问题
Can any one explain why the below code will not compile but the second one does?
Do not compile
private void doNotCompile() {
List<Integer> out;
out = IntStream
.range(1, 10)
.filter(e -> e % 2 == 0)
.map(e -> Integer.valueOf(2 * e))
.collect(Collectors.toList());
System.out.println(out);
}
Compilation errors on the collect line
- The method collect(Supplier, ObjIntConsumer, BiConsumer) in the type IntStream is not applicable for the arguments (Collector>)
- Type mismatch: cannot convert from Collector> to Supplier
Compiles
private void compiles() {
List<Integer> in;
in = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9);
List<Integer> out;
out = in.stream()
.filter(e -> e % 2 == 0)
.map(e -> 2 * e)
.collect(Collectors.toList());
System.out.println(out);
}
回答1:
IntStream
doesn't have a collect
method that accepts a Collector
. If you want a List<Integer>
, you have to box the IntStream
into a Stream<Integer>
:
out = IntStream
.range(1, 10)
.filter(e -> e % 2 == 0)
.map(e -> 2 * e)
.boxed()
.collect(Collectors.toList());
An alternative to .map().boxed()
is mapToObj()
:
out = IntStream
.range(1, 10)
.filter(e -> e % 2 == 0)
.mapToObj(e -> 2 * e)
.collect(Collectors.toList ());
or you can use the IntStream
collect
method:
out = IntStream
.range(1, 10)
.filter(e -> e % 2 == 0)
.map(e -> 2 * e)
.collect(ArrayList<Integer>::new, ArrayList::add, ArrayList::addAll);
回答2:
The collect
method of IntStream
is different in the non-compilable case.
public <R> R collect(Supplier<R> supplier, ObjIntConsumer<R> accumulator, BiConsumer<R,R> combiner)
Therefore it does not accept the parameter you supplied to collect
You could get around this by casting List<Integer>
to the IntStream
回答3:
In the first example, you are operating over a stream of primitive integers. Primitive integers can't go into a List
, basically because generics in Java are less than ideal. Java language designers are working on potentially fixing this.
In the meantime, to solve this you need to box these primitive ints into an Integer wrapper first. See Eran's answer for a code sample.
In the second example, you're already iterating over Integer
s so It Just Works™.
I thought I was boxing those int to Integers when I did Integer.valueOf in the mapper
The map function of IntStream takes a IntUnaryOperator which is a functional interface that takes a primitive int and returns a primitive int.
The Integer
you get from valueOf
is unboxed to match the functional interface.
来源:https://stackoverflow.com/questions/46827927/java-stream-collectors-tolist-wont-compile