxed()
.collect(Collectors.toCollection(TreeSet::new));
set.stream()
.filter(i -> i > 5)
.sorted()
.forEach(set::remove);
throws a ConcurrentModificationException. The reason is that the implementation optimizes this scenario, as the source is already sorted. In principle, it could do the same optimization to your original example, as forEach is explicitly performing the action in no specified order, hence, the sorting is unnecessary.
There are other optimizations imaginable, e.g. sorted().findFirst() could get converted to a “find the minimum†operation, without the need to copy the element into a new storage for sorting.
So the bottom line is, when relying on unspecified behavior, what may happen to work today, may break tomorrow, when new optimizations are added.
answered yesterday
Holger
150k20206390
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
add a comment |Â
up vote
27
down vote
accepted
The API documentation makes no such guarantee “that subsequent operations no longer operate on the backing collectionâ€Â, hence, you should never rely on such a behavior of a particular implementation.
Your example happens to do the desired thing by accident; there’s not even a guarantee that the List created by collect(Collectors.toList()) supports the remove operation.
To show a counter-example
Set<Integer> set = IntStream.range(0, 10).boxed()
.collect(Collectors.toCollection(TreeSet::new));
set.stream()
.filter(i -> i > 5)
.sorted()
.forEach(set::remove);
throws a ConcurrentModificationException. The reason is that the implementation optimizes this scenario, as the source is already sorted. In principle, it could do the same optimization to your original example, as forEach is explicitly performing the action in no specified order, hence, the sorting is unnecessary.
There are other optimizations imaginable, e.g. sorted().findFirst() could get converted to a “find the minimum†operation, without the need to copy the element into a new storage for sorting.
So the bottom line is, when relying on unspecified behavior, what may happen to work today, may break tomorrow, when new optimizations are added.
answered yesterday
Holger
150k20206390
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
add a comment |Â
up vote
27
down vote
accepted
up vote
27
down vote
accepted
The API documentation makes no such guarantee “that subsequent operations no longer operate on the backing collectionâ€Â, hence, you should never rely on such a behavior of a particular implementation.
Your example happens to do the desired thing by accident; there’s not even a guarantee that the List created by collect(Collectors.toList()) supports the remove operation.
To show a counter-example
Set<Integer> set = IntStream.range(0, 10).boxed()
.collect(Collectors.toCollection(TreeSet::new));
set.stream()
.filter(i -> i > 5)
.sorted()
.forEach(set::remove);
throws a ConcurrentModificationException. The reason is that the implementation optimizes this scenario, as the source is already sorted. In principle, it could do the same optimization to your original example, as forEach is explicitly performing the action in no specified order, hence, the sorting is unnecessary.
There are other optimizations imaginable, e.g. sorted().findFirst() could get converted to a “find the minimum†operation, without the need to copy the element into a new storage for sorting.
So the bottom line is, when relying on unspecified behavior, what may happen to work today, may break tomorrow, when new optimizations are added.
answered yesterday
Holger
150k20206390
The API documentation makes no such guarantee “that subsequent operations no longer operate on the backing collectionâ€Â, hence, you should never rely on such a behavior of a particular implementation.
Your example happens to do the desired thing by accident; there’s not even a guarantee that the List created by collect(Collectors.toList()) supports the remove operation.
To show a counter-example
Set<Integer> set = IntStream.range(0, 10).boxed()
.collect(Collectors.toCollection(TreeSet::new));
set.stream()
.filter(i -> i > 5)
.sorted()
.forEach(set::remove);
throws a ConcurrentModificationException. The reason is that the implementation optimizes this scenario, as the source is already sorted. In principle, it could do the same optimization to your original example, as forEach is explicitly performing the action in no specified order, hence, the sorting is unnecessary.
There are other optimizations imaginable, e.g. sorted().findFirst() could get converted to a “find the minimum†operation, without the need to copy the element into a new storage for sorting.
So the bottom line is, when relying on unspecified behavior, what may happen to work today, may break tomorrow, when new optimizations are added.
answered yesterday
Holger
150k20206390
edited yesterday
answered yesterday
Holger
150k20206390
answered yesterday
Holger
150k20206390
answered yesterday
Holger
150k20206390
150k20206390
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
add a comment |Â
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
9
9
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
@Nikolas well, that blog is focusing on observed behavior and looking at the code, where it should have referred to the specification instead. But at least the conclusion is basically correct: “we can only suggest that you never actually do modify a backing collection while consuming a streamâ€Â.
– Holger
yesterday
add a comment |Â
up vote
6
down vote
Well sorted has to be a full copying barrier for the stream pipeline, after all your source could be not sorted; but this is not documented as such, thus do not rely on it.
This is not just about sorted per-se, but what other optimization can be done to the stream pipeline, so that sorted could be entirely skipped. For example:
List<Integer> sortedList = IntStream.range(0, 10)
.boxed()
.collect(Collectors.toList());
StreamSupport.stream(() -> sortedList.spliterator(), Spliterator.SORTED, false)
.sorted()
.forEach(sortedList::remove); // fails with CME, thus no copying occurred
Of course, sorted needs to be a full barrier and stop to do an entire sort, unless, of course, it can be skipped, thus the documentation makes no such promises, so that we don't run in weird surprises.
distinct on the other hand does not have to be a full barrier, all distinct does is check one element at a time, if it is unique; so after a single element is checked (and it is unique) it is passed to the next stage, thus without being a full barrier. Either way, this is not documented also...
answered yesterday
Eugene
59.7k980139
add a comment |Â
up vote
6
down vote
Well sorted has to be a full copying barrier for the stream pipeline, after all your source could be not sorted; but this is not documented as such, thus do not rely on it.
This is not just about sorted per-se, but what other optimization can be done to the stream pipeline, so that sorted could be entirely skipped. For example:
List<Integer> sortedList = IntStream.range(0, 10)
.boxed()
.collect(Collectors.toList());
StreamSupport.stream(() -> sortedList.spliterator(), Spliterator.SORTED, false)
.sorted()
.forEach(sortedList::remove); // fails with CME, thus no copying occurred
Of course, sorted needs to be a full barrier and stop to do an entire sort, unless, of course, it can be skipped, thus the documentation makes no such promises, so that we don't run in weird surprises.
distinct on the other hand does not have to be a full barrier, all distinct does is check one element at a time, if it is unique; so after a single element is checked (and it is unique) it is passed to the next stage, thus without being a full barrier. Either way, this is not documented also...
answered yesterday
Eugene
59.7k980139
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Well sorted has to be a full copying barrier for the stream pipeline, after all your source could be not sorted; but this is not documented as such, thus do not rely on it.
This is not just about sorted per-se, but what other optimization can be done to the stream pipeline, so that sorted could be entirely skipped. For example:
List<Integer> sortedList = IntStream.range(0, 10)
.boxed()
.collect(Collectors.toList());
StreamSupport.stream(() -> sortedList.spliterator(), Spliterator.SORTED, false)
.sorted()
.forEach(sortedList::remove); // fails with CME, thus no copying occurred
Of course, sorted needs to be a full barrier and stop to do an entire sort, unless, of course, it can be skipped, thus the documentation makes no such promises, so that we don't run in weird surprises.
distinct on the other hand does not have to be a full barrier, all distinct does is check one element at a time, if it is unique; so after a single element is checked (and it is unique) it is passed to the next stage, thus without being a full barrier. Either way, this is not documented also...
answered yesterday
Eugene
59.7k980139
Well sorted has to be a full copying barrier for the stream pipeline, after all your source could be not sorted; but this is not documented as such, thus do not rely on it.
This is not just about sorted per-se, but what other optimization can be done to the stream pipeline, so that sorted could be entirely skipped. For example:
List<Integer> sortedList = IntStream.range(0, 10)
.boxed()
.collect(Collectors.toList());
StreamSupport.stream(() -> sortedList.spliterator(), Spliterator.SORTED, false)
.sorted()
.forEach(sortedList::remove); // fails with CME, thus no copying occurred
Of course, sorted needs to be a full barrier and stop to do an entire sort, unless, of course, it can be skipped, thus the documentation makes no such promises, so that we don't run in weird surprises.
distinct on the other hand does not have to be a full barrier, all distinct does is check one element at a time, if it is unique; so after a single element is checked (and it is unique) it is passed to the next stage, thus without being a full barrier. Either way, this is not documented also...
answered yesterday
Eugene
59.7k980139
edited yesterday
answered yesterday
Eugene
59.7k980139
answered yesterday
Eugene
59.7k980139
answered yesterday
Eugene
59.7k980139
59.7k980139
add a comment |Â
add a comment |Â
up vote
2
down vote
You shouldn't have brought up the cases with a terminal operation forEach(list::remove) because list::remove is an interfering function and it violates the "non-interference" principle for terminal actions.
It's vital to follow the rules before wondering why an incorrect code snippet causes unexpected (or undocumented) behaviour.
I believe that list::remove is the root of the problem here. You wouldn't have noticed the difference between the operations for this scenario if you'd written a proper action for forEach.
answered yesterday
Andrew Tobilko
20.5k83873
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed thatlist::removeis inappropriate here. I wanted to highlight it.
– Andrew Tobilko
yesterday
add a comment |Â
up vote
2
down vote
You shouldn't have brought up the cases with a terminal operation forEach(list::remove) because list::remove is an interfering function and it violates the "non-interference" principle for terminal actions.
It's vital to follow the rules before wondering why an incorrect code snippet causes unexpected (or undocumented) behaviour.
I believe that list::remove is the root of the problem here. You wouldn't have noticed the difference between the operations for this scenario if you'd written a proper action for forEach.
answered yesterday
Andrew Tobilko
20.5k83873
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed thatlist::removeis inappropriate here. I wanted to highlight it.
– Andrew Tobilko
yesterday
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You shouldn't have brought up the cases with a terminal operation forEach(list::remove) because list::remove is an interfering function and it violates the "non-interference" principle for terminal actions.
It's vital to follow the rules before wondering why an incorrect code snippet causes unexpected (or undocumented) behaviour.
I believe that list::remove is the root of the problem here. You wouldn't have noticed the difference between the operations for this scenario if you'd written a proper action for forEach.
answered yesterday
Andrew Tobilko
20.5k83873
You shouldn't have brought up the cases with a terminal operation forEach(list::remove) because list::remove is an interfering function and it violates the "non-interference" principle for terminal actions.
It's vital to follow the rules before wondering why an incorrect code snippet causes unexpected (or undocumented) behaviour.
I believe that list::remove is the root of the problem here. You wouldn't have noticed the difference between the operations for this scenario if you'd written a proper action for forEach.
answered yesterday
Andrew Tobilko
20.5k83873
edited yesterday
answered yesterday
Andrew Tobilko
20.5k83873
answered yesterday
Andrew Tobilko
20.5k83873
answered yesterday
Andrew Tobilko
20.5k83873
20.5k83873
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed thatlist::removeis inappropriate here. I wanted to highlight it.
– Andrew Tobilko
yesterday
add a comment |Â
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed thatlist::removeis inappropriate here. I wanted to highlight it.
– Andrew Tobilko
yesterday
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
I am aware the first paragraph of your answer well, Andrew. Here I’d use Iterator or collect to a List to remove it all from the source. The second paragraph makes sence.
– Nikolas
yesterday
@Nikolas no one pointed that
list::remove is inappropriate here. I wanted to highlight it.– Andrew Tobilko
yesterday
@Nikolas no one pointed that
list::remove is inappropriate here. I wanted to highlight it.– Andrew Tobilko
yesterday
add a comment |Â
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