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 **ool = java.lang.Boolean.TRUE **ool: 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(**ool) <console>:9: error: type mismatch; found : java.lang.Boolean required: scala.Boolean sboolMethod(**ool) ^ scala> implicit def **ool2sbool(bool: java.lang.Boolean): scala.Boolean = bool.booleanValue **ool2sbool: (bool: java.lang.Boolean)Boolean scala> sboolMethod(**ool) got scala.Boolean true
问题是:为什么没有从 java.lang.Boolean 到 scala.Boolean 的默认隐式转换?该问题还代表 java.lang.Long 与 scala.Long 以及可能的其他标准类型(还没有尝试过所有这些类型).
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,有这样的转换,大概是为了帮助与 Java 的互操作性.(Scala 自己不需要它,因为它透明地对基元进行装箱和拆箱,这可能是它之前没有被包含在内的原因.)
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.)