java.lang.Boolean的到scala.Boolean问题 [英] java.lang.Boolean to scala.Boolean question

问题描述

georgii@gleontiev:~$ scala
Welcome to Scala version 2.8.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_24).
Type in expressions to have them evaluated.
Type :help for more information.

scala> val jbool = java.lang.Boolean.TRUE    
jbool: java.lang.Boolean = true

scala> val sbool = true         
sbool: Boolean = true

scala> def sboolMethod(sbool: Boolean) = print("got scala.Boolean " + sbool)              
sboolMethod: (sbool: Boolean)Unit

scala> sboolMethod(sbool)
got scala.Boolean true

scala> sboolMethod(jbool)
<console>:9: error: type mismatch;
 found   : java.lang.Boolean
 required: scala.Boolean
       sboolMethod(jbool)
                   ^

scala> implicit def jbool2sbool(bool: java.lang.Boolean): scala.Boolean = bool.booleanValue
jbool2sbool: (bool: java.lang.Boolean)Boolean

scala> sboolMethod(jbool)                                                                  
got scala.Boolean true

现在的问题是：为什么没有从 java.lang.Boolean中的到 scala.Boolean ？这个问题也代表 java.lang.Long中的 VS scala.Long ，可能等标准类型（没试过所有的人）。

The question is: why isn't there a default implicit conversion from java.lang.Boolean to scala.Boolean? The question also stands for java.lang.Long vs scala.Long and probably other standard types (haven't tried all of them).

推荐答案

在2.9中，有这样一个转换，presumably帮助与Java的互操作性。 （斯卡拉并不需要它自身，因为它透明盒和unboxes元，这也许就是为什么它不更早包括在内。）

In 2.9, there is such a conversion, presumably to aid interoperability with Java. (Scala doesn't need it on its own, because it transparently boxes and unboxes primitives, which is perhaps why it wasn't included earlier.)

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