[英]Scala, can't implement generic java method
I'd like to implement a java method that uses generics in scala (2.9.2). 我想实现一个在scala(2.9.2)中使用泛型的java方法。 But I'm failing...
但是我失败了
Java interface method: Java接口方法:
public <T extends Number> void setAttribute(Key<T> key, Number value);
Scala code that want to implement that method: 要实现该方法的Scala代码:
def setAttribute[T <: Number](key: Key[T], value: Number) = {
setAttributeLocal(key, value) }
private def setAttributeLocal[T](key: Key[T], value: T) = {
val stringValue = ConvertUtils.convert(value, classOf[String]).asInstanceOf[String]
session = session + (key.getValue() -> stringValue)
}
Key looks like: 密钥如下所示:
public class Key<T>
But this doesn't compile. 但这无法编译。
[error] found : mypackage.Key[T]
[error] required: mypackage.Key[java.lang.Number]
[error] Note: T <: java.lang.Number, but Java-defined class Key is invariant in type T.
[error] You may wish to investigate a wildcard type such as `_ <: java.lang.Number`. (SLS 3.2.10)
[error] setAttributeLocal(key, value)
I can't figure out what's the problem. 我不知道是什么问题。 Any suggestions/idea?
有什么建议/想法吗?
greez GarfieldKlon 格里兹·加菲尔德·科隆
It appears the compiler is unhappy with your call to setAttributeLocal
. 看来编译器对您对
setAttributeLocal
的调用不满意。 setAttributeLocal
requires a Key[Number]
, but you are providing a Key[_ <: T]
. setAttributeLocal
需要一个Key[Number]
,但是您要提供一个Key[_ <: T]
。 In Java-Land this means you're trying to pass a Key<? extends Number>
在Java-Land中,这意味着您尝试传递
Key<? extends Number>
Key<? extends Number>
off as a Key<Number>
. Key<? extends Number>
为Key<Number>
。
The suggestion is to have setAttributeLocal
accept Key<? extends Number>
建议是让
setAttributeLocal
接受Key<? extends Number>
Key<? extends Number>
or Key[_ <: Number]
, depending on whether it is Java- or Scala-defined. Key<? extends Number>
或Key[_ <: Number]
,具体取决于它是Java定义还是Scala定义的。
Something looks a bit off here. 这里看起来有些不对劲。
Have you tried: 你有没有尝试过:
def setAttribute[T <: Number](key: Key[T], value: T) =
setAttributeLocal(key, value)
It seems strange/bad to preserve the type T for the key, but not use it on the value. 保留键的类型T,但不对值使用它,这似乎很奇怪/不好。 My guess is that's where you're getting an invariant error.
我的猜测是,您会收到一个不变的错误。 You're trying to assing the value of type
Number
to a key of type T
and the compiler isn't sure if it can't pass Number
for T
(while it knows it can pass T
for Number
). 您试图将
Number
类型的值赋给T
类型的键,并且编译器不确定它是否不能为T
传递Number
(虽然它知道可以为Number
传递T
)。
Can we see more code? 我们可以看到更多代码吗?
As @jsuereth already pointed out, there is a discrepancy between the signatures of setAttribute
and setAttributeLocal
, namely, that the former accepts a Key[T <: Number]
but fixes the value that goes with the key to exactly be a Number
, whereas the latter is more flexible and allows key and value to be T <: Number
. 正如@jsuereth已经指出的那样,
setAttribute
和setAttributeLocal
的签名之间存在差异,即前者接受Key[T <: Number]
但将与该键一起的值固定为完全为Number
,而后者更灵活,并允许键和值是T <: Number
。 This looks rather odd and you might want to reconsider that decision. 这看起来很奇怪,您可能需要重新考虑这个决定。 It also leads to the problem explained by @Ben Schulz.
这也导致了@Ben Schulz解释的问题。
In any case, the compiler (2.9.1) is happy with the following setup: 无论如何,编译器(2.9.1)对以下设置感到满意:
MyI.java
: MyI.java
:
public interface MyI {
public <T extends Number> void setAttribute(Key<T> key, Number value);
}
Key.java
: Key.java
:
public interface Key<T> {
}
Test.scala
: Test.scala
:
object Test extends App {
def setAttribute[T <: Number](key: Key[T], value: Number) = {
setAttributeLocal(key, value) }
private def setAttributeLocal[T](key: Key[T], value: Number) = {
/* Notice that value is now also of type Number. */
}
}
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