Why do I have to chain Stream operations in Java? [duplicate]
up vote
23
down vote
favorite
This question already has an answer here:
When is a Java 8 Stream considered to be consumed?
2 answers
I think all of the resources I have studied one way or another emphasize that a stream can be consumed only once, and the consumption is done by so-called terminal operations (which is very clear to me).
Just out of curiosity I tried this:
import java.util.stream.IntStream;
class App
public static void main(String args)
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
which ends up throwing a Runtime Exception:
Exception in thread "main" java.lang.IllegalStateException: stream has already been operated upon or closed
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:229)
at java.util.stream.IntPipeline.reduce(IntPipeline.java:456)
at java.util.stream.IntPipeline.sum(IntPipeline.java:414)
at App.main(scratch.java:10)
This is usual, I am missing something, but still want to ask: As far as I know map
is an intermediate (and lazy) operation and does nothing on the Stream by itself. Only when the terminal operation sum
(which is an eager operation) is called, the Stream gets consumed and operated on.
But why do I have to chain them?
What is the difference between
is.map(i -> i + 1);
is.sum();
and
is.map(i -> i + 1).sum();
?
java java-8 java-stream
marked as duplicate by Oleksandr, DaveyDaveDave, Federico Peralta Schaffner
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Nov 12 at 14:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
23
down vote
favorite
This question already has an answer here:
When is a Java 8 Stream considered to be consumed?
2 answers
I think all of the resources I have studied one way or another emphasize that a stream can be consumed only once, and the consumption is done by so-called terminal operations (which is very clear to me).
Just out of curiosity I tried this:
import java.util.stream.IntStream;
class App
public static void main(String args)
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
which ends up throwing a Runtime Exception:
Exception in thread "main" java.lang.IllegalStateException: stream has already been operated upon or closed
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:229)
at java.util.stream.IntPipeline.reduce(IntPipeline.java:456)
at java.util.stream.IntPipeline.sum(IntPipeline.java:414)
at App.main(scratch.java:10)
This is usual, I am missing something, but still want to ask: As far as I know map
is an intermediate (and lazy) operation and does nothing on the Stream by itself. Only when the terminal operation sum
(which is an eager operation) is called, the Stream gets consumed and operated on.
But why do I have to chain them?
What is the difference between
is.map(i -> i + 1);
is.sum();
and
is.map(i -> i + 1).sum();
?
java java-8 java-stream
marked as duplicate by Oleksandr, DaveyDaveDave, Federico Peralta Schaffner
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Nov 12 at 14:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
1
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56
add a comment |
up vote
23
down vote
favorite
up vote
23
down vote
favorite
This question already has an answer here:
When is a Java 8 Stream considered to be consumed?
2 answers
I think all of the resources I have studied one way or another emphasize that a stream can be consumed only once, and the consumption is done by so-called terminal operations (which is very clear to me).
Just out of curiosity I tried this:
import java.util.stream.IntStream;
class App
public static void main(String args)
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
which ends up throwing a Runtime Exception:
Exception in thread "main" java.lang.IllegalStateException: stream has already been operated upon or closed
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:229)
at java.util.stream.IntPipeline.reduce(IntPipeline.java:456)
at java.util.stream.IntPipeline.sum(IntPipeline.java:414)
at App.main(scratch.java:10)
This is usual, I am missing something, but still want to ask: As far as I know map
is an intermediate (and lazy) operation and does nothing on the Stream by itself. Only when the terminal operation sum
(which is an eager operation) is called, the Stream gets consumed and operated on.
But why do I have to chain them?
What is the difference between
is.map(i -> i + 1);
is.sum();
and
is.map(i -> i + 1).sum();
?
java java-8 java-stream
This question already has an answer here:
When is a Java 8 Stream considered to be consumed?
2 answers
I think all of the resources I have studied one way or another emphasize that a stream can be consumed only once, and the consumption is done by so-called terminal operations (which is very clear to me).
Just out of curiosity I tried this:
import java.util.stream.IntStream;
class App
public static void main(String args)
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
which ends up throwing a Runtime Exception:
Exception in thread "main" java.lang.IllegalStateException: stream has already been operated upon or closed
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:229)
at java.util.stream.IntPipeline.reduce(IntPipeline.java:456)
at java.util.stream.IntPipeline.sum(IntPipeline.java:414)
at App.main(scratch.java:10)
This is usual, I am missing something, but still want to ask: As far as I know map
is an intermediate (and lazy) operation and does nothing on the Stream by itself. Only when the terminal operation sum
(which is an eager operation) is called, the Stream gets consumed and operated on.
But why do I have to chain them?
What is the difference between
is.map(i -> i + 1);
is.sum();
and
is.map(i -> i + 1).sum();
?
This question already has an answer here:
When is a Java 8 Stream considered to be consumed?
2 answers
java java-8 java-stream
java java-8 java-stream
edited Nov 12 at 2:32
nullpointer
39.1k1074149
39.1k1074149
asked Nov 12 at 2:04
Koray Tugay
8,63926111219
8,63926111219
marked as duplicate by Oleksandr, DaveyDaveDave, Federico Peralta Schaffner
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Nov 12 at 14:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
1
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56
add a comment |
2
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
1
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56
2
2
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
1
1
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56
add a comment |
3 Answers
3
active
oldest
votes
up vote
45
down vote
accepted
When you do this:
int sum = IntStream.of(1, 2, 3, 4).map(i -> i + 1).sum();
Every chained method is being invoked on the return value of the previous method in the chain.
So map
is invoked on what IntStream.of(1, 2, 3, 4)
returns and sum
on what map(i -> i + 1)
returns.
You don't have to chain stream methods, but it's more readable and less error-prone than using this equivalent code:
IntStream is = IntStream.of(1, 2, 3, 4);
is = is.map(i -> i + 1);
int sum = is.sum();
Which is not the same as the code you've shown in your question:
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
As you see, you're disregarding the reference returned by map
. This is the cause of the error.
EDIT (as per the comments, thanks to @IanKemp for pointing this out): Actually, this is the external cause of the error. If you stop to think about it, map
must be doing something internally to the stream itself, otherwise, how would then the terminal operation trigger the transformation passed to map
on each element? I agree in that intermediate operations are lazy, i.e. when invoked, they do nothing to the elements of the stream. But internally, they must configure some state into the stream pipeline itself, so that they can be applied later.
Despite I'm not aware of the full details, what happens is that, conceptually, map
is doing at least 2 things:
It's creating and returning a new stream that holds the function passed as an argument somewhere, so that it can be applied to elements later, when the terminal operation is invoked.
It is also setting a flag to the old stream instance, i.e. the one which it has been called on, indicating that this stream instance no longer represents a valid state for the pipeline. This is because the new, updated state which holds the function passed to
map
is now encapsulated by the instance it has returned. (I believe that this decision might have been taken by the jdk team to make errors appear as early as possible, i.e. by throwing an early exception instead of letting the pipeline go on with an invalid/old state that doesn't hold the function to be applied, thus letting the terminal operation return unexpected results).
Later on, when a terminal operation is invoked on this instance flagged as invalid, you're getting that IllegalStateException
. The two items above configure the deep, internal cause of the error.
Another way to see all this is to make sure that a Stream
instance is operated only once, by means of either an intermediate or a terminal operation. Here you are violating this requirement, because you are calling map
and sum
on the same instance.
In fact, javadocs for Stream
state it clearly:
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw
IllegalStateException
if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
1
@Koray To add to that, adding a debug point at both your linesis.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipelineis
which is what you are not reusing and I believe sinceis
is already marked aslinkedOrConsumed
in your case, you see an ISE.
– nullpointer
Nov 12 at 2:27
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
This feels very similar to the age oldstr.toLowerCase();
vss = str.toLowerCase();
stumbling block in Java
– DeadChex
Nov 12 at 6:48
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,map()
is an intermediate operation, which means that applying it and then throwing away the result viais.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operationint sum = is.sum();
on that underlying stream should still succeed.
– Ian Kemp
Nov 12 at 9:24
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
|
show 5 more comments
up vote
15
down vote
Imagine the IntStream is a wrapper around your data stream with an
immutable list of operations. These operations are not executed until you need the final result (sum in your case).
Since the list is immutable, you need a new instance of IntStream with a list that contains the previous items plus the new one, which is what '. map' returns.
This means that if you don't chain, you will operate on the old instance, which does not have that operation.
The stream library also keeps some internal tracking of what's going on, that's why it's able to throw the exception in the sum
step.
If you don't want to chain, you can use a variable for each step:
IntStream is = IntStream.of(1, 2, 3, 4);
IntStream is2 = is.map(i -> i + 1);
int sum = is2.sum();
add a comment |
up vote
3
down vote
Intermediate operations return a new stream. They are always lazy; executing an intermediate operation such as filter() does not actually perform any filtering, but instead creates a new stream that, when traversed, contains the elements of the initial stream that match the given predicate.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html under "Stream Operations and Pipelines"
At the lowest level, all streams are driven by a spliterator.
Taken from the same link under "Low-level stream construction"
Traversal and splitting exhaust elements; each Spliterator is useful for only a single bulk computation.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/Spliterator.html
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
45
down vote
accepted
When you do this:
int sum = IntStream.of(1, 2, 3, 4).map(i -> i + 1).sum();
Every chained method is being invoked on the return value of the previous method in the chain.
So map
is invoked on what IntStream.of(1, 2, 3, 4)
returns and sum
on what map(i -> i + 1)
returns.
You don't have to chain stream methods, but it's more readable and less error-prone than using this equivalent code:
IntStream is = IntStream.of(1, 2, 3, 4);
is = is.map(i -> i + 1);
int sum = is.sum();
Which is not the same as the code you've shown in your question:
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
As you see, you're disregarding the reference returned by map
. This is the cause of the error.
EDIT (as per the comments, thanks to @IanKemp for pointing this out): Actually, this is the external cause of the error. If you stop to think about it, map
must be doing something internally to the stream itself, otherwise, how would then the terminal operation trigger the transformation passed to map
on each element? I agree in that intermediate operations are lazy, i.e. when invoked, they do nothing to the elements of the stream. But internally, they must configure some state into the stream pipeline itself, so that they can be applied later.
Despite I'm not aware of the full details, what happens is that, conceptually, map
is doing at least 2 things:
It's creating and returning a new stream that holds the function passed as an argument somewhere, so that it can be applied to elements later, when the terminal operation is invoked.
It is also setting a flag to the old stream instance, i.e. the one which it has been called on, indicating that this stream instance no longer represents a valid state for the pipeline. This is because the new, updated state which holds the function passed to
map
is now encapsulated by the instance it has returned. (I believe that this decision might have been taken by the jdk team to make errors appear as early as possible, i.e. by throwing an early exception instead of letting the pipeline go on with an invalid/old state that doesn't hold the function to be applied, thus letting the terminal operation return unexpected results).
Later on, when a terminal operation is invoked on this instance flagged as invalid, you're getting that IllegalStateException
. The two items above configure the deep, internal cause of the error.
Another way to see all this is to make sure that a Stream
instance is operated only once, by means of either an intermediate or a terminal operation. Here you are violating this requirement, because you are calling map
and sum
on the same instance.
In fact, javadocs for Stream
state it clearly:
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw
IllegalStateException
if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
1
@Koray To add to that, adding a debug point at both your linesis.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipelineis
which is what you are not reusing and I believe sinceis
is already marked aslinkedOrConsumed
in your case, you see an ISE.
– nullpointer
Nov 12 at 2:27
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
This feels very similar to the age oldstr.toLowerCase();
vss = str.toLowerCase();
stumbling block in Java
– DeadChex
Nov 12 at 6:48
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,map()
is an intermediate operation, which means that applying it and then throwing away the result viais.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operationint sum = is.sum();
on that underlying stream should still succeed.
– Ian Kemp
Nov 12 at 9:24
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
|
show 5 more comments
up vote
45
down vote
accepted
When you do this:
int sum = IntStream.of(1, 2, 3, 4).map(i -> i + 1).sum();
Every chained method is being invoked on the return value of the previous method in the chain.
So map
is invoked on what IntStream.of(1, 2, 3, 4)
returns and sum
on what map(i -> i + 1)
returns.
You don't have to chain stream methods, but it's more readable and less error-prone than using this equivalent code:
IntStream is = IntStream.of(1, 2, 3, 4);
is = is.map(i -> i + 1);
int sum = is.sum();
Which is not the same as the code you've shown in your question:
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
As you see, you're disregarding the reference returned by map
. This is the cause of the error.
EDIT (as per the comments, thanks to @IanKemp for pointing this out): Actually, this is the external cause of the error. If you stop to think about it, map
must be doing something internally to the stream itself, otherwise, how would then the terminal operation trigger the transformation passed to map
on each element? I agree in that intermediate operations are lazy, i.e. when invoked, they do nothing to the elements of the stream. But internally, they must configure some state into the stream pipeline itself, so that they can be applied later.
Despite I'm not aware of the full details, what happens is that, conceptually, map
is doing at least 2 things:
It's creating and returning a new stream that holds the function passed as an argument somewhere, so that it can be applied to elements later, when the terminal operation is invoked.
It is also setting a flag to the old stream instance, i.e. the one which it has been called on, indicating that this stream instance no longer represents a valid state for the pipeline. This is because the new, updated state which holds the function passed to
map
is now encapsulated by the instance it has returned. (I believe that this decision might have been taken by the jdk team to make errors appear as early as possible, i.e. by throwing an early exception instead of letting the pipeline go on with an invalid/old state that doesn't hold the function to be applied, thus letting the terminal operation return unexpected results).
Later on, when a terminal operation is invoked on this instance flagged as invalid, you're getting that IllegalStateException
. The two items above configure the deep, internal cause of the error.
Another way to see all this is to make sure that a Stream
instance is operated only once, by means of either an intermediate or a terminal operation. Here you are violating this requirement, because you are calling map
and sum
on the same instance.
In fact, javadocs for Stream
state it clearly:
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw
IllegalStateException
if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
1
@Koray To add to that, adding a debug point at both your linesis.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipelineis
which is what you are not reusing and I believe sinceis
is already marked aslinkedOrConsumed
in your case, you see an ISE.
– nullpointer
Nov 12 at 2:27
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
This feels very similar to the age oldstr.toLowerCase();
vss = str.toLowerCase();
stumbling block in Java
– DeadChex
Nov 12 at 6:48
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,map()
is an intermediate operation, which means that applying it and then throwing away the result viais.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operationint sum = is.sum();
on that underlying stream should still succeed.
– Ian Kemp
Nov 12 at 9:24
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
|
show 5 more comments
up vote
45
down vote
accepted
up vote
45
down vote
accepted
When you do this:
int sum = IntStream.of(1, 2, 3, 4).map(i -> i + 1).sum();
Every chained method is being invoked on the return value of the previous method in the chain.
So map
is invoked on what IntStream.of(1, 2, 3, 4)
returns and sum
on what map(i -> i + 1)
returns.
You don't have to chain stream methods, but it's more readable and less error-prone than using this equivalent code:
IntStream is = IntStream.of(1, 2, 3, 4);
is = is.map(i -> i + 1);
int sum = is.sum();
Which is not the same as the code you've shown in your question:
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
As you see, you're disregarding the reference returned by map
. This is the cause of the error.
EDIT (as per the comments, thanks to @IanKemp for pointing this out): Actually, this is the external cause of the error. If you stop to think about it, map
must be doing something internally to the stream itself, otherwise, how would then the terminal operation trigger the transformation passed to map
on each element? I agree in that intermediate operations are lazy, i.e. when invoked, they do nothing to the elements of the stream. But internally, they must configure some state into the stream pipeline itself, so that they can be applied later.
Despite I'm not aware of the full details, what happens is that, conceptually, map
is doing at least 2 things:
It's creating and returning a new stream that holds the function passed as an argument somewhere, so that it can be applied to elements later, when the terminal operation is invoked.
It is also setting a flag to the old stream instance, i.e. the one which it has been called on, indicating that this stream instance no longer represents a valid state for the pipeline. This is because the new, updated state which holds the function passed to
map
is now encapsulated by the instance it has returned. (I believe that this decision might have been taken by the jdk team to make errors appear as early as possible, i.e. by throwing an early exception instead of letting the pipeline go on with an invalid/old state that doesn't hold the function to be applied, thus letting the terminal operation return unexpected results).
Later on, when a terminal operation is invoked on this instance flagged as invalid, you're getting that IllegalStateException
. The two items above configure the deep, internal cause of the error.
Another way to see all this is to make sure that a Stream
instance is operated only once, by means of either an intermediate or a terminal operation. Here you are violating this requirement, because you are calling map
and sum
on the same instance.
In fact, javadocs for Stream
state it clearly:
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw
IllegalStateException
if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
When you do this:
int sum = IntStream.of(1, 2, 3, 4).map(i -> i + 1).sum();
Every chained method is being invoked on the return value of the previous method in the chain.
So map
is invoked on what IntStream.of(1, 2, 3, 4)
returns and sum
on what map(i -> i + 1)
returns.
You don't have to chain stream methods, but it's more readable and less error-prone than using this equivalent code:
IntStream is = IntStream.of(1, 2, 3, 4);
is = is.map(i -> i + 1);
int sum = is.sum();
Which is not the same as the code you've shown in your question:
IntStream is = IntStream.of(1, 2, 3, 4);
is.map(i -> i + 1);
int sum = is.sum();
As you see, you're disregarding the reference returned by map
. This is the cause of the error.
EDIT (as per the comments, thanks to @IanKemp for pointing this out): Actually, this is the external cause of the error. If you stop to think about it, map
must be doing something internally to the stream itself, otherwise, how would then the terminal operation trigger the transformation passed to map
on each element? I agree in that intermediate operations are lazy, i.e. when invoked, they do nothing to the elements of the stream. But internally, they must configure some state into the stream pipeline itself, so that they can be applied later.
Despite I'm not aware of the full details, what happens is that, conceptually, map
is doing at least 2 things:
It's creating and returning a new stream that holds the function passed as an argument somewhere, so that it can be applied to elements later, when the terminal operation is invoked.
It is also setting a flag to the old stream instance, i.e. the one which it has been called on, indicating that this stream instance no longer represents a valid state for the pipeline. This is because the new, updated state which holds the function passed to
map
is now encapsulated by the instance it has returned. (I believe that this decision might have been taken by the jdk team to make errors appear as early as possible, i.e. by throwing an early exception instead of letting the pipeline go on with an invalid/old state that doesn't hold the function to be applied, thus letting the terminal operation return unexpected results).
Later on, when a terminal operation is invoked on this instance flagged as invalid, you're getting that IllegalStateException
. The two items above configure the deep, internal cause of the error.
Another way to see all this is to make sure that a Stream
instance is operated only once, by means of either an intermediate or a terminal operation. Here you are violating this requirement, because you are calling map
and sum
on the same instance.
In fact, javadocs for Stream
state it clearly:
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw
IllegalStateException
if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
edited Nov 12 at 17:02
answered Nov 12 at 2:18
Federico Peralta Schaffner
21.8k43369
21.8k43369
1
@Koray To add to that, adding a debug point at both your linesis.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipelineis
which is what you are not reusing and I believe sinceis
is already marked aslinkedOrConsumed
in your case, you see an ISE.
– nullpointer
Nov 12 at 2:27
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
This feels very similar to the age oldstr.toLowerCase();
vss = str.toLowerCase();
stumbling block in Java
– DeadChex
Nov 12 at 6:48
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,map()
is an intermediate operation, which means that applying it and then throwing away the result viais.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operationint sum = is.sum();
on that underlying stream should still succeed.
– Ian Kemp
Nov 12 at 9:24
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
|
show 5 more comments
1
@Koray To add to that, adding a debug point at both your linesis.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipelineis
which is what you are not reusing and I believe sinceis
is already marked aslinkedOrConsumed
in your case, you see an ISE.
– nullpointer
Nov 12 at 2:27
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
This feels very similar to the age oldstr.toLowerCase();
vss = str.toLowerCase();
stumbling block in Java
– DeadChex
Nov 12 at 6:48
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,map()
is an intermediate operation, which means that applying it and then throwing away the result viais.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operationint sum = is.sum();
on that underlying stream should still succeed.
– Ian Kemp
Nov 12 at 9:24
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
1
1
@Koray To add to that, adding a debug point at both your lines
is.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipeline is
which is what you are not reusing and I believe since is
is already marked as linkedOrConsumed
in your case, you see an ISE.– nullpointer
Nov 12 at 2:27
@Koray To add to that, adding a debug point at both your lines
is.map(i -> i + 1); int sum = is.sum();
, you can notice the change in the IntPipeline is
which is what you are not reusing and I believe since is
is already marked as linkedOrConsumed
in your case, you see an ISE.– nullpointer
Nov 12 at 2:27
1
1
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
@nullpointer I'm not sure. Memorization sounds too similar to memoization, which is a completely different thing... Maybe tracking is a better term?
– Federico Peralta Schaffner
Nov 12 at 2:35
6
6
This feels very similar to the age old
str.toLowerCase();
vs s = str.toLowerCase();
stumbling block in Java– DeadChex
Nov 12 at 6:48
This feels very similar to the age old
str.toLowerCase();
vs s = str.toLowerCase();
stumbling block in Java– DeadChex
Nov 12 at 6:48
1
1
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,
map()
is an intermediate operation, which means that applying it and then throwing away the result via is.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operation int sum = is.sum();
on that underlying stream should still succeed.– Ian Kemp
Nov 12 at 9:24
"As you see, you're disregarding the reference returned by map. This is the cause of the error." - that doesn't make sense to me. As the asker has noted,
map()
is an intermediate operation, which means that applying it and then throwing away the result via is.map(i -> i + 1);
should have no effect on the underlying stream, and thus calling the terminal operation int sum = is.sum();
on that underlying stream should still succeed.– Ian Kemp
Nov 12 at 9:24
1
1
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
I agree with @IanKemp that this answer doesn't really answer the question. As the intermediate operations return a new stream, one would think that the old stream stays untouched. The interface suggests immutability.
– xehpuk
Nov 12 at 13:42
|
show 5 more comments
up vote
15
down vote
Imagine the IntStream is a wrapper around your data stream with an
immutable list of operations. These operations are not executed until you need the final result (sum in your case).
Since the list is immutable, you need a new instance of IntStream with a list that contains the previous items plus the new one, which is what '. map' returns.
This means that if you don't chain, you will operate on the old instance, which does not have that operation.
The stream library also keeps some internal tracking of what's going on, that's why it's able to throw the exception in the sum
step.
If you don't want to chain, you can use a variable for each step:
IntStream is = IntStream.of(1, 2, 3, 4);
IntStream is2 = is.map(i -> i + 1);
int sum = is2.sum();
add a comment |
up vote
15
down vote
Imagine the IntStream is a wrapper around your data stream with an
immutable list of operations. These operations are not executed until you need the final result (sum in your case).
Since the list is immutable, you need a new instance of IntStream with a list that contains the previous items plus the new one, which is what '. map' returns.
This means that if you don't chain, you will operate on the old instance, which does not have that operation.
The stream library also keeps some internal tracking of what's going on, that's why it's able to throw the exception in the sum
step.
If you don't want to chain, you can use a variable for each step:
IntStream is = IntStream.of(1, 2, 3, 4);
IntStream is2 = is.map(i -> i + 1);
int sum = is2.sum();
add a comment |
up vote
15
down vote
up vote
15
down vote
Imagine the IntStream is a wrapper around your data stream with an
immutable list of operations. These operations are not executed until you need the final result (sum in your case).
Since the list is immutable, you need a new instance of IntStream with a list that contains the previous items plus the new one, which is what '. map' returns.
This means that if you don't chain, you will operate on the old instance, which does not have that operation.
The stream library also keeps some internal tracking of what's going on, that's why it's able to throw the exception in the sum
step.
If you don't want to chain, you can use a variable for each step:
IntStream is = IntStream.of(1, 2, 3, 4);
IntStream is2 = is.map(i -> i + 1);
int sum = is2.sum();
Imagine the IntStream is a wrapper around your data stream with an
immutable list of operations. These operations are not executed until you need the final result (sum in your case).
Since the list is immutable, you need a new instance of IntStream with a list that contains the previous items plus the new one, which is what '. map' returns.
This means that if you don't chain, you will operate on the old instance, which does not have that operation.
The stream library also keeps some internal tracking of what's going on, that's why it's able to throw the exception in the sum
step.
If you don't want to chain, you can use a variable for each step:
IntStream is = IntStream.of(1, 2, 3, 4);
IntStream is2 = is.map(i -> i + 1);
int sum = is2.sum();
edited Nov 12 at 2:27
nullpointer
39.1k1074149
39.1k1074149
answered Nov 12 at 2:17
Pascal Ludwig
6311614
6311614
add a comment |
add a comment |
up vote
3
down vote
Intermediate operations return a new stream. They are always lazy; executing an intermediate operation such as filter() does not actually perform any filtering, but instead creates a new stream that, when traversed, contains the elements of the initial stream that match the given predicate.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html under "Stream Operations and Pipelines"
At the lowest level, all streams are driven by a spliterator.
Taken from the same link under "Low-level stream construction"
Traversal and splitting exhaust elements; each Spliterator is useful for only a single bulk computation.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/Spliterator.html
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
add a comment |
up vote
3
down vote
Intermediate operations return a new stream. They are always lazy; executing an intermediate operation such as filter() does not actually perform any filtering, but instead creates a new stream that, when traversed, contains the elements of the initial stream that match the given predicate.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html under "Stream Operations and Pipelines"
At the lowest level, all streams are driven by a spliterator.
Taken from the same link under "Low-level stream construction"
Traversal and splitting exhaust elements; each Spliterator is useful for only a single bulk computation.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/Spliterator.html
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
add a comment |
up vote
3
down vote
up vote
3
down vote
Intermediate operations return a new stream. They are always lazy; executing an intermediate operation such as filter() does not actually perform any filtering, but instead creates a new stream that, when traversed, contains the elements of the initial stream that match the given predicate.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html under "Stream Operations and Pipelines"
At the lowest level, all streams are driven by a spliterator.
Taken from the same link under "Low-level stream construction"
Traversal and splitting exhaust elements; each Spliterator is useful for only a single bulk computation.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/Spliterator.html
Intermediate operations return a new stream. They are always lazy; executing an intermediate operation such as filter() does not actually perform any filtering, but instead creates a new stream that, when traversed, contains the elements of the initial stream that match the given predicate.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html under "Stream Operations and Pipelines"
At the lowest level, all streams are driven by a spliterator.
Taken from the same link under "Low-level stream construction"
Traversal and splitting exhaust elements; each Spliterator is useful for only a single bulk computation.
Taken from https://docs.oracle.com/javase/8/docs/api/java/util/Spliterator.html
edited Nov 12 at 2:43
answered Nov 12 at 2:22
4dc0
42539
42539
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
add a comment |
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
1
1
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
Yes ... but that doesn't explain what happens if you ignore the result.
– Stephen C
Nov 12 at 2:28
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
you were right, i updated my answer
– 4dc0
Nov 12 at 2:44
add a comment |
2
A stream should be operated on (invoking an intermediate or terminal stream operation) only once. This rules out, for example, "forked" streams, where the same source feeds two or more pipelines, or multiple traversals of the same stream. A stream implementation may throw IllegalStateException if it detects that the stream is being reused. However, since some stream operations may return their receiver rather than a new stream object, it may not be possible to detect reuse in all cases.
– nullpointer
Nov 12 at 2:16
1
all streams are driven by a spliterator, and each spliterator is only useful for a single bulk computation
– 4dc0
Nov 12 at 2:56