Why isn't Collections.sort() optimized for ArrayList, but is for LinkedList?
Why does Collections.sort()
creates an extra object array and performs Tim sort on the array and then finally copies the sorted array back into List
object? I know this call is optimized for LinkedList
, but won't we lose out on performance for ArrayList
?
We could have avoided 2n
number of operations in converting it into an object array and adding them back to the list. I know that these extra operations wouldn't affect the Big-O of the whole sorting operation, but I believe it could've been further optimised for ArrayList
.
Am I missing something here? I'm just trying to understand why the architecture is laid out as such. Thanks.
https://hg.openjdk.java.net/jdk8/jdk8/jdk/file/687fd7c7986d/src/share/classes/java/util/Collections.java#l164
java algorithm sorting timsort
add a comment |
Why does Collections.sort()
creates an extra object array and performs Tim sort on the array and then finally copies the sorted array back into List
object? I know this call is optimized for LinkedList
, but won't we lose out on performance for ArrayList
?
We could have avoided 2n
number of operations in converting it into an object array and adding them back to the list. I know that these extra operations wouldn't affect the Big-O of the whole sorting operation, but I believe it could've been further optimised for ArrayList
.
Am I missing something here? I'm just trying to understand why the architecture is laid out as such. Thanks.
https://hg.openjdk.java.net/jdk8/jdk8/jdk/file/687fd7c7986d/src/share/classes/java/util/Collections.java#l164
java algorithm sorting timsort
Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42
add a comment |
Why does Collections.sort()
creates an extra object array and performs Tim sort on the array and then finally copies the sorted array back into List
object? I know this call is optimized for LinkedList
, but won't we lose out on performance for ArrayList
?
We could have avoided 2n
number of operations in converting it into an object array and adding them back to the list. I know that these extra operations wouldn't affect the Big-O of the whole sorting operation, but I believe it could've been further optimised for ArrayList
.
Am I missing something here? I'm just trying to understand why the architecture is laid out as such. Thanks.
https://hg.openjdk.java.net/jdk8/jdk8/jdk/file/687fd7c7986d/src/share/classes/java/util/Collections.java#l164
java algorithm sorting timsort
Why does Collections.sort()
creates an extra object array and performs Tim sort on the array and then finally copies the sorted array back into List
object? I know this call is optimized for LinkedList
, but won't we lose out on performance for ArrayList
?
We could have avoided 2n
number of operations in converting it into an object array and adding them back to the list. I know that these extra operations wouldn't affect the Big-O of the whole sorting operation, but I believe it could've been further optimised for ArrayList
.
Am I missing something here? I'm just trying to understand why the architecture is laid out as such. Thanks.
https://hg.openjdk.java.net/jdk8/jdk8/jdk/file/687fd7c7986d/src/share/classes/java/util/Collections.java#l164
java algorithm sorting timsort
java algorithm sorting timsort
edited Jan 13 at 10:49
Slaw
7,5533932
7,5533932
asked Jan 13 at 10:41
Vineeth ChittetiVineeth Chitteti
711522
711522
Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42
add a comment |
Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42
Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42
Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42
add a comment |
1 Answer
1
active
oldest
votes
You are looking at an older JDK version. At least since since JDK 1.8.0_162 Collections.sort()
calls List
's sort(Comparator<? super E> c)
. And while the default implementation creates an array from the List
and sorts that array, ArrayList
overrides that default implementation, and sorts the backing array directly.
Collections
's sort
:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}
List
's sort
:
default void sort(Comparator<? super E> c) {
Object a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
ArrayList
's sort
:
public void sort(Comparator<? super E> c) {
final int expectedModCount = modCount;
Arrays.sort((E) elementData, 0, size, c);
if (modCount != expectedModCount) {
throw new ConcurrentModificationException();
}
modCount++;
}
In other words, this has been addressed in JDK versions newer than the one you are looking at.
You can look at the source here (thanks to eckes's link).
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does showCollections.sort
using the defaultList.sort
implementation directly.
– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on thetoArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.
– Joe C
Jan 13 at 10:56
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are looking at an older JDK version. At least since since JDK 1.8.0_162 Collections.sort()
calls List
's sort(Comparator<? super E> c)
. And while the default implementation creates an array from the List
and sorts that array, ArrayList
overrides that default implementation, and sorts the backing array directly.
Collections
's sort
:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}
List
's sort
:
default void sort(Comparator<? super E> c) {
Object a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
ArrayList
's sort
:
public void sort(Comparator<? super E> c) {
final int expectedModCount = modCount;
Arrays.sort((E) elementData, 0, size, c);
if (modCount != expectedModCount) {
throw new ConcurrentModificationException();
}
modCount++;
}
In other words, this has been addressed in JDK versions newer than the one you are looking at.
You can look at the source here (thanks to eckes's link).
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does showCollections.sort
using the defaultList.sort
implementation directly.
– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on thetoArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.
– Joe C
Jan 13 at 10:56
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
add a comment |
You are looking at an older JDK version. At least since since JDK 1.8.0_162 Collections.sort()
calls List
's sort(Comparator<? super E> c)
. And while the default implementation creates an array from the List
and sorts that array, ArrayList
overrides that default implementation, and sorts the backing array directly.
Collections
's sort
:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}
List
's sort
:
default void sort(Comparator<? super E> c) {
Object a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
ArrayList
's sort
:
public void sort(Comparator<? super E> c) {
final int expectedModCount = modCount;
Arrays.sort((E) elementData, 0, size, c);
if (modCount != expectedModCount) {
throw new ConcurrentModificationException();
}
modCount++;
}
In other words, this has been addressed in JDK versions newer than the one you are looking at.
You can look at the source here (thanks to eckes's link).
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does showCollections.sort
using the defaultList.sort
implementation directly.
– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on thetoArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.
– Joe C
Jan 13 at 10:56
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
add a comment |
You are looking at an older JDK version. At least since since JDK 1.8.0_162 Collections.sort()
calls List
's sort(Comparator<? super E> c)
. And while the default implementation creates an array from the List
and sorts that array, ArrayList
overrides that default implementation, and sorts the backing array directly.
Collections
's sort
:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}
List
's sort
:
default void sort(Comparator<? super E> c) {
Object a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
ArrayList
's sort
:
public void sort(Comparator<? super E> c) {
final int expectedModCount = modCount;
Arrays.sort((E) elementData, 0, size, c);
if (modCount != expectedModCount) {
throw new ConcurrentModificationException();
}
modCount++;
}
In other words, this has been addressed in JDK versions newer than the one you are looking at.
You can look at the source here (thanks to eckes's link).
You are looking at an older JDK version. At least since since JDK 1.8.0_162 Collections.sort()
calls List
's sort(Comparator<? super E> c)
. And while the default implementation creates an array from the List
and sorts that array, ArrayList
overrides that default implementation, and sorts the backing array directly.
Collections
's sort
:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}
List
's sort
:
default void sort(Comparator<? super E> c) {
Object a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
ArrayList
's sort
:
public void sort(Comparator<? super E> c) {
final int expectedModCount = modCount;
Arrays.sort((E) elementData, 0, size, c);
if (modCount != expectedModCount) {
throw new ConcurrentModificationException();
}
modCount++;
}
In other words, this has been addressed in JDK versions newer than the one you are looking at.
You can look at the source here (thanks to eckes's link).
edited Jan 13 at 11:14
answered Jan 13 at 10:44
EranEran
282k37457543
282k37457543
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does showCollections.sort
using the defaultList.sort
implementation directly.
– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on thetoArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.
– Joe C
Jan 13 at 10:56
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
add a comment |
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does showCollections.sort
using the defaultList.sort
implementation directly.
– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on thetoArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.
– Joe C
Jan 13 at 10:56
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
1
1
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does show
Collections.sort
using the default List.sort
implementation directly.– Slaw
Jan 13 at 10:48
The link to the source file (supposedly for JDK8, though I don't fully understand the repo's organization) does show
Collections.sort
using the default List.sort
implementation directly.– Slaw
Jan 13 at 10:48
Not what I see; I see sorting done on the
toArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.– Joe C
Jan 13 at 10:56
Not what I see; I see sorting done on the
toArray()
on line 230 (perhaps you're looking at the synchronized wrapper?). That said, OpenJDK11 is showing what you're saying in its source.– Joe C
Jan 13 at 10:56
1
1
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
@Slaw the linked source file seems to be jdk8-b132. My answer contains code of JDK 1.8.0_162
– Eran
Jan 13 at 10:57
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
Here's the JDK9 version - hg.openjdk.java.net/jdk9/jdk9/jdk/file/d966fc5a7be5/src/…. I'll link to JDK 1.8.0_162 if I can find it
– Eran
Jan 13 at 11:00
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
@Eran hg.openjdk.java.net/jdk8u/jdk8u/jdk/file/c00bdbbd9a77/src/share/…
– eckes
Jan 13 at 11:09
add a comment |
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Please PM here if the question is a possible duplicate. I will be glad to know that there is already an answer to this question.
– Vineeth Chitteti
Jan 13 at 10:42