我无法理解
之间的区别Stream<Optional<Integer>> optionalStream = Stream.of(
Optional.of(1),
Optional.empty(),
Optional.of(5));
List<Optional<Integer>> optionalList = optionalStream.collect(Collectors.toList());
哪种方法很好,并且:
List<Optional<Integer>> optionalList1 = Stream.of(
Optional.of(1),
Optional.empty(),
Optional.of(5)).collect(Collectors.toList());
我收到错误的地方
Error:(138, 40) java: incompatible types: inference variable T has incompatible bounds
equality constraints: java.util.Optional<java.lang.Integer>
lower bounds: java.util.Optional<? extends java.lang.Object>
答案 0 :(得分:4)
我稍微减少了一些示例,并尝试使用-XDverboseResolution=all
编译以输出有关类型推断的信息:
final class One {
void one() {
Stream<Optional<Integer>> optionalStream = Stream.of(Optional.empty());
List<Optional<Integer>> optionalList = optionalStream.collect(Collectors.toList());
}
}
final class Two {
void two() {
List<Optional<Integer>> optionalList1 =
Stream.of(Optional.empty()).collect(Collectors.toList());
}
}
对于Two
,看起来Stream.of
的延迟实例化甚至在查看后续collect
之前就完成了:
...
Two.java:9: Note: Deferred instantiation of method <T>of(T)
Stream.of(Optional.empty()).collect(Collectors.toList());
^
instantiated signature: (Optional<Object>)Stream<Optional<Object>>
target-type: <none>
where T is a type-variable:
T extends Object declared in method <T>of(T)
Two.java:9: Note: resolving method collect in type Stream to candidate 0
Stream.of(Optional.empty()).collect(Collectors.toList());
...
(“解析方法collect
是第一次提及collect
)
没有target-type
来限制它;实例化的签名显示它是Stream<Optional<Object>>
。
如果查看One
的相应输出:
...
One.java:8: Note: Deferred instantiation of method <T>of(T)
Stream<Optional<Integer>> optionalStream = Stream.of(Optional.empty());
^
instantiated signature: (Optional<Integer>)Stream<Optional<Integer>>
target-type: Stream<Optional<Integer>>
where T is a type-variable:
T extends Object declared in method <T>of(T)
...
它是正确的,因为它知道目标类型。
我不能确切地说为什么延迟实例化在Two
的此时发生,因为我对应用类型推断的方式不太熟悉。
我认为是因为Stream.of
的调用不被视为多元表达,但我无法真正说服自己为什么(请参阅编辑历史记录以了解一些不连贯的翻译)。 / p>
我建议的修正方法是对Optional.empty()
应用类型提示,即Optional.<Integer>empty()
。这具有在推理中更早获得Optional
的实际类型的效果,因此在延迟实例化时已知事件,尽管目标类型仍未知:
final class Three {
void three() {
List<Optional<Integer>> optionalList1 =
Stream.of(Optional.<Integer>empty()).collect(Collectors.toList());
}
}
...
Three.java:9: Note: resolving method of in type Stream to candidate 1
Stream.of(Optional.<Integer>empty()).collect(Collectors.toList());
^
phase: BASIC
with actuals: Optional<Integer>
with type-args: no arguments
candidates:
#0 not applicable method found: <T#1>of(T#1...)
(cannot infer type-variable(s) T#1
(argument mismatch; Optional<Integer> cannot be converted to T#1[]))
#1 applicable method found: <T#2>of(T#2)
(partially instantiated to: (Optional<Integer>)Stream<Optional<Integer>>)
where T#1,T#2 are type-variables:
T#1 extends Object declared in method <T#1>of(T#1...)
T#2 extends Object declared in method <T#2>of(T#2)
Three.java:9: Note: Deferred instantiation of method <T>of(T)
Stream.of(Optional.<Integer>empty()).collect(Collectors.toList());
^
instantiated signature: (Optional<Integer>)Stream<Optional<Integer>>
target-type: <none>
where T is a type-variable:
T extends Object declared in method <T>of(T)
...
答案 1 :(得分:3)
Stream.of(...)
或Optional.empty()
是通用方法。如果您不提供type参数,则会推断出它。对于Optional.empty()
,您将获得Optional<Object>
,因此Stream.of(Optional.of(1), Optional.empty(), Optional.of(5))
会产生Stream<Optional<? extends Object>>
。
您可以通过在Optional.<Integer>empty()
或Stream.<Optional<Integer>>of(...)
中提供类型参数来解决此问题。我更喜欢第一个。