scala case类私有应用方法(repl bug？)

Question

在Scala2.10.0 REPL中

Welcome to Scala version 2.10.0 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_13).
Type in expressions to have them evaluated. 
Type :help for more information.

scala> case class A private(i:Int)
defined class A

scala> A(1)
res0: A = A(1)

但如果编译

$ scala -version
Scala code runner version 2.10.0 -- Copyright 2002-2012, LAMP/EPFL
$ cat Main.scala 
package foo

case class A private (i:Int)

object Main extends App{
  println(A(1))
}

$ scalac Main.scala 
Main.scala:6: error: constructor A in class A cannot be accessed in object Main
  println(A(1))
          ^
one error found

A.apply(1) 是编译错误.

这个Scala2.10.0 REPL错误？

FYI Scala2.9.2 REPL如下

Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_13).
Type in expressions to have them evaluated.
Type :help for more information.

scala> case class A private(i:Int)
defined class A

scala> A(1)
<console>:10: error: constructor A in class A cannot be accessed in object $iw
              A(1)
              ^

Answer 1

这绝对看起来像一个 REPL 错误。

请注意，构造函数被正确标记为private（换句话说，new A(1)没有按预期进行编译），只有工厂 ( A.apply) 被错误地公开。