Indeed it's the case that a varargs method can't determine whether it was called
with several explicit arguments or whether it was called with an array. However,
that doesn't have any bearing on whether or not Set.of rejects duplicates.
The model for Set.of is to support a collection-literal-like syntax where the
programmer can write an arbitrary number of elements in the source code for
inclusion in the set. Here's an example (though it uses Map.ofEntries instead of
Set.of, the same rationale applies):
Map<String, TokenType> tokens = Map.ofEntries(
entry("@", AT),
entry("|", VERTICAL_BAR),
entry("#", HASH),
entry("%", PERCENT),
entry(":", COLON),
entry("^", CARET),
entry("&", AMPERSAND),
entry("|", EXCLAM),
entry("?", QUESTION),
entry("$", DOLLAR),
entry("::", PAAMAYIM_NEKUDOTAYIM),
entry("=", EQUALS),
entry(";", SEMICOLON)
);
This errors out instead of silently dropping one of the entries.
As an optimization, the API provides several fixed-arg overloads of Set.of. With few
arguments, the fixed-arg methods are called. If more arguments are added, at a
certain point it transparently switches to the varargs form. "Transparently" means
that you can't tell (without counting the arguments) whether a fixed-arg or varargs
form of Set.of will be called. You don't want the duplicate rejection semantics to
change if you add or remove an argument that happens to cross the fixed/varargs
threshold. Thus, Set.of rejects duplicates, whether in fixed or varargs form.
Set.copyOf(Arrays.asList(...)) is the best way to deduplicate an explicit list of
elements into a set.
s'marks
On 2/1/21 3:01 PM, Aaron Scott-Boddendijk wrote:
Dave,
|| Dave said...
|| Okay, I understand this reasoning, but when you want to construct a Set
from an array, you might be tempted to use Set.of(...) because it looks
like it supports an
|| array and indeed, you can do Set.of(new int[] {1, 2 }) I believe?
||
|| Maybe this is just a quirk because of how varargs work.
| Rémi said...
| Set.of(int[]) will call Set.of(E) with E being an int[].
| but
| Set.of(new Integer[] { ... }) calls Set.of(...).
|
| Yes, exactly, it's a known issue with varargs, you have no way to say, i
don't want this varargs to be called with an array.
I think the confusion is the interaction of boxing and varargs.
List<Integer> list = List.of(1, 2);
is actually, once auto-boxing is applied by the compiler, executed as...
List<Integer> list = List.of(Integer.valueOf(1), Integer.valueOf(2));
So the equivalent explicit array form should use `Integer[]` not `int[]`...
Integer[] numbers = new Integer[] {1, 2};
List<Integer> list = List.of(numbers);
Interestingly, if you actually wanted a `List<Integer[]>` you would then
need to say
Integer[] numbers = new Integer[] {1, 2};
List<Integer> list = List.<Integer[]>of(numbers);
Which is explicitly telling the compiler what the type arguments are for
this invocation of the generic method 'of'' (rather than allowing it to use
type-inference)
Regarding the use of `Set.copyOf(Arrays.asList(...))`. I do wonder about
improving the ceremony (because I agree that we want an obvious way of
getting immutable Sets from non-unique inputs) by following the pattern
presented in Optional (`Optional.of` and `Optional.ofNullable`) and
providing `Set.of` and `Set.ofMaybeUnique` (better name needed -
'ofOptionallyUnique'?) - to which the implementation could just be
`Set.copyOf(Arrays.asList(args))` (unless a more efficient path proves
valuable).
`Arrays.asList(...array...)` is not all that expensive. It is _not_ an
ArrayList but a much simpler type with rather trivial implementations for
most methods (and 'always throws' implementations for methods that are
unsupported). So not only does it mean that there's no copying occuring to
make the list but it's even possible that JIT manages enough specialisation
and inlining to elide the allocation entirely (though in practice this
doesn't occur as often as we might like).
--
Aaron Scott-Boddendijk
On Mon, Feb 1, 2021 at 10:35 AM <fo...@univ-mlv.fr> wrote:
De: "dfranken jdk" <dfranken....@gmail.com>
À: "Remi Forax" <fo...@univ-mlv.fr>
Cc: "core-libs-dev" <core-libs-dev@openjdk.java.net>
Envoyé: Dimanche 31 Janvier 2021 13:54:44
Objet: Re: Why does Set.of disallow duplicate elements?
BQ_BEGIN
Okay, I understand this reasoning, but when you want to construct a Set
from an array, you might be tempted to use Set.of(...) because it looks
like it supports an array and indeed, you can do Set.of(new int[] {1, 2 })
I believe?
BQ_END
Set.of(int[]) will call Set.of(E) with E being an int[].
but
Set.of(new Integer[] { ... }) calls Set.of(...).
BQ_BEGIN
Maybe this is just a quirk because of how varargs work.
BQ_END
Yes, exactly, it's a known issue with varargs, you have no way to say, i
don't want this varargs to be called with an array.
BQ_BEGIN
I wondered if there was a canonical way to create a Set from an array, but
couldn't find it, maybe I am missing something?
I did notice Arrays.asList exists (which makes sense because it creates an
ArrayList backed by the array), but not Arrays.asSet.
BQ_END
asList() reuse the same backing array, you can not do that with asSet() or
contains() will be in O(n) in the worst case.
BQ_BEGIN
So the way I would create a Set from an array would be either
Arrays.stream(myArr).collect(Collectors.toUnmodifiableSet()) or new
HashSet<>(Arrays.asList(myArray)) or Set.copyOf(Arrays.asList(myArray)).
BQ_END
yes, the last one is the easy way to create an unmodifiable set from an
array.
BQ_BEGIN
I'm not saying the way it is currently implemented is wrong, it's just
something which can suprise developers as it surprised me. :)
BQ_END
Arrays are currently second class citizen in Java, because they are always
modifiable and always covariant (String[] can be seen as a Object[]).
We have talked several times to introduce new variants of arrays,
non-modifiable one, non-covariant one, etc under the name Array 2.0, but
Valhalla generics is a blocker for that project.
Once Valhalla is done, it may be a follow up.
BQ_BEGIN
Kind regards,
Dave
BQ_END
regards,
Rémi
BQ_BEGIN
Op za 30 jan. 2021 om 21:30 schreef Remi Forax < [ mailto:
fo...@univ-mlv.fr | fo...@univ-mlv.fr ] >:
BQ_BEGIN
Set.of() is the closest way we've got to a literal Set without having
introduced a special syntax for that in the language.
The idea is that if you conceptually want to write
Set<String> set = { "hello", "world" };
instead, you write
Set<String> set = Set.of("hello", "world");
In that context, it makes sense to reject Set constructed with the same
element twice because this is usually a programming error.
So
Set.of("hello", "hello")
throws an IAE.
If you want a Set from a collection of elements, you can use
Set.copyOf(List.of("hello", "hello"))
regards,
Rémi
----- Mail original -----
De: "dfranken jdk" < [ mailto:dfranken....@gmail.com |
dfranken....@gmail.com ] >
À: "core-libs-dev" < [ mailto:core-libs-dev@openjdk.java.net |
core-libs-dev@openjdk.java.net ] >
Envoyé: Samedi 30 Janvier 2021 19:30:06
Objet: Why does Set.of disallow duplicate elements?
Dear users,
While looking at the implementation of Set.of(...) I noticed that
duplicate elements are not allowed, e.g. Set.of(1, 1) will throw an
IllegalArgumentException. Why has it been decided to do this?
My expectation was that duplicates would simply be removed.
If I do for instance new HashSet<>(<collection containing duplicates>)
it works and duplicates are removed. To me, it looks a bit inconsistent
to have duplicates removed for a collection passed in the constructor,
but not for a collection (even though it is a vararg array) passed to a
static factory method.
Kind regards,
Dave Franken
BQ_END
BQ_END
