Scala 2.9 中限制类构造函数的可见性
您好，
如何使 Foo 构造函数仅对此包（单元测试+伴随对象）可见？
我不想在这 2 个文件之外实例化 Foo...
Foo.scala
package project.foo
class Foo(val value: String)
object Foo {
def generate: Foo = new Foo("test")
}
FooSpec.scala
package project.foo
import org.spec2.mutable._
class FooSpec extends Specification {
"Foo" should {
"be constructed with a string" {
val foo = new Foo("test")
foo.value must be "test"
}
}
}
我正在使用 Scala 2.9
Greetings,
How can I make the Foo constructor visible only to this package (unit test + companion object) ?
I don't want to be able to instantiate Foo outside of this 2 files...
Foo.scala
package project.foo
class Foo(val value: String)
object Foo {
def generate: Foo = new Foo("test")
}
FooSpec.scala
package project.foo
import org.spec2.mutable._
class FooSpec extends Specification {
"Foo" should {
"be constructed with a string" {
val foo = new Foo("test")
foo.value must be "test"
}
}
}
I'm using Scala 2.9
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试试这个：
那么
Foo的构造函数只能从foo包中访问。您可以在此处阅读有关 Scala 可见性的更多信息（特别是范围内私有和范围内受保护）< /a>.
Try this:
Then the constructor of
Foois only accessible from thefoopackage.You can read more about Scala's visibility (look especially for scoped private and scoped protected) here.