问题描述
我有一个使用以下代码的课程:
I have a class with this code:
package shop.orders.services.email
private[services] class EmailService {...}
然后在另一个包中，我使用该类:
Then in a different package, I use that class:
package shop.ui
import shop.orders.services.email.EmailService
class PaymentConfirmation extends WithFacesContext {
var emailService: EmailService = null
查看生成的字节码，没有任何访问修饰符的迹象，这很有意义，因为Java不支持这种访问限制.因此，如果我创建一个包含类似块一的代码的库，然后尝试针对该库编译块二，会发生什么情况-由于信息丢失，编译器不会失败.还是包含在清单中?
Looking at the generated bytecode, there is no sign of any access modifier, which makes sense, as Java does not support such access restrictions. So what happens if I create a library containing code like block one, and attempt to compile block two against the library - there is no chance that the compiler will fail, since the information is lost. Or is it contained in something like a manifest?
我正在使用Scala 2.9.2.
I'm using Scala 2.9.2.
推荐答案
您可以从Java引用EmailService，但不能从Scala引用EmailService，因为Scala将类的签名存储为scala.reflect.ScalaSignature注释. Scala编译器将失败，并显示以下错误:
You could reference EmailService from Java, but not from Scala, because Scala stores the signature of the class as a scala.reflect.ScalaSignature annotation. The Scala compiler will fail with the following error:
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